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	<title>Science And Technology</title>
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	<link>http://oscience.info</link>
	<description>The ultimate resource for Science and Technology</description>
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		<title>Elephants can&#8217;t jump!</title>
		<link>http://oscience.info/science-facts/elephants-cant-jump/</link>
		<comments>http://oscience.info/science-facts/elephants-cant-jump/#comments</comments>
		<pubDate>Thu, 03 May 2012 09:35:35 +0000</pubDate>
		<dc:creator>subash</dc:creator>
				<category><![CDATA[Biology Facts]]></category>
		<category><![CDATA[Science Facts]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6758</guid>
		<description><![CDATA[Elephants are large land mammals. They are the largest living land animals. One amazing fact about elephants is that Elephants are the only animals in planet that cannot jump even though they have no less bones in their feet than any normal mammal! The most elephants can do is balance in their back foots! Elephants [...]]]></description>
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</script></div><p>Elephants are large land mammals. They are the largest living land animals.</p>
<p>One amazing fact about elephants is that Elephants are the only animals in planet that cannot jump even though they have no less bones in their feet than any normal mammal! The most elephants can do is balance in their back foots!</p>
<div id="attachment_6759" class="wp-caption aligncenter" style="width: 310px"><a href="http://oscience.info/science-facts/elephants-cant-jump/attachment/elephants_cant_jump/" rel="attachment wp-att-6759"><img class="size-medium wp-image-6759" title="Elephants can't jump" src="http://oscience.info/wp-content/uploads/Elephants_cant_jump-300x219.jpg" alt="Elephants can't jump" width="300" height="219" /></a><p class="wp-caption-text">Elephants can&#39;t jump</p></div>
<p>Elephants need very less muscular energy to  stand up due to their straight legs and large foots , But to jump elephants will need a lot of energy because of their gigantic size.</p>
<p>The bones in an Elephant&#8217;s leg are not well developed to let it jump! And an Elephant stays standing most of it&#8217;s life. It only lies down when it is sick ; with the exception of Indian Elephants.</p>
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		<title>Alaskan Wood Frog and it&#8217;s amazing capability</title>
		<link>http://oscience.info/science-facts/alaskan-wood-frog-and-its-amazing-capability/</link>
		<comments>http://oscience.info/science-facts/alaskan-wood-frog-and-its-amazing-capability/#comments</comments>
		<pubDate>Thu, 03 May 2012 09:13:58 +0000</pubDate>
		<dc:creator>subash</dc:creator>
				<category><![CDATA[Biology Facts]]></category>
		<category><![CDATA[Science Facts]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6756</guid>
		<description><![CDATA[Alaskan Wood Frog is a small creature ; usually less than three inches long; that is found in the North pole of the Alaska and Canada.It is a Cold Blooded creature , which means unlike the warm clodded animal it don&#8217;t have to migrate to other places during winter in Arctic region. &#160; An amazing [...]]]></description>
			<content:encoded><![CDATA[<p>Alaskan Wood Frog is a small creature ; usually less than three inches long; that is found in the North pole of the Alaska and Canada.It is a Cold Blooded creature , which means unlike the warm clodded animal it don&#8217;t have to migrate to other places during winter in Arctic region.</p>
<div id="attachment_6757" class="wp-caption aligncenter" style="width: 310px"><a href="http://oscience.info/science-facts/alaskan-wood-frog-and-its-amazing-capability/attachment/alaskan_wood_frog_fact/" rel="attachment wp-att-6757"><img class="size-medium wp-image-6757" title="Alaskan wood frog fact" src="http://oscience.info/wp-content/uploads/alaskan_wood_frog_fact-300x180.jpg" alt="Alaskan wood frog fact" width="300" height="180" /></a><p class="wp-caption-text">Alaskan wood frog fact</p></div>
<p>&nbsp;</p>
<p>An amazing fact about the Alaskan Wood Frog is that they stay frozen for month during winter , and then revive itself back to normal life after the winter is gone. During the frozen state more than 70% of the body fluid of the frog froze into ice and even the heart of the frog gets frozen.</p>
<p>After after all those freezing days , Scientists are still unknowns to how it gets back into life.</p>
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		<title>Photo Detector Circuit</title>
		<link>http://oscience.info/infos/photo-detector-circuit/</link>
		<comments>http://oscience.info/infos/photo-detector-circuit/#comments</comments>
		<pubDate>Mon, 30 Apr 2012 06:26:48 +0000</pubDate>
		<dc:creator>subash</dc:creator>
				<category><![CDATA[Infos]]></category>
		<category><![CDATA[Tech Infos]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6748</guid>
		<description><![CDATA[This post contains the design and circuit of a simple photo detector or a simple circuit to detect light energy .This circuit can be used for various purposes like controlling street lights ,  detecting humans or opaque objects and can also be used to count numbers of items with some extra aided digital circuits. Photo Detector [...]]]></description>
			<content:encoded><![CDATA[<p>This post contains the design and circuit of a simple photo detector or a simple circuit to detect light energy .This circuit can be used for various purposes like controlling street lights ,  detecting humans or opaque objects and can also be used to count numbers of items with some extra aided digital circuits.</p>
<h2>Photo Detector Circuit:</h2>
<p>The circuit diagram for the Photo Detector is given below: ( Click on the image to enlarge it )</p>
<div id="attachment_6754" class="wp-caption aligncenter" style="width: 310px"><a href="http://oscience.info/infos/photo-detector-circuit/attachment/photo_detector_circuit/" rel="attachment wp-att-6754"><img class="size-medium wp-image-6754" title="photo detector circuit" src="http://oscience.info/wp-content/uploads/photo_detector_circuit-300x256.jpg" alt="photo detector circuit" width="300" height="256" /></a><p class="wp-caption-text">photo detector circuit</p></div>
<p>&nbsp;</p>
<p>The components used in the circuit are:</p>
<p>1&gt; LDR.</p>
<p>2&gt; Potentiometer, Value: 50 Kilo ohms</p>
<p>3&gt; Resistor, value: 1 Kilo ohms</p>
<p>4&gt; Relay, Value: 6 Volts</p>
<p>5&gt; Transistor, value: BC 548 see below for the orientation of emitter , base and collector.</p>
<p>&nbsp;</p>
<p>You can choose the sensitivity level of the circuit by rotation the potentiometer. And you can use the switch of the relay to switch on and off any any device according to the intensity of the light falling on the LDR.</p>
<p>&nbsp;</p>
<h3>LDR:</h3>
<p>A simple LDR or Light Dependent Resistor is used in the circuit as the light sensor.</p>
<div id="attachment_6749" class="wp-caption aligncenter" style="width: 434px"><img class="size-full wp-image-6749" title="LDR" src="http://oscience.info/wp-content/uploads/LDR.jpg" alt="LDR" width="424" height="392" /><p class="wp-caption-text">LDR</p></div>
<p>LDR: It is a special type of resistor whose resistance decreases with increasing intensity of light falling on it. Or the more light falls on the LDR the more electrical energy it conducts through it.  It is made up of a photo sensitive material &#8220;cadmium sulfide&#8221; thus it exhibits such electrical properties.</p>
<p>&nbsp;</p>
<h3>Potentiometer or variable resistance:</h3>
<div id="attachment_6751" class="wp-caption aligncenter" style="width: 387px"><img class="size-full wp-image-6751" title="Potentiometer" src="http://oscience.info/wp-content/uploads/Potentiometer1.jpg" alt="Potentiometer" width="377" height="236" /><p class="wp-caption-text">Potentiometer</p></div>
<p>Potentiometer or a variable resistance is a three terminal resistance , where the resistance of the middle terminal with the remaining terminal varies when the slider or the rotor is rotated. Here in the circuit we are using a Potentiometer in series with the LDR to make a voltage divider from which the voltage is provided to the transistor for switching purpose.</p>
<p>&nbsp;</p>
<h3>Transistor:</h3>
<div id="attachment_6752" class="wp-caption aligncenter" style="width: 461px"><img class="size-full wp-image-6752" title="transistor" src="http://oscience.info/wp-content/uploads/transistor.jpg" alt="transistor" width="451" height="270" /><p class="wp-caption-text">transistor</p></div>
<p>Transistor is a semiconductor device used to amplify or switch electrical signals.</p>
<p>In our case we are using the NPN transistor numbered or labeled &#8220;BC548&#8243; for amplifying purpose which amplifies the signal in it&#8217;s base and provides the signal to relay for switching on any device.</p>
<p>&nbsp;</p>
<p>A transistor are of Type NPN and PNP , here we are using NPN transistor &#8220;BC 548&#8243; .</p>
<p>A transistor have three terminals which are named &#8220;Emitter&#8221; , &#8220;Base&#8221; and &#8220;collector&#8221; the orientation of Emitter , base and collector of transistor bc 548 is given below:</p>
<div id="attachment_6753" class="wp-caption aligncenter" style="width: 291px"><img class="size-full wp-image-6753" title="bc548 transistor" src="http://oscience.info/wp-content/uploads/bc548_transistor.jpg" alt="bc548 transistor" width="281" height="249" /><p class="wp-caption-text">bc548 transistor</p></div>
<p>So , place the BC5448 transistor facing the flat face toward you , then number the terminal from left to right 1 , 2 and 3.<br />
The the first or &#8220;1&#8243; terminal is Collector , the middle or &#8220;2&#8243; terminal is Base and the last or terminal &#8220;3&#8243; is Emitter.</p>
<h3>Relay:</h3>
<p>A relay is simply a switch which is electrically operated. In our case we are using a 6 V relay.</p>
<p>&nbsp;</p>
<p>After the circuit is ready , you can use it for any purpose. This Photo Detector Circuit can be used for simply switching on the lights when the surrounding is dark and switch it off when the surrounding is bright.  You can also output the amplified signal from the collector terminal of the transistor to any other device instead of relay and use this circuit as photo detector circuit aid on to any other device or circuit.</p>
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		<title>How to solve Unknown encoder ‘libmp3lame’ problem in linux.</title>
		<link>http://oscience.info/infos/how-to-solve-unknown-encoder-libmp3lame-problem-in-linux/</link>
		<comments>http://oscience.info/infos/how-to-solve-unknown-encoder-libmp3lame-problem-in-linux/#comments</comments>
		<pubDate>Sun, 15 Apr 2012 06:02:18 +0000</pubDate>
		<dc:creator>subash</dc:creator>
				<category><![CDATA[Infos]]></category>
		<category><![CDATA[Tech Infos]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6721</guid>
		<description><![CDATA[Unknown encoder ‘libmp3lame’ problem: WinFF is a cross platform GUI for the command line video converter. WinFF is great to work with ; if you want to convert videos and musics from almost all format to any other format. I also tried converting a flac audio file to mp3 , so that I can use the [...]]]></description>
			<content:encoded><![CDATA[<h3>Unknown encoder ‘libmp3lame’ problem:</h3>
<p>WinFF is a cross platform GUI for the command line video converter. WinFF is great to work with ; if you want to convert videos and musics from almost all format to any other format.</p>
<div id="attachment_6723" class="wp-caption aligncenter" style="width: 468px"><img class="size-full wp-image-6723" title="winff ubuntu" src="http://oscience.info/wp-content/uploads/winff-ubuntu.jpg" alt="winff ubuntu" width="458" height="485" /><p class="wp-caption-text">winff ubuntu</p></div>
<p>I also tried converting a flac audio file to mp3 , so that I can use the file in my musics player.</p>
<p>I tried converting the media from ubuntu 11.10 , But I got the following error:</p>
<blockquote><p><em>Unknown encoder ‘libmp3lame’</em></p></blockquote>
<p>&nbsp;</p>
<p>I tried researching the issue , and found that the MP3 encoder &#8220;libmp3lame&#8221; is not supported by pre-compiled ffmpeg which I downloaded from ubuntu software center.</p>
<h2>To solve the problem:</h2>
<p>One way to solve the problem might be to download the sources of ffmpeg or winff and then compile and install it from the source instead of installing from the ubuntu software center.</p>
<p>But there is also another easy way to solve the problem ; you just need to install &#8220;libavcodec-extra-52&#8243; or &#8220;libavcodec-extra-53&#8243;</p>
<p>You need to install one of the packeges listed above.</p>
<p>So go to the terminal ( press ctrl + alt + t ) and type the following command:</p>
<blockquote><p>sudo apt-get install libavcodec-extra-52</p></blockquote>
<p>and if it does not works or you get &#8220;package not found&#8221; error then try:</p>
<blockquote><p>sudo apt-get install libavcodec-extra-53</p></blockquote>
<p>&nbsp;</p>
<p>This will solve the error  Unknown encoder ‘libmp3lame’.</p>
<p>&nbsp;</p>
]]></content:encoded>
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		<title>Reputation Management “for Dummies” Crash Course</title>
		<link>http://oscience.info/infos/reputation-management-for-dummies-crash-course/</link>
		<comments>http://oscience.info/infos/reputation-management-for-dummies-crash-course/#comments</comments>
		<pubDate>Tue, 10 Apr 2012 14:18:57 +0000</pubDate>
		<dc:creator>subash</dc:creator>
				<category><![CDATA[Infos]]></category>
		<category><![CDATA[Tech Infos]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6715</guid>
		<description><![CDATA[&#160; If you have been the victim of a hoax report or some other false negative campaign, you might want to check your phone for taps and start watching your back for a “tail.” The old movie clichés immediately come to mind, but the assailant is typically operating in cyberspace. If this fact surprises you, [...]]]></description>
			<content:encoded><![CDATA[<p>&nbsp;</p>
<p>If you have been the victim of a hoax report or some other false negative campaign, you might want to check your phone for taps and start watching your back for a “tail.” The old movie clichés immediately come to mind, but the assailant is typically operating in cyberspace. If this fact surprises you, you’re going to need a “Reputation Management for Dummies” crash course. Here are a few links you can start with to start learning more about how to fix online reputation problems:</p>
<p>&nbsp;</p>
<p><a href="http://lifehacker.com/357460/manage-your-online-reputation" rel="nofollow" target="_blank">Lifehacker</a> – For basic tips and tricks.<br />
<a href="http://www.internetreputationmanagement.com/" rel="nofollow" target="_blank">Internet Reputation Management</a> – For professional level help and for existing problems.<br />
<a href="http://mashable.com/2011/05/22/how-to-facebook-page/" rel="nofollow" target="_blank">Facebook</a> on Mashable &#8211; How to setup a Facebook page</p>
<p>In the modern world, companies that play by the rules and do their best for their customers still have companies out to sabotage them. That’s right, other companies – AKA ‘competitors’. The good news is that therer are plenty of resources out there to help you craft your online reputation.</p>
<p>&nbsp;</p>
<p>The concept is simple: negative campaigns rely on holes in the content of your website, missing brand related content online, or a lack of strength of any of these (strength as in what Google thinks is relevant).</p>
<p>&nbsp;</p>
<p>Knowing that you are vulnerable in this department, a competitor will plan the assault and watch what happens. If you don’t figure out what is happening soon enough, it could take months – if not years – to repair the damage. Your response must be swift – and your hand steady.</p>
<p>&nbsp;</p>
<p>The companies that protect your reputation online can prevent this sort of attack from happening in the first place. They make sure your online presence reflects the true character of your operation, whether that be a retirement home or a political campaign. If you don’t have someone watching your back online, it could lead to your company getting blindsided in ways you never imagined.</p>
<p>&nbsp;</p>
<p>If nothing else, ask your Chief Marketing Officer to see the companies reputation management strategy, if that’s you… well.</p>
]]></content:encoded>
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		<title>Elecric Field</title>
		<link>http://oscience.info/physics/electric-field/elecric-field/</link>
		<comments>http://oscience.info/physics/electric-field/elecric-field/#comments</comments>
		<pubDate>Thu, 05 Apr 2012 08:13:34 +0000</pubDate>
		<dc:creator>sandeep</dc:creator>
				<category><![CDATA[Electric Field]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6711</guid>
		<description><![CDATA[Frictional Electricity: Frictional electricity is the electricity caused due to friction. When two neutral bodies (e.g. glass rod and silk) are rubbed together, both gets charged and the charges produced on them are opposite in nature. The reason of both bodies being electrified is that some of electrons of one body are transferred to the [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Frictional Electricity: </strong></p>
<p>Frictional electricity is the electricity caused due to friction. When two neutral bodies (e.g. glass rod and silk) are rubbed together, both gets charged and the charges produced on them are opposite in nature. The reason of both bodies being electrified is that some of electrons of one body are transferred to the other.<br />
Types of charges: Charges are of two types:</p>
<p>(i) Positive charges</p>
<p>(ii) Negative charges</p>
<p>Electrons possess negative charge. Therefore when a body gains electrons, it becomes negatively charged but when it gives off electrons, it becomes positively charged.<br />
Thus positively charge means deficiency of electrons and negative charge means excess of electrons.<br />
Interaction between charged bodies:<br />
It is found experimentally</p>
<p>(i) That the similarly charged bodies repel; while oppositely charged bodies attract. In other words similar charges repel while opposite charges attract.</p>
<p>(ii) A charged body also attracts a light neutral body.</p>
<p>Thus if a body (A) is attracted by a charged (or electrified) body; then the body (A) may be oppositely charged or neutral.<br />
This implies that attraction is not the sure test of electrification of body.<br />
On the other hand, if a body is repelled by a charged body; it must be similarly charged. This implies that the repulsion is the sure test of electrification.</p>
<p><strong>Conservation of charge: </strong></p>
<p>The net charge on a system remains constant. This means that the charge can neither be created nor destroyed, but it may merely be transferred from one body to another. When the two objects are rubbed, the electrons of one object are transferred to the other.</p>
<p><strong>Unit of charge:</strong></p>
<p>In S.I. system the unit of charge is Coulomb abbreviated as C.<br />
In C.G.S. system the unit of charge is often written as e.s.u. of charge or stat-coulomb.</p>
<p>1 coul = 3 x 10<sup>9</sup> stat-coul<br />
In practice coulomb is big unit; therefore small units microcoulomb is often used.<br />
1 microcoulomb (<img src='http://s.wordpress.com/latex.php?latex=%5Cmu%20C%20%3D%2010%5E%7B-6%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mu C = 10^{-6}' title='\mu C = 10^{-6}' class='latex' />)</p>
<p>Dimensions of charge Q are [Q]: [AT].<br />
<strong><em>Elementary Unit :</em></strong> As charge is produced due to transfer of electrons; the magnitude of charge on electron is taken as elementary unit of charge (or elementary charge or, fundamental charge).<br />
1 elementary unit, e = charge on electron</p>
<p>= 1.6 x 10<sup>-19</sup> Coulomb</p>
<p><strong><em>Quantum Nature of charge or principle oi atomicity of charge</em></strong><em> :</em></p>
<p><em> A physically existing charge Q is always on integral multiple of elementary charge (e) i.e. Q = ne.</em><br />
where n may be positive or negative integer.<br />
Accordingly charge can take values e, 2e, 3e: &#8212;&#8212;-, but it cannot take values 1.5e, 2.2e, 3.8e &#8212;-~ etc.<br />
This is called principle of quantization (or atomicity) of charge.</p>
<p>If charge is 1 coulomb, then</p>
<p>n = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7BQ%7D%7Be%7D%20%3D%20%5Cdfrac%7B1%20coul%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%20coul%7D%20%3D%206.25%20%5Ctimes%2010%5E18&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{Q}{e} = \dfrac{1 coul}{1.6 \times 10^{-19} coul} = 6.25 \times 10^18' title='\dfrac{Q}{e} = \dfrac{1 coul}{1.6 \times 10^{-19} coul} = 6.25 \times 10^18' class='latex' /></p>
<p>Thus, 1 coulomb charge is equivalent to deficiency of 6.25 x 10<sup>18</sup> electrons</p>
<p><strong>Coulomb’s Law:</strong></p>
<p>The force of attraction or repulsion between two point charges (q1 and q2) at<br />
finite separation (r) is directly proportional to the product of charges and inversely proportional to the square of distance between the charges i.e.</p>
<img src='http://s.wordpress.com/latex.php?latex=F_m%20%5Cpropto%20%5Cdfrac%7Bq1%20q2%7D%7Br%5E2%7D%20%5Ctext%7Bor%7D%20F%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%7D%20.%20%5Cdfrac%7Bq1%20q2%7D%7Br%5E2%7D%20%5Cdots%20%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F_m \propto \dfrac{q1 q2}{r^2} \text{or} F = \dfrac{1}{4 \pi \varepsilon} . \dfrac{q1 q2}{r^2} \dots (1)' title='F_m \propto \dfrac{q1 q2}{r^2} \text{or} F = \dfrac{1}{4 \pi \varepsilon} . \dfrac{q1 q2}{r^2} \dots (1)' class='latex' />
<p>where <img src='http://s.wordpress.com/latex.php?latex=%5Cvarepsilon&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\varepsilon' title='\varepsilon' class='latex' /> is the permittivity of medium between the charges.<br />
If charges are separated by free space (or air), then <img src='http://s.wordpress.com/latex.php?latex=%5Cvarepsilon&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\varepsilon' title='\varepsilon' class='latex' />is replaced by <img src='http://s.wordpress.com/latex.php?latex=%5Cvarepsilon%20_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\varepsilon _0' title='\varepsilon _0' class='latex' />, where <img src='http://s.wordpress.com/latex.php?latex=%5Cvarepsilon%20_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\varepsilon _0' title='\varepsilon _0' class='latex' /> is the permittivity of free space (or air); so</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Ctext%7Bforce%7D%20F%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cvarepsilon%20_0%7D.%20%5Cdfrac%7Bq_1q_2%7D%7Br%5E2%7D%20%5Ccdots%20%282%29%20%5C%5C%20%5Ctext%7BThe%20permittivity%20of%20free%20space%20%7D%20%5C%5C%20%5Cvarepsilon%20_0%20%3D%208.86%20%5Ctimes%2010%5E%7B-12%7D%20%7Bcoul%7D%20%5E2%20%2F%20N-%20m%20%5E2%20%5C%5C%20%5Ctext%7Band%7D%20%5C%5C%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Ctimes%203.14%20%5Ctimes%208.86%20%5Ctimes%2010%5E%7B-12%7D%7D%20%5C%5C%20%3D%209%20%5Ctimes%2010%5E9%20N-%20m%5E2%20%5Ctext%7Bcoul%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{force} F = \dfrac{1}{4 \varepsilon _0}. \dfrac{q_1q_2}{r^2} \cdots (2) \\ \text{The permittivity of free space } \\ \varepsilon _0 = 8.86 \times 10^{-12} {coul} ^2 / N- m ^2 \\ \text{and} \\ \dfrac{1}{4 \pi \varepsilon _0} = \dfrac{1}{4 \times 3.14 \times 8.86 \times 10^{-12}} \\ = 9 \times 10^9 N- m^2 \text{coul}^2' title='\text{force} F = \dfrac{1}{4 \varepsilon _0}. \dfrac{q_1q_2}{r^2} \cdots (2) \\ \text{The permittivity of free space } \\ \varepsilon _0 = 8.86 \times 10^{-12} {coul} ^2 / N- m ^2 \\ \text{and} \\ \dfrac{1}{4 \pi \varepsilon _0} = \dfrac{1}{4 \times 3.14 \times 8.86 \times 10^{-12}} \\ = 9 \times 10^9 N- m^2 \text{coul}^2' class='latex' />
<p><strong>Dielectric constant:</strong></p>
<p>The ratio of permittivity of medium ( <img src='http://s.wordpress.com/latex.php?latex=%5Cvarepsilon&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\varepsilon' title='\varepsilon' class='latex' /> ) to the permittivity of free space (<img src='http://s.wordpress.com/latex.php?latex=%5Cvarepsilon%20_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\varepsilon _0' title='\varepsilon _0' class='latex' />) is called the dielectric constant or relative permittivity of medium</p>
<p>Dielectric constant K = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Cvarepsilon%7D%7B%5Cvarepsilon%20_0%7D%20%5Cdots%20%283%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{\varepsilon}{\varepsilon _0} \dots (3)' title='\dfrac{\varepsilon}{\varepsilon _0} \dots (3)' class='latex' /></p>
<p>Alternatively the dielectric constant of a  medium is the ratio of force between two charged particles in air and in medium</p>
<p>i.e.     K = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7BF_%7B%5Ctext%7Bair%7D%7D%7D%7BF_%7B%5Ctext%7Bmedium%7D%7D%7D%20%5Cdots%20%284%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{F_{\text{air}}}{F_{\text{medium}}} \dots (4)' title='\dfrac{F_{\text{air}}}{F_{\text{medium}}} \dots (4)' class='latex' /></p>
<p>Equation (1) may also be expressed as</p>
<img src='http://s.wordpress.com/latex.php?latex=F%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%7D%20.%20%5Cdfrac%7Bq1%20q2%7D%7Br%5E2%7D%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7Bq1%20q2%7D%7BK%20r%5E2%7D%20%5Cdots%20%285%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F = \dfrac{1}{4 \pi \varepsilon} . \dfrac{q1 q2}{r^2} = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{q1 q2}{K r^2} \dots (5)' title='F = \dfrac{1}{4 \pi \varepsilon} . \dfrac{q1 q2}{r^2} = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{q1 q2}{K r^2} \dots (5)' class='latex' />
<p>Remarks: 1.</p>
<p>The force is a vector quantity, while calculating the force from Coulomb’s law, the sign of charge is not retained in formula; but the signs of charges indicate direction of force which is seen by inspection with the rule that the charge on which force is to be calculated is assumed to have the tendency of motion while the other charge due to which force is to be calculated is assumed at rest, unless otherwise stated.<br />
If the force on a charge q<sub>o</sub> due to a number of charges to be found, the forces on charge due to individual charges are calculated and then summed vectorially to find the resultant.</p>
<p><strong>Electric Field Lines of Force: </strong></p>
<p>The electric field is a region in which a charged particle experiences a force.<br />
The electric field strength ( <img src='http://s.wordpress.com/latex.php?latex=%5Coverrightarrow%7BE%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\overrightarrow{E}' title='\overrightarrow{E}' class='latex' /> ) at any point is defined as the force experienced by per unit infinitesimal positive charge when placed at that point i.e.</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Coverrightarrow%7BE%7D%20%3D%20%5Clim_%7Bq_0%20%5Cto%200%7D%20%5Cfrac%7B%5Coverrightarrow%7BF%7D%7D%7Bq_0%7D%20%5Cdots%20%286%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\overrightarrow{E} = \lim_{q_0 \to 0} \frac{\overrightarrow{F}}{q_0} \dots (6)' title='\overrightarrow{E} = \lim_{q_0 \to 0} \frac{\overrightarrow{F}}{q_0} \dots (6)' class='latex' />
<p>where q<sub>o</sub> is infinitesimal positive test charge.<br />
The electric field strength is a vector quantity. Its, direction is away from a positive charge and towards the negative charge. Its unit is Newton coulomb or volt/meter.<br />
Michael Faraday always thought a vector field in terms of lines of force. The lines of force form a convenient way of visualizing electric field patterns. The magnitude of electric field strength at any point is measured by the number of electric lines of force passing per unit small area around that point normally<em> </em>and the direction of field at any point is given by the tangent to the line of force at the point<em>.</em></p>
<p><strong>Properties of lines of force :</strong><br />
1. The lines of force diverge out from a positive charge and converge at a negative charge. More correctly the lines of force are always directed from higher to lower potential.<br />
2. The electric lines of force contract lengthwise and expand laterally.<br />
3. Two lines of force never intersect. If they are assumed to intersect, there will be two directions of electric field at the point of intersection: which is impossible.</p>
<p><strong> Calculation of electric field strength</strong></p>
<p><strong><em>Electric field due to point charge:</em></strong></p>
<p>Let q be a point charge producing electric field. The force on a positive test charge q<sub>o</sub> at separation r from charge q is</p>
<img src='http://s.wordpress.com/latex.php?latex=F%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7Bq%20q_0%7D%7Br%5E2%7D%20%5C%5C%20%5Ctherefore%20%5C%2C%20%5C%2C%20%5Ctext%7BField%20strength%20%7D%20E%20%3D%20%5Cdfrac%7BF%7D%7Bq_0%7D%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%5Cdfrac%7Bq%7D%7Br%5E2%7D%20%5Cdots%20%287%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{q q_0}{r^2} \\ \therefore \, \, \text{Field strength } E = \dfrac{F}{q_0} = \dfrac{1}{4 \pi \varepsilon _0} .\dfrac{q}{r^2} \dots (7)' title='F = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{q q_0}{r^2} \\ \therefore \, \, \text{Field strength } E = \dfrac{F}{q_0} = \dfrac{1}{4 \pi \varepsilon _0} .\dfrac{q}{r^2} \dots (7)' class='latex' />
<p><strong><em>Electric Field due to a group of charges:</em></strong></p>
<p>To find the field due to a group of point charges q1, q2,&#8230;..qn, the field <img src='http://s.wordpress.com/latex.php?latex=%20%5Coverrightarrow%7BE1%7D%2C%20%5Coverrightarrow%7BE1%7D%2C%20%5Cdots%20%5Coverrightarrow%7BEn%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt=' \overrightarrow{E1}, \overrightarrow{E1}, \dots \overrightarrow{En}' title=' \overrightarrow{E1}, \overrightarrow{E1}, \dots \overrightarrow{En}' class='latex' /> at given point due to individual point charges are found and then vectorially added to find the resultant field <img src='http://s.wordpress.com/latex.php?latex=%5Coverrightarrow%7BE1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\overrightarrow{E1}' title='\overrightarrow{E1}' class='latex' />  i.e</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Coverrightarrow%7BE%7D%20%3D%20%5Coverrightarrow%7BE1%7D%20%2B%20%5Coverrightarrow%7BE1%7D%20%2B%20%5Cdots%20%5Coverrightarrow%7BEn%7D%20%5Cdots%20%288%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\overrightarrow{E} = \overrightarrow{E1} + \overrightarrow{E1} + \dots \overrightarrow{En} \dots (8)' title='\overrightarrow{E} = \overrightarrow{E1} + \overrightarrow{E1} + \dots \overrightarrow{En} \dots (8)' class='latex' />
<p><strong>Electric Flux: </strong></p>
<p>The total number of electric lines of force through a given area normally is called the electric flux. The electric flux through a surface element <img src='http://s.wordpress.com/latex.php?latex=%5Ctriangle%20%5Coverrightarrow%7BS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\triangle \overrightarrow{S}' title='\triangle \overrightarrow{S}' class='latex' /> is</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Ctriangle%20%5Cphi%20%3D%20%5Coverrightarrow%7BE%7D%20.%20%5Ctriangle%20%5Coverrightarrow%7BS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\triangle \phi = \overrightarrow{E} . \triangle \overrightarrow{S}' title='\triangle \phi = \overrightarrow{E} . \triangle \overrightarrow{S}' class='latex' />
<p>Net electric flux through the whole surface S is</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cphi%20%3D%20%5Csum%20E%20.%20%5Ctriangle%20%5Coverrightarrow%7BS%7D%20%3D%20%5Cint%20%5Coverrightarrow%7BE%7D.%20%5Coverrightarrow%7BdS%7D%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%28%20%5Ctext%7Ba%20scalar%20%7D%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\phi = \sum E . \triangle \overrightarrow{S} = \int \overrightarrow{E}. \overrightarrow{dS} \, \, \, \, ( \text{a scalar } )' title='\phi = \sum E . \triangle \overrightarrow{S} = \int \overrightarrow{E}. \overrightarrow{dS} \, \, \, \, ( \text{a scalar } )' class='latex' />
<p><strong>Gauss’s Theorem: </strong></p>
<p>The net electric flux through a closed surface is equal to <img src='http://s.wordpress.com/latex.php?latex=1%20%2F%20%5Cvarepsilon%20_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1 / \varepsilon _0' title='1 / \varepsilon _0' class='latex' />times the net charge within the surface . i.e.</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cphi%20%3D%20%5Cdfrac%7B1%7D%7B%5Cvarepsilon%20_0%7D%20%5Ctimes%20%5Ctext%7Bcharge%20enclosed%20by%20surface%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\phi = \dfrac{1}{\varepsilon _0} \times \text{charge enclosed by surface}' title='\phi = \dfrac{1}{\varepsilon _0} \times \text{charge enclosed by surface}' class='latex' />
<p>Mathematically <img src='http://s.wordpress.com/latex.php?latex=%5Cint_s%20%5Coverrightarrow%7BE%7D%20.%20%5Coverrightarrow%7BdS%7D%20%3D%20%5Cdfrac%7B1%7D%7B%5Cvarepsilon%20_0%7D%20%5Ctimes%20Q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\int_s \overrightarrow{E} . \overrightarrow{dS} = \dfrac{1}{\varepsilon _0} \times Q' title='\int_s \overrightarrow{E} . \overrightarrow{dS} = \dfrac{1}{\varepsilon _0} \times Q' class='latex' /><br />
where Q is net charge enclosed by the surface.</p>
<p><strong>Electric potential: </strong></p>
<p>The electric potential at any point in an electric field is defined as the work done required to move per unit small positive test charge from infinity to that point. If W is the work done in bringing the charge q0 from infinity to given point P, then potential at P</p>
<img src='http://s.wordpress.com/latex.php?latex=V%20%3D%20%5Clim_%7Bq_0%20%5Cto%200%7D%20%5Cfrac%7BW%7D%7Bq_0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V = \lim_{q_0 \to 0} \frac{W}{q_0}' title='V = \lim_{q_0 \to 0} \frac{W}{q_0}' class='latex' />
<p>Alternatively the electric field at any point in an electric field is defined as the negative line integral for the electric field vector <img src='http://s.wordpress.com/latex.php?latex=%5Coverrightarrow%7BE%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\overrightarrow{E}' title='\overrightarrow{E}' class='latex' /> from a infinitely away from all charges (giving rise to ( <img src='http://s.wordpress.com/latex.php?latex=%5Coverrightarrow%7BE%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\overrightarrow{E}' title='\overrightarrow{E}' class='latex' />), to that point i.e.</p>
<img src='http://s.wordpress.com/latex.php?latex=V%20%3D-%20%5Cint_%7B%5Cinfty%7D%5E%7Br%7D%20%5Coverrightarrow%7BE%7D%20.%20%5Coverrightarrow%7Bdr%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V =- \int_{\infty}^{r} \overrightarrow{E} . \overrightarrow{dr}' title='V =- \int_{\infty}^{r} \overrightarrow{E} . \overrightarrow{dr}' class='latex' />
<p>The potential difference between two points is the work done in bringing per unit small positive change from one point to another. The unit of electric potential is joule/coul or volt.</p>
<p>The electric potential due to a point charge q at separation r is</p>
<img src='http://s.wordpress.com/latex.php?latex=V%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%20%7D.%20%5Cdfrac%7Bq%7D%7Br%7D%20%28%20%5Ctext%7Bsign%20of%20charge%20is%20to%20be%20retained%7D%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V = \dfrac{1}{4 \pi \varepsilon _0 }. \dfrac{q}{r} ( \text{sign of charge is to be retained} )' title='V = \dfrac{1}{4 \pi \varepsilon _0 }. \dfrac{q}{r} ( \text{sign of charge is to be retained} )' class='latex' />
<p>The electric potential due to a group of charges q<sub>1</sub>,q<sub>2</sub>,….q<sub>n</sub> is</p>
<p>V = V<sub>1</sub> + V<sub>2</sub> + V<sub>3</sub> + …..+ V<sub>N</sub></p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%7D%20.%20%28%20%5Cdfrac%7Bq_1%7D%7Br_1%7D%20%2B%20%5Cdfrac%7Bq_2%7D%7Br_2%7D%20%2B%20%5Cdfrac%7Bq_n%7D%7Br_n%7D%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%7D%20%5Csum%20%5Cdfrac%7Bq_i%7D%7Br_i%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon} . ( \dfrac{q_1}{r_1} + \dfrac{q_2}{r_2} + \dfrac{q_n}{r_n} = \dfrac{1}{4 \pi \varepsilon} \sum \dfrac{q_i}{r_i}' title='\dfrac{1}{4 \pi \varepsilon} . ( \dfrac{q_1}{r_1} + \dfrac{q_2}{r_2} + \dfrac{q_n}{r_n} = \dfrac{1}{4 \pi \varepsilon} \sum \dfrac{q_i}{r_i}' class='latex' />
<p>where r <sub>I </sub>; is the distance of point from charge q <sub>i</sub>.<br />
The relation between electric field and potential is</p>
<img src='http://s.wordpress.com/latex.php?latex=E%20%3D%20-%20%5Cdfrac%7B%5Ctriangle%20V%7D%7B%5Ctriangle%20r%7D%20%3D%20%5Cdfrac%7BV%7D%7Bd%7D%20%28%20%5Ctext%7Bnumerically%7D%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E = - \dfrac{\triangle V}{\triangle r} = \dfrac{V}{d} ( \text{numerically} )' title='E = - \dfrac{\triangle V}{\triangle r} = \dfrac{V}{d} ( \text{numerically} )' class='latex' />
<p><strong>Electric potential energy: </strong></p>
<p>The electric potential energy of a system of point charges is defined as the work sane in assembling this system of charges from an infinite distance apart from one another: It is assumed that all the charges are at test when they are at infinite distance apart from one another (i.e. their initial kinetic energy is zero).<br />
The electric potential energy of two point charges q<sub>1</sub> and q<sub>2</sub> at separation r<sub>12</sub> is</p>
<img src='http://s.wordpress.com/latex.php?latex=U_12%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7Bq_1q_2%7D%7Br_%7B12%7D%7D%20%5Cdots%20%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='U_12 = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{q_1q_2}{r_{12}} \dots (1)' title='U_12 = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{q_1q_2}{r_{12}} \dots (1)' class='latex' />
<p>In general for a system of n charges, the electric potential energy is</p>
<img src='http://s.wordpress.com/latex.php?latex=U%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20%5Csum_%7Bj%20%3D%201%7D%5En%20%5Csum_%7Bj%3E1%7D%5En%20%5Cdfrac%7Bq_i%20q_j%7D%7Br_%7Bij%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='U = \dfrac{1}{4 \pi \varepsilon _0} \sum_{j = 1}^n \sum_{j&gt;1}^n \dfrac{q_i q_j}{r_{ij}}' title='U = \dfrac{1}{4 \pi \varepsilon _0} \sum_{j = 1}^n \sum_{j&gt;1}^n \dfrac{q_i q_j}{r_{ij}}' class='latex' />
<p><strong>Electron volt: </strong></p>
<p>A small unit of energy: When a charge q is accelerated through a potential difference of V volts, the energy gained by charge = qV.<br />
When one electronic charge (= 1.6 X 10<sup>-19</sup> coul) is accelerated through a potential difference of one volt, then energy is called 1 electron volt. Electron volt abbreviated as eV is a small unit of energy in Particle Physics.</p>
<p>1 ev = 1.6 X 10<sup>-19</sup> coul</p>
<p><strong>Electric dipole</strong>:</p>
<p>Two equal and opposite charges separated by a finite distance constitute an electric dipole. If –q and +q are charges at distance 2l apart, then dipole moment</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Coverrightarrow%7Bp%7D%20%3D%20%5Coverrightarrow%7Bq%7D%20.%202%20%5Coverrightarrow%7Bl%7D%20%5Cdots%20%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\overrightarrow{p} = \overrightarrow{q} . 2 \overrightarrow{l} \dots (1)' title='\overrightarrow{p} = \overrightarrow{q} . 2 \overrightarrow{l} \dots (1)' class='latex' />
<p><strong>An electric dipole in a uniform field:</strong></p>
<p>When an electric dipole is placed in a uniform electric field E, at an angle <img src='http://s.wordpress.com/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\theta' title='\theta' class='latex' /> with the direction of field, the charges + q and –q  experience forces qE and qE in opposite directions, so that force on dipole = 0.</p>
<p><img class="aligncenter size-full wp-image-6712" title="" src="http://oscience.info/wp-content/uploads/Electric-dipole.jpg" alt="" width="251" height="138" /></p>
<p>Its direction is directed from –q to +q. The torque on a dipole in uniform electric field</p>
<p>t = <img src='http://s.wordpress.com/latex.php?latex=q%20E%20%28%202l%20%5Csin%20%5Ctheta%20%29%20%3D%20%28%20q%20.%202l%20%29%20E%20%5Csin%20%5Ctheta%20%5Cdots%20%281%29%20%5C%5C%20pE%20%5Csin%20%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='q E ( 2l \sin \theta ) = ( q . 2l ) E \sin \theta \dots (1) \\ pE \sin \theta' title='q E ( 2l \sin \theta ) = ( q . 2l ) E \sin \theta \dots (1) \\ pE \sin \theta' class='latex' /></p>
<p>Remark: If the electric field is non-uniform, then electric dipole placed in the field experiences both force and torque.<br />
Potential energy of electric dipole, <img src='http://s.wordpress.com/latex.php?latex=U%20%3D%20-pE%20%5Ccos%20%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='U = -pE \cos \theta' title='U = -pE \cos \theta' class='latex' /><br />
Work done in rotating the dipole from angle <img src='http://s.wordpress.com/latex.php?latex=%5Ctheta%20_1%20to%20%5Ctheta%20_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\theta _1 to \theta _2' title='\theta _1 to \theta _2' class='latex' />; is equal to the difference of potential energy of dipole is two orientations.</p>
<p>i.e. <img src='http://s.wordpress.com/latex.php?latex=U%20%3D%20U_2%20-%20U_1%20%3D%20pE%20%5Ccos%20%5Ctheta%20_2%20-%20%28%20-%20p%20E%20%5Ccos%20%5Ctheta%20_1%20%29%20%5C%5C%20pE%20%28%20%5Ccos%20%5Ctheta%20_1%20-%20%5Ccos%20%5Ctheta%20_2%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='U = U_2 - U_1 = pE \cos \theta _2 - ( - p E \cos \theta _1 ) \\ pE ( \cos \theta _1 - \cos \theta _2 )' title='U = U_2 - U_1 = pE \cos \theta _2 - ( - p E \cos \theta _1 ) \\ pE ( \cos \theta _1 - \cos \theta _2 )' class='latex' /></p>
<p><strong>Potential due to a short dipole:</strong></p>
<p>(i) At the axis : On a distance r from dipole</p>
<p>The electric field <img src='http://s.wordpress.com/latex.php?latex=E%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7B2p%7D%7Br%5E3%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{2p}{r^3}' title='E = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{2p}{r^3}' class='latex' /></p>
<p>The potential <img src='http://s.wordpress.com/latex.php?latex=V%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7Bp%7D%7Br%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{p}{r^2}' title='V = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{p}{r^2}' class='latex' /></p>
<p>(ii) At the equator: At a distance r from dipole</p>
<p>The electric field E = <img src='http://s.wordpress.com/latex.php?latex=E%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7Bp%7D%7Br%5E3%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{p}{r^3}' title='E = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{p}{r^3}' class='latex' /> parallel to dipole along +q to –q.<br />
The potential V = 0.</p>
<p>Electric potential at general point P (r, <img src='http://s.wordpress.com/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\theta' title='\theta' class='latex' /> );<br />
due to short dipole is</p>
<p><img class="aligncenter size-medium wp-image-6713" title="" src="http://oscience.info/wp-content/uploads/electric-191x300.jpg" alt="" width="191" height="300" /></p>
<p><strong>Electric field and potential in some special cases:</strong><strong></p>
<p></strong></p>
<p>#Due to a charged spherical conductor of charge Q, radius R at distance r:</p>
<p>Field outside point , E ( r &gt; R ) = <img src='http://s.wordpress.com/latex.php?latex=E%20%3D%20%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%7D%7Br%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{r^2}' title='E = \dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{r^2}' class='latex' /></p>
<p><strong>Surface point</strong>, E (r = R) = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%7D%7BR%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R^2}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R^2}' class='latex' /></p>
<p>Interior point , E ( r &lt; R ) = 0</p>
<p><strong>Potential </strong></p>
<p>Outside point , V ( r &gt; R ) = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%7D%7BR%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R}' class='latex' /></p>
<p>Potential at interior point is same as on the surface<br />
i.e. V ( r = R ) = V ( r &lt; R ) = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%7D%7BR%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R}' class='latex' /></p>
<p>#Due to a <strong><em>uniformly non-conducting charged of sphere</em></strong> of charge Q and radius R at distance r<br />
<strong>Field.</strong></p>
<p>Outside point, E ( r &gt; R ) =  <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%7D%7Br%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{r^2}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{r^2}' class='latex' /></p>
<p>Surface point, E ( r = R ) =  <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%7D%7BR%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R^2}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R^2}' class='latex' /><br />
Interior point, E (r &lt; R) =  <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%20r%7D%7BR%5E3%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q r}{R^3}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q r}{R^3}' class='latex' /><br />
<strong>Potential</strong></p>
<p>V ( r &gt; R ) =  <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%7D%7Br%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{r}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{r}' class='latex' /></p>
<p>V( r = R ) =  <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%7D%7BR%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q}{R}' class='latex' /></p>
<p>V ( r &lt; R ) =  <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7BQ%20%28%203%20R%5E2%20-%20r%5E2%20%29%7D%7B2R%5E3%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q ( 3 R^2 - r^2 )}{2R^3}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{Q ( 3 R^2 - r^2 )}{2R^3}' class='latex' /></p>
<p># For a long line charge of linear charge density q, electric field at separation r,</p>
<p>E = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7B2q%7D%7Br%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{2q}{r}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{2q}{r}' class='latex' /></p>
<p>Potential between two points at distance r<sub>1</sub> and r<sub>2</sub> from the wire is</p>
<p>V  =  <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%202q%20%5Clog%20_e%20%5Cdfrac%7Br_2%7D%7Br%5E1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} 2q \log _e \dfrac{r_2}{r^1}' title='\dfrac{1}{4 \pi \varepsilon _0} 2q \log _e \dfrac{r_2}{r^1}' class='latex' /></p>
<p># Near a thin flat sheet of charge of surface charge density <img src='http://s.wordpress.com/latex.php?latex=%5Csigma%20%5C%5C%20E%20%3D%20%5Csigma%20%2F%202%20%5Cvarepsilon%20_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sigma \\ E = \sigma / 2 \varepsilon _0' title='\sigma \\ E = \sigma / 2 \varepsilon _0' class='latex' />,\.</p>
<p>#Near a conductor of any shape, <img src='http://s.wordpress.com/latex.php?latex=E%20%3D%20%5Csigma%20%2F%20%5Cvarepsilon%20_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E = \sigma / \varepsilon _0' title='E = \sigma / \varepsilon _0' class='latex' /></p>
<p>#The electric field at the axis of a uniformly charged ring of charge q, and radius R at a distance x from centre is</p>
<p>E = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B1%7D%7B4%20%5Cpi%20%5Cvarepsilon%20_0%7D%20.%20%5Cdfrac%7Bqx%7D%7B%28%20R%5E2%20%2B%20x%5E2%20%29%5E%7B%5Cdfrac%7B3%7D%7B2%7D%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{qx}{( R^2 + x^2 )^{\dfrac{3}{2}}}' title='\dfrac{1}{4 \pi \varepsilon _0} . \dfrac{qx}{( R^2 + x^2 )^{\dfrac{3}{2}}}' class='latex' /></p>
<p><strong>Kinetic energy of a charged particle</strong> accelerated through a potential difference of V volts is</p>
<p>E<sub>K</sub> = qV</p>
<p>If v is the velocity gained, then</p>
<p>½ mv<sup>2</sup> = qV</p>
<p>or v = <img src='http://s.wordpress.com/latex.php?latex=%5Csqrt%7B2%20%5Cdfrac%7BqV%7D%7Bm%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{2 \dfrac{qV}{m}}' title='\sqrt{2 \dfrac{qV}{m}}' class='latex' /></p>
<p><strong>Path of charged particle in Transverse Electric Field:</strong></p>
<p>Let the electric field E be along Y-axis and the charged particle q, mass m enter the field along X-axis with speed u. The electric force on the particle is along Y-axis given by<br />
<img class="aligncenter size-medium wp-image-6714" title="" src="http://oscience.info/wp-content/uploads/transverse-277x300.jpg" alt="" width="277" height="300" /></p>
<p>The initial velocity of the particle is u along X-axis and it is zero along Y-axis. Therefore the deflection of charged particle along Y-axis after time is</p>
<p>y = u<sub>y</sub> t + ½ a<sub>y</sub> r <sup>2</sup></p>
<img src='http://s.wordpress.com/latex.php?latex=%3D%200%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%20.%20%5Cdfrac%7BqE%7D%7Bm%7Dr%5E2%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20.%20%5Cdfrac%7BqE%7D%7Bm%7Dr%5E2%20%5Cdots%20%283%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= 0 + \dfrac{1}{2} . \dfrac{qE}{m}r^2 = \dfrac{1}{2} . \dfrac{qE}{m}r^2 \dots (3)' title='= 0 + \dfrac{1}{2} . \dfrac{qE}{m}r^2 = \dfrac{1}{2} . \dfrac{qE}{m}r^2 \dots (3)' class='latex' />
<p>As there is no acceleration along X-axis, therefore the distance traversed by particle in time t along X—axis is<br />
x = u t &#8230;..(4)</p>
<p>Eliminating from (3) and (4), we get</p>
<img src='http://s.wordpress.com/latex.php?latex=y%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20.%20%5Cdfrac%7BqE%7D%7Bm%7D%20%28%20%5Cdfrac%7Bx%7D%7Bu%7D%20%29%5E2%20i.e.%20y%20%3D%20%5Cdfrac%7BqE%7D%7B2mu%5E2%7D%20x%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = \dfrac{1}{2} . \dfrac{qE}{m} ( \dfrac{x}{u} )^2 i.e. y = \dfrac{qE}{2mu^2} x^2' title='y = \dfrac{1}{2} . \dfrac{qE}{m} ( \dfrac{x}{u} )^2 i.e. y = \dfrac{qE}{2mu^2} x^2' class='latex' />
<p>This shows that the path of charged particle in transverse electric field is a parabola.</p>
<p><strong>Energy associated with Electric Field:</strong></p>
<p>The electric energy stored in electrified per unit volume.</p>
<p><img src='http://s.wordpress.com/latex.php?latex=u_e%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cvarepsilon%20E%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_e = \dfrac{1}{2} \varepsilon E^2' title='u_e = \dfrac{1}{2} \varepsilon E^2' class='latex' /><br />
This is called electric energy density.</p>
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		<title>Doppler&#8217;s effect</title>
		<link>http://oscience.info/physics/dopplers-effect/dopplers-effect/</link>
		<comments>http://oscience.info/physics/dopplers-effect/dopplers-effect/#comments</comments>
		<pubDate>Thu, 05 Apr 2012 07:01:02 +0000</pubDate>
		<dc:creator>sandeep</dc:creator>
				<category><![CDATA[Doppler's effect]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6705</guid>
		<description><![CDATA[Doppler’s Effect: The apparent change in frequency of a sounding body due to relative motion between source and observer is called Doppler’s effect. Doppler’s effect takes place both in sound and light. In sound it depends on whether the source or observer or both are in motion; while in light it depends only on the [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Doppler’s Effect: </strong></p>
<p>The apparent change in frequency of a sounding body due to relative motion between source and observer is called Doppler’s effect. Doppler’s effect takes place both in sound and light. In sound it depends on whether the source or observer or both are in motion; while in light it depends only on the fact that whether the distance between source and observer is decreasing or increasing.</p>
<p><strong><span style="text-decoration: underline;">Doppler’s Effect in Sound</span></strong><strong></strong></p>
<p><em>(<strong>i) Source in motion and observer at rest. </strong></em></p>
<p>If source is approaching the observer with speed v<sub>s</sub>, then n waves emitted per second are spread in distance v &#8211; v<sub>s</sub> instead of v.<br />
Therefore apparent wavelength</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Clambda%27%20%3D%20%5Cdfrac%7Bv%20-%20v_s%7D%7Bn%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda&#039; = \dfrac{v - v_s}{n}' title='\lambda&#039; = \dfrac{v - v_s}{n}' class='latex' />
<p>or the apparent frequency n’ is given by</p>
<img src='http://s.wordpress.com/latex.php?latex=n%27%20%3D%20%5Cdfrac%7Bv%7D%7B%5Clambda%27%7D%20%3D%20%5Cdfrac%7Bv%7D%7Bv%20-%20v_s%7D%20n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n&#039; = \dfrac{v}{\lambda&#039;} = \dfrac{v}{v - v_s} n' title='n&#039; = \dfrac{v}{\lambda&#039;} = \dfrac{v}{v - v_s} n' class='latex' />
<p>Similarly, if source is moving away from the observer; then</p>
<img src='http://s.wordpress.com/latex.php?latex=n%27%20%3D%20%5Cdfrac%7Bv%7D%7Bv%20%2B%20v_s%7D%20n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n&#039; = \dfrac{v}{v + v_s} n' title='n&#039; = \dfrac{v}{v + v_s} n' class='latex' />
<p><strong><span style="text-decoration: underline;">(ii) Source at rest and observer in motion.</span></strong></p>
<p>If observer is moving away from the source with speed v<sub>o</sub>, then the observer does not receive the waves in distance v<sub>0</sub>, therefore the apparent frequency</p>
<img src='http://s.wordpress.com/latex.php?latex=n%27%20%3D%20n%20-%20%5Ctriangle%20n%20%5Cdfrac%7Bv_0%7D%7B%5Clambda%7D%20%3D%20n%20-%20%5Cdfrac%7Bv_0%7D%7Bv%2Fn%7D%20%5C%5C%20%3D%20%28%201%20-%20%5Cdfrac%7Bv_0%7D%7Bv%7D%20%29%20n%20%3D%20%5Cdfrac%7Bv%20-%20v_0%7D%7Bv%7D%20n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n&#039; = n - \triangle n \dfrac{v_0}{\lambda} = n - \dfrac{v_0}{v/n} \\ = ( 1 - \dfrac{v_0}{v} ) n = \dfrac{v - v_0}{v} n' title='n&#039; = n - \triangle n \dfrac{v_0}{\lambda} = n - \dfrac{v_0}{v/n} \\ = ( 1 - \dfrac{v_0}{v} ) n = \dfrac{v - v_0}{v} n' class='latex' />
<p>Similarly, If observer its approaching the source, then apparent frequency</p>
<img src='http://s.wordpress.com/latex.php?latex=n%27%20%3D%20n%20%2B%20%5Ctriangle%20n%20%3D%20%5Cdfrac%7Bv%20%2B%20v_0%7D%7Bv%7D%20n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n&#039; = n + \triangle n = \dfrac{v + v_0}{v} n' title='n&#039; = n + \triangle n = \dfrac{v + v_0}{v} n' class='latex' />
<p>(<strong><span style="text-decoration: underline;">iii) Source and observer are both in motion. </span></strong></p>
<p>When source and observer are both in motion along the same direction, then apparent frequency due to motion of source alone is increased to<br />
<img class="aligncenter size-medium wp-image-6706" title="" src="http://oscience.info/wp-content/uploads/source-300x52.jpg" alt="" width="300" height="52" /><br />
while that due to motion of observer is decreased to</p>
<img src='http://s.wordpress.com/latex.php?latex=n%27%20%3D%20%5Cdfrac%7Bv%20-%20v_0%7D%7Bv%7D%20n_1%20%3D%20%5Cdfrac%7Bv%20-%20v_0%7D%7Bv%7D%20.%20%5Cdfrac%7Bv%7D%7Bv%20-%20v_s%7D%20n%20%3D%20%5Cdfrac%7Bv%20-%20v_0%7D%7Bv%20-%20v_s%7D%20n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n&#039; = \dfrac{v - v_0}{v} n_1 = \dfrac{v - v_0}{v} . \dfrac{v}{v - v_s} n = \dfrac{v - v_0}{v - v_s} n' title='n&#039; = \dfrac{v - v_0}{v} n_1 = \dfrac{v - v_0}{v} . \dfrac{v}{v - v_s} n = \dfrac{v - v_0}{v - v_s} n' class='latex' />
<p>Remark :</p>
<p>(i) If wind also blows with speed <img src='http://s.wordpress.com/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\omega' title='\omega' class='latex' /> then v should be replaced by v + <img src='http://s.wordpress.com/latex.php?latex=%5Comega%20%5Ccos%20%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\omega \cos \theta' title='\omega \cos \theta' class='latex' /> where <img src='http://s.wordpress.com/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\theta' title='\theta' class='latex' /> is the angle between direction of propagation of sound and direction of wind. If wind blows along the direction of propagation of sound  <img src='http://s.wordpress.com/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\theta' title='\theta' class='latex' /> = 0 or <img src='http://s.wordpress.com/latex.php?latex=%5Ccos%20%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\cos \theta' title='\cos \theta' class='latex' /> = 1 but if it blows in opposite direction <img src='http://s.wordpress.com/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\theta' title='\theta' class='latex' /> = 180<sup>o</sup> or <img src='http://s.wordpress.com/latex.php?latex=%5Ccos%20%5Ctheta%20%3D%20-%201&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\cos \theta = - 1' title='\cos \theta = - 1' class='latex' /></p>
<p>(ii) In the solution of problems the formula for apparent frequency is chosen as</p>
<img src='http://s.wordpress.com/latex.php?latex=n%27%20%3D%20%5Cdfrac%7Bv%20-%20v_0%7D%7Bv%20-%20v_s%7D%20n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n&#039; = \dfrac{v - v_0}{v - v_s} n' title='n&#039; = \dfrac{v - v_0}{v - v_s} n' class='latex' />
<p>If wind blows with velocity <img src='http://s.wordpress.com/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\omega' title='\omega' class='latex' /> in a direction at angle <img src='http://s.wordpress.com/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\theta' title='\theta' class='latex' /> with <img src='http://s.wordpress.com/latex.php?latex=%5Coverrightarrow%7Bv%7D%20%5C%5C%20n%27%20%3D%20%5Cdfrac%7Bv%20%2B%20%5Comega%20%5Ccos%20%5Ctheta%20-%20v_0%7D%7Bv%20%2B%20%5Comega%20%5Ccos%20%5Ctheta%20-%20v_s%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\overrightarrow{v} \\ n&#039; = \dfrac{v + \omega \cos \theta - v_0}{v + \omega \cos \theta - v_s}' title='\overrightarrow{v} \\ n&#039; = \dfrac{v + \omega \cos \theta - v_0}{v + \omega \cos \theta - v_s}' class='latex' /></p>
<p>If direction of source or observer changes, the formula is modified taking proper sign for velocities.</p>
<p><strong>Doppler’s Effect in Light. </strong></p>
<p>If v is actual frequency of a light source and v’ the apparent frequency then for approach</p>
<p><img src='http://s.wordpress.com/latex.php?latex=v%27%20%3D%20%5Csqrt%7B1%20%2B%20v%20%2F%20c%7D%7B1%20-%20v%20%2F%20c%7D%20v%20%5C%2C%20i.e.%20v%27%20%3E%20v%20%5C%5C%20%5Ctext%7Band%7D%20%5Ctriangle%20v%20%3D%20v%27%20-%20v%20%3D%20%5Cdfrac%7Bv%7D%7Bc%7D%20v%20if%20v%20%3C%3C%20c%20%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v&#039; = \sqrt{1 + v / c}{1 - v / c} v \, i.e. v&#039; &gt; v \\ \text{and} \triangle v = v&#039; - v = \dfrac{v}{c} v if v &lt;&lt; c ,' title='v&#039; = \sqrt{1 + v / c}{1 - v / c} v \, i.e. v&#039; &gt; v \\ \text{and} \triangle v = v&#039; - v = \dfrac{v}{c} v if v &lt;&lt; c ,' class='latex' /><br />
v being relative velocity of source and observer and c = speed of light in vacuum = 3 x 10<sup>8</sup> m/s</p>
<p><sup> </sup>For receding</p>
<img src='http://s.wordpress.com/latex.php?latex=v%27%20%3D%20%5Csqrt%7B%20%28%20%5Cdfrac%7B1%20%2B%20v%2Fc%7D%7B1%20-%20v%2Fc%7D%20%29%20%7D%20v%20%5C%2C%20i.e.%20v%27%20%3Cc%20%5C%5C%20%5Ctext%7Band%7D%20%5Ctriangle%20v%20%3D%20v%20-%20v%27%20%3D%20%5Cdfrac%7Bv%7D%7Bc%7D%20%5Clambda%20%5Ctext%7Band%7D%20%5Clambda%27%20%5Ctext%7Bare%20actual%20and%20observed%20wavelength%20%2C%20then%7D%20%5C%5C%20%5Ctriangle%20%5Clambda%20%3D%20%5Clambda%20-%20%5Clambda%27%20%3D%20%5Cdfrac%7Bv%7D%7Bc%7D%20%5Clambda%20%5Ctext%7Bfor%20approval%7D%20%5C%5C%20%5Ctriangle%20%5Clambda%20%3D%20%5Clambda%27%20-%20%5Clambda%20%3D%20%5Cdfrac%7Bv%7D%7Bc%7D%20%5Clambda%20%5Ctext%7Bfor%20recording%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v&#039; = \sqrt{ ( \dfrac{1 + v/c}{1 - v/c} ) } v \, i.e. v&#039; &lt;c \\ \text{and} \triangle v = v - v&#039; = \dfrac{v}{c} \lambda \text{and} \lambda&#039; \text{are actual and observed wavelength , then} \\ \triangle \lambda = \lambda - \lambda&#039; = \dfrac{v}{c} \lambda \text{for approval} \\ \triangle \lambda = \lambda&#039; - \lambda = \dfrac{v}{c} \lambda \text{for recording}' title='v&#039; = \sqrt{ ( \dfrac{1 + v/c}{1 - v/c} ) } v \, i.e. v&#039; &lt;c \\ \text{and} \triangle v = v - v&#039; = \dfrac{v}{c} \lambda \text{and} \lambda&#039; \text{are actual and observed wavelength , then} \\ \triangle \lambda = \lambda - \lambda&#039; = \dfrac{v}{c} \lambda \text{for approval} \\ \triangle \lambda = \lambda&#039; - \lambda = \dfrac{v}{c} \lambda \text{for recording}' class='latex' />
<p><strong>Characteristics of Musical Sound. </strong></p>
<p>There are three characteristics of sound :</p>
<p><strong>(i) Loudness or intensity:</strong></p>
<p>The intensity of sound is defined as the amount of sound energy crossing per unit area around a point per second. The intensity I is given by</p>
<img src='http://s.wordpress.com/latex.php?latex=I%20%3D%202%20%5Cpi%20%5E2%20n%5E2%20a%5E2%20%5Crho%20v%20%5Cdots%20%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I = 2 \pi ^2 n^2 a^2 \rho v \dots (1)' title='I = 2 \pi ^2 n^2 a^2 \rho v \dots (1)' class='latex' />
<p>when   n=frequency of wave, a = amplitude, <img src='http://s.wordpress.com/latex.php?latex=%5Crho&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\rho' title='\rho' class='latex' />= density of medium, v = speed of wave<br />
The level of sound is measured in bels and decibels.</p>
<p>Number of bels = <img src='http://s.wordpress.com/latex.php?latex=%5Clog_10%20%5Cdfrac%7BI%7D%7BI_0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\log_10 \dfrac{I}{I_0}' title='\log_10 \dfrac{I}{I_0}' class='latex' /></p>
<p>Number of decibels = <img src='http://s.wordpress.com/latex.php?latex=10%20%5Clog_10%20%5Cdfrac%7BI%7D%7BI_0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='10 \log_10 \dfrac{I}{I_0}' title='10 \log_10 \dfrac{I}{I_0}' class='latex' /><br />
where I<sub>0</sub> is reference intensity = 10<sup>-12</sup> watt/m<sup>2</sup>.</p>
<p>The loudness of sound is the degree of sensation produced in the ear and it depends on intensity  as well as upon the sensitiveness of ear. The loudness<br />
L and intensity I are related as</p>
<p>L = k log 1                                   &#8230;(2)</p>
<p><img class="aligncenter size-medium wp-image-6707" title="" src="http://oscience.info/wp-content/uploads/noise-level-300x246.jpg" alt="" width="300" height="246" /></p>
<p>(ii) Pitch or Frequency:</p>
<p>The pitch is the characteristic which distinguishes between a shrill (or sharp) sound and a grave (or flat) sound. A sound of high pitch is said to be shrill and of low pitch a grave sound. The pitch does not depend on intensity and loudness but depends on the frequency. The pitch of a sound changes due to Doppler’s effect.</p>
<p>(iii) Quality or Timbre:</p>
<p>The quality of sound enables us to distinguish between two sounds having same loudness and pitch. The quality of sound depends on the presence of overtones. Due to quality of of sound one can recognise the voice of his friend without seeing him.</p>
<p>&nbsp;</p>
<p><strong>Musical Scale:</strong></p>
<p>A series of notes sounded in succession in such a way that their frequencies have a definite ratios and which produce a pleasant on the ear is called a musical scale.<br />
The ratio between two frequencies of two notes of musical scale is called a musical interval<strong><em>.</em></strong></p>
<p><em>Major Diatonic Scale:</em> The most common musical scale is Diatonic Scale. It consists of eight<br />
notes in ascending order of frequencies, comprising &#8211; an octave i.e. the frequency ratio of eighth and first note is 2 : 1. Conventionally the fundamental frequency of first note is taken as 256 Hz and that of the last note as 512 Hz. The following table illustrates the frequencies of intermediate notes along with their western and Indian names.</p>
<p><img class="aligncenter size-medium wp-image-6708" title="Major diatonic scale" src="http://oscience.info/wp-content/uploads/Major-diatonic-scale-300x265.jpg" alt="" width="300" height="265" /><br />
Clearly the frequency ratio of adjacent notes in diatonic scale bear simple ratios i.e. either 9/8 or 10/9 or 16/15.</p>
<p>Key Note:</p>
<p>The note C of lowest frequency is called the key note and is usually taken to be of frequency 256 Hz or 2<sup>8</sup> Hz.</p>
<p>Major Tone:</p>
<p>The interval between two notes having frequency ratio 9/8 is called major tone.</p>
<p>Minor Tone:</p>
<p>The interval between two notes having frequency ratio 10/9 is called a minor tone.</p>
<p>Half Tone:</p>
<p>The interval between two notes having frequency ratio 16/15 is called a half tone.<br />
The major diatonic scale has the following characteristics.<br />
(i) The interval between any note of the diatonic scale with the key note bears a simple ratio.<br />
(ii) The interval between any two adjacent notes of the diatomic scale also bears a simple ratio.<br />
(iii) The interval between any two notes is the equal to the sum (in musical scale the product of intervals is called the sum) of the intervals between the intermediate notes.<br />
For example take intervals E and B.<br />
The interval between E and B = <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B480%7D%7B320%7D%20%3D%20%5Cdfrac%7B3%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{480}{320} = \dfrac{3}{2}' title='\dfrac{480}{320} = \dfrac{3}{2}' class='latex' /><br />
The sum of intermediate intervals is <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B16%7D%7B5%7D%20%5Ctimes%20%5Cdfrac%7B9%7D%7B8%7D%20%5Ctimes%20%5Cdfrac%7B10%7D%7B9%7D%20%5Ctimes%20%5Cdfrac%7B9%7D%7B8%7D%20%5Ctimes%20%5Cdfrac%7B3%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{16}{5} \times \dfrac{9}{8} \times \dfrac{10}{9} \times \dfrac{9}{8} \times \dfrac{3}{2}' title='\dfrac{16}{5} \times \dfrac{9}{8} \times \dfrac{10}{9} \times \dfrac{9}{8} \times \dfrac{3}{2}' class='latex' /></p>
<p><strong>Practical Difficulty in Diatonic Scale:</strong></p>
<p>The first note of diatonic scale is called the key note (or tonic) note of the scale. In diatonic scale C is key note. The characteristic of music is that a singer often changes his key (or tones) note he may start with C or D or E as key note. If he starts with D or E, then another scale with above given intervals has to be constructed, otherwise the sound will appear unpleasant. To remove the difficulty, equally tempered scale has been constructed.</p>
<p><strong>Equally Tempered Scale: </strong></p>
<p>It consists of 13 keys and 12 intervals. The intervals are equal and<br />
each interval is 2 <sup>1/2.</sup><br />
In this scale few new notes have been introduced. They are Cs, Ds, Fs, Gs and As. The main advantage of this scale in that the interval is same between the consecutive notes and so a singer can conveniently use any key as his tonic. Clearly the intervals in equally tempered scale form a geometric progression. In the Harmonium 13 keys are provided. 8 white keys and 5 black keys (fig.) 8 white keys</p>
<p><img class="aligncenter size-medium wp-image-6709" title="" src="http://oscience.info/wp-content/uploads/Tempered-scale-300x152.jpg" alt="" width="300" height="152" /></p>
<p>represent appropriately the diatonic scale. If a singer starts with C as key note, he can play only with white keys forming diatonic scale, but if he uses any other key&#8217; as key note, black keys have to be used. Usually 3 or 4 octaves are provided in harmonium or piano.<br />
<strong>4. Acoustics of Buildings.</strong><br />
Acoustic of buildings deals with the design and construction of halls, auditoria etc. with respect to sound waves. If the speech and music delivered in a hall or an auditorium is audible to listeners at all positions without any disturbance, the hall or auditorium is said to be accoustically good.<br />
W.C. Sabine in 1911, studied the problem and laid down the following essential features or demand for a hall or auditorium for its being acoustically good:</p>
<p>(i) The sound must be large enough at each position of the hall.<br />
(ii) There should be no echo so that the quality of sound should remain unchanged.<br />
(iii) The successive syllables spoken should be clear without any echo.<br />
(iv)There should be no noise at all.</p>
<p>Let us now consider each feature one by one.</p>
<p><strong>Sufficient loudness: </strong></p>
<p>At each position of the hall, sufficient loudness is essential. Reflecting surfaces like walls and ceiling of the hall are quite helpful in providing proper loudness.<br />
The walls-behind the speaker and ceiling -reflect the sound of speaker and help in increasing the loudness. The reflection of sound to the back is very useful in increasing the sound intensity there. Therefore, it is useful to make back and side walls in the vicinity of the speaker in the form of reflector with the speaker at the focus. By this method the sound may be heard uniformly in the hall by a large gathering, but this method has following two shortcomings.<br />
(i) The area of speaker becomes-limited.<br />
(ii) All incidental noise in the hall is focused at speaker.</p>
<p><strong>Echoes: </strong></p>
<p>It is found that if the interval between the direct sound and reflected sound is less than 1/15 sec, the reflected sound is helpful in increasing the loudness but those sounds arriving later cause confusions and should be weakened as far as possible. In big halls with high ceiling, the surfaces which may give echo should be covered with sound absorbing material to prevent reflection from them.</p>
<p><strong> Zones of maximum and minimum sound:</strong></p>
<p>In a hall there may be zones of maximum and minimum sound. This is due to curved reflecting surfaces of a hall which produce undesirable concentration of sound in some regions of hall and reduce intensity at other places. In an acoustically good hall, this is undesirable. Therefore, the curved surfaces should be avoided or covered with sound absorbing materials.</p>
<p><strong>Proper reverberation:</strong></p>
<p>It is common experience that sound of a speaker persists unduly in the hall. The sound reaches the listener directly and then after successive reflections from the walls, ceiling etc. The listener therefore receives a series of sounds of diminishing intensity. The persistence of sound is called reverberation.</p>
<p>Professor Sabine measured intensity of sound in a hall with time. If a graph is plotted between intensity and time, it is as shown in fig. When sound is started, it grows to a maximum value and when it is stopped it decays to zero in exponential manner: (i.e. first rapidly and then slowly). Clearly if a syllable persists for a time during which the next is uttered, it will overlap. The most important parameter to be controlled is reverberation time which is roughly the time during which the sound persists in a hall. More precisely reverberation time is nearly the time during which the sound intensity; falls to the minimum audible Value (called audibility threshold) from an initial value which is 10<sup>6</sup> time the audibility threshold. Thus, the reverberation time<br />
is the time taken for the sound to fall 10<sup>-6</sup> time of its maximum steady value. If the period of reverberation is large, then at following sequence of short sounds such as syllables in speech will ova lap. The greater is the time of reverberation of a half the greater will be the confusion due to mixing different syllables. As a matter of fact excessive reverberation is the common defect in acoustics large buildings. By decrasing time of reverberation, the loudness of the hall also decreases.<br />
If the reverberation is very small, the loudness is insufficient. In such a hall the speaker finds no response from the hall. Such a hall is said to be dead for the speaker. In such a hall the sound falls below audibility very soon.<br />
Thus reverberation period should neither be too large nor small, but it must possess a suitable value which may be good for speaker and the audience. This suitable value of time is called optimum reverberation time for the hall. Sabine showed that the optimum period of reverberation depends on the size of the hall and the quality of sound.<br />
After measuring reverberation time (T) in various rooms halls of different sizes, Sabine gave an empirical relation.</p>
<img src='http://s.wordpress.com/latex.php?latex=T%20%3D%20%5Cdfrac%7BCV%7D%7BA%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T = \dfrac{CV}{A}' title='T = \dfrac{CV}{A}' class='latex' />
<p>where C is a constant, V is volume of hall in m<sup>3</sup> and A is total absorption of sound in a hall.</p>
<p><img src='http://s.wordpress.com/latex.php?latex=A%20%3D%20%5Csum%20a%20s&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A = \sum a s' title='A = \sum a s' class='latex' /><br />
where a’s are absorption coefficients of surfaces of surface areas of different materials present in the hall.<strong> </strong>Clearly reverberation time can be reduced if the total absorbing area A is increased.<br />
That is why the reverberation period of an will be more than that of a crowded hall.<br />
For a given hall; the reverberation time can be controlled by inserting or removing sound absorbing materials at the walls.<br />
The period is greater for music than for speech. For example, for a hall of 10<sup>4</sup> m<sup>3</sup>, the reverberation period is between 1.0 and 1.5 sec for speech and between 1.5 and 2.0 sec for music. Below this value the intensity of sound is weak and above this value, the syllables overlap.</p>
<p><strong> Insulation of hall from noise:</strong></p>
<p>The building should be insulated properly from external and internal noises. To eliminate external noise the doors and windows should be tightly fitted. Double doors and windows with separate frames for each having insulating material between them should be used.<br />
The inside noise of typewriters, machinery etc. is reduced by placing them on absorbing materials like pads, wood, felt etc. and using curtains of absorbing materials.</p>
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		<title>Principle of superposition</title>
		<link>http://oscience.info/physics/principle-of-superposition/</link>
		<comments>http://oscience.info/physics/principle-of-superposition/#comments</comments>
		<pubDate>Wed, 04 Apr 2012 19:07:01 +0000</pubDate>
		<dc:creator>sandeep</dc:creator>
				<category><![CDATA[Physics]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6677</guid>
		<description><![CDATA[Principle of superposition When two or more waves propagating in a medium arrive at the same point simultaneously, a new wave is produced. This phenomenon is called superposition of According to Young the net displacement at any point of the medium is equal to the algebraic sum of displacements of individual waves arriving at that [...]]]></description>
			<content:encoded><![CDATA[<p><strong>Principle of superposition</strong></p>
<p>When two or more waves propagating in a medium arrive at the same point simultaneously, a new wave is produced. This phenomenon is called superposition of According to Young the net displacement at any point of the medium is equal to the algebraic sum of displacements of individual waves arriving at that point simultaneously. This is called the principle of superposition and holds good as long as the amplitude of the waves is not too large. This principle is of extreme importance and can be applied to many types of waves e.g. sound waves, light waves, wave pulses etc. The superposition of harmonic waves gives rise to interference, beats and standing waves.<br />
To recognize the types of waves, the following<br />
equations must be known.</p>
<p><span style="text-decoration: underline;">Equation of a straight line </span></p>
<div id="attachment_6678" class="wp-caption aligncenter" style="width: 310px"><img class="size-full wp-image-6678" title="Equation of straight line" src="http://oscience.info/wp-content/uploads/Equation-of-straight-line.jpg" alt="" width="300" height="272" /><p class="wp-caption-text">Equation of straight line</p></div>
<p>y = mx + c</p>
<p><span style="text-decoration: underline;">Equation of circle of radius</span> ‘a’<strong> </strong>is x<sup>2</sup> + y<sup>2</sup> = a<sup>2</sup></p>
<div id="attachment_6681" class="wp-caption aligncenter" style="width: 310px"><img class="size-medium wp-image-6681" title="Equation of circle" src="http://oscience.info/wp-content/uploads/Equation-of-circle1-300x289.jpg" alt="" width="300" height="289" /><p class="wp-caption-text">Equation of circle</p></div>
<p><span style="text-decoration: underline;">Equation of ellipse </span></p>
<p>&nbsp;</p>
<div id="attachment_6683" class="wp-caption aligncenter" style="width: 310px"><img class="size-medium wp-image-6683" title="Equation of ellipse" src="http://oscience.info/wp-content/uploads/Equation-of-ellipse1-300x267.jpg" alt="" width="300" height="267" /><p class="wp-caption-text">Equation of ellipse</p></div>
<p><span style="text-decoration: underline;"><br />
</span></p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D%20%2B%20%5Cdfrac%7By%5E2%7D%7Bb%5E2%7D%20%3D%201&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1' title='\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1' class='latex' />
<p>Where ,</p>
<p>a = semi-major axis</p>
<p>b = semi-minor axis</p>
<p><span style="text-decoration: underline;">Equation of parabola </span>is</p>
<p>y<sup>2</sup> = 4 a x ( fig .a )</p>
<div id="attachment_6687" class="wp-caption aligncenter" style="width: 310px"><img class="size-medium wp-image-6687" title="Equation of parabola" src="http://oscience.info/wp-content/uploads/Equation-of-parabola2-300x163.jpg" alt="" width="300" height="163" /><p class="wp-caption-text">Equation of parabola</p></div>
<p>This is parabola symmetrically about x-axis and</p>
<p>x<sup>2</sup> = 4 a y (fig . b)</p>
<p>This is parabola symmetrical about y &#8211; axis</p>
<p>Beats:</p>
<p>When two waves of slightly different frequencies travel along the same straight line and along the same direction, they superimpose in such a way that the resultant intensity alternatively increases or decreases. This phenomenon of waxing and waning of sound is called beats. One waxing and one waning forms one beat.<br />
For the sake of simplicity we assume the two waves of slightly different frequencies n<sub>1</sub> and n<sub>2 </sub>(at x =0 ) are represented as</p>
<img src='http://s.wordpress.com/latex.php?latex=y_1%20%3D%20a%20sin%202%20%5Cpi%20n_1%20t%20%5Ctext%7Band%7D%20%5C%2C%20y_2%20%3D%20a%20%5Csin%202%20%5Cpi%20n_2%20t&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y_1 = a sin 2 \pi n_1 t \text{and} \, y_2 = a \sin 2 \pi n_2 t' title='y_1 = a sin 2 \pi n_1 t \text{and} \, y_2 = a \sin 2 \pi n_2 t' class='latex' />
<p>From Young’s principle of superposition</p>
<p><img src='http://s.wordpress.com/latex.php?latex=y%20%3D%20y_1%20%2B%20y_2%20%3D%20a%20%5Csin%202%20%5Cpi%20n_1%20t%20%2B%20a%20%5Csin%202%20%5Cpi%20n_2%20t%20%5C%5C%202a%20%5Csin%20%5Cdfrac%7B2%20%5Cpi%20%28%20n_1%20%2B%20n_2%20%29%20t%7D%7B2%7D%20%5Ccos%20%5Cdfrac%7B2%20%5Cpi%20%28%20n_1%20-%20n_2%20%29%20t%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = y_1 + y_2 = a \sin 2 \pi n_1 t + a \sin 2 \pi n_2 t \\ 2a \sin \dfrac{2 \pi ( n_1 + n_2 ) t}{2} \cos \dfrac{2 \pi ( n_1 - n_2 ) t}{2}' title='y = y_1 + y_2 = a \sin 2 \pi n_1 t + a \sin 2 \pi n_2 t \\ 2a \sin \dfrac{2 \pi ( n_1 + n_2 ) t}{2} \cos \dfrac{2 \pi ( n_1 - n_2 ) t}{2}' class='latex' /> …..(1)</p>
<p>Let n<sub>1</sub> = n and n<sub>2</sub> = n + <img src='http://s.wordpress.com/latex.php?latex=%5Ctriangle%20n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\triangle n' title='\triangle n' class='latex' /> such that</p>
<p><img src='http://s.wordpress.com/latex.php?latex=%5Ctriangle%20n%20%3C%3C%20n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\triangle n &lt;&lt; n' title='\triangle n &lt;&lt; n' class='latex' /> ; then equation (1) gives</p>
<p><img src='http://s.wordpress.com/latex.php?latex=y%20%3D%202a%20%5Csin%202%20%5Cpi%20n%20t%20%5Ccos%20%5Cpi%20%28n_1%20-%20n_2%20%29t%20%5C%5C%20%5Ctext%7Bor%7D%20%5C%2C%20%5C%2C%20%5C%2C%20y%20%3D%202a%20%5Ccos%20%5Cpi%20%28%20n_1%20-%20n_2%20%29%20t%20%5Csin%202%20%5Cpi%20n%20t%20%5Cdots%20%282a%29%20%5C%5C%20%5C%2C%20%3D%20A%20%5Csin%202%20%5Cpi%20n%20t%20%5C%5C%20%5Ctext%7Bwhere%7D%20A%20%3D%202a%20%5Ccos%20%5Cpi%20%28%20n_1%20-%20n_2%20%29t&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = 2a \sin 2 \pi n t \cos \pi (n_1 - n_2 )t \\ \text{or} \, \, \, y = 2a \cos \pi ( n_1 - n_2 ) t \sin 2 \pi n t \dots (2a) \\ \, = A \sin 2 \pi n t \\ \text{where} A = 2a \cos \pi ( n_1 - n_2 )t' title='y = 2a \sin 2 \pi n t \cos \pi (n_1 - n_2 )t \\ \text{or} \, \, \, y = 2a \cos \pi ( n_1 - n_2 ) t \sin 2 \pi n t \dots (2a) \\ \, = A \sin 2 \pi n t \\ \text{where} A = 2a \cos \pi ( n_1 - n_2 )t' class='latex' /> is the amplitude of resultant and obviously depends on time.</p>
<p>The amplitude A = 2a <img src='http://s.wordpress.com/latex.php?latex=%5Ccos%20%5Cpi%20%28%20n_1%20-%20n_2%20%29%20t&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\cos \pi ( n_1 - n_2 ) t' title='\cos \pi ( n_1 - n_2 ) t' class='latex' /> ….(3)</p>
<p>is maximum for <img src='http://s.wordpress.com/latex.php?latex=%5Ccos%20%5Cpi%20%28%20n_1%20-%20n_2%20%29%20t%20%3D%20%5Cpm%201%20%5C%5C%20%5Ctext%7Bor%7D%20%5Cpi%20%28n_1%20-%20n_2%20%29t%20%3D%20r%20%5Cpi%20&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\cos \pi ( n_1 - n_2 ) t = \pm 1 \\ \text{or} \pi (n_1 - n_2 )t = r \pi ' title='\cos \pi ( n_1 - n_2 ) t = \pm 1 \\ \text{or} \pi (n_1 - n_2 )t = r \pi ' class='latex' /> where r is an integer.</p>
<p>The instant of maximum are given by</p>
<p><img src='http://s.wordpress.com/latex.php?latex=t%20%3D%20%5Cdfrac%7Br%7D%7Bn_1%20-%20n_2%7D%20%3D%200%20%2C%20%5Cdfrac%7B1%7D%7Bn_1%20-%20n_2%7D%20%2C%20%5Cdfrac%7B2%7D%7Bn_1%20-%20n_2%7D%20%2C%20%5Cdfrac%7B3%7D%7Bn_1%20-%20n_2%7D%2C%20%5Cdots%20&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t = \dfrac{r}{n_1 - n_2} = 0 , \dfrac{1}{n_1 - n_2} , \dfrac{2}{n_1 - n_2} , \dfrac{3}{n_1 - n_2}, \dots ' title='t = \dfrac{r}{n_1 - n_2} = 0 , \dfrac{1}{n_1 - n_2} , \dfrac{2}{n_1 - n_2} , \dfrac{3}{n_1 - n_2}, \dots ' class='latex' /> (4)</p>
<p>Obviously time interval between two consecutive maxima</p>
<img src='http://s.wordpress.com/latex.php?latex=%3D%20%5Cdfrac%7B1%7D%7Bn_1%20-%20n_2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= \dfrac{1}{n_1 - n_2}' title='= \dfrac{1}{n_1 - n_2}' class='latex' />
<p><span style="text-decoration: underline;">Frequency of maxima</span> = ( n<sub>1</sub> – n<sub>2</sub> ) sec <sup>-1</sup></p>
<p>The amplitude is minimum for <img src='http://s.wordpress.com/latex.php?latex=%5Ccos%20%5Cpi%20%28%20n_1%20-%20n_2%20%29%20t%20%3D%200%20%5C%5C%20%5Ctext%7Bor%7D%20%5Cpi%20%28%20n_1%20-%20n_2%20%29%20t%20%3D%20%28%202r%20%2B%201%20%29%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%20%2C%20r%20%3D%200%2C1%2C2%2C3%2C%20%5Cdots%20%5C%5C%20%5Ctext%7BThe%20instant%20of%20minima%20are%20given%20by%20%7D%20%5C%5C%20t%20%3D%20%28%20r%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%20%29%20%28%20%5Cdfrac%7B1%7D%7Bn_1%20-%20n_2%20%29%7D%20%5C%5C%20%3D%20%5Cdfrac%7B1%7D%7B2%20%28%20n_1%20-%20n_2%20%29%7D%20%2C%20%5Cdfrac%7B3%7D%7B2%20%28%20n_1%20-%20n_2%20%29%7D%20%2C%20%5Cdfrac%7B5%7D%7B2%20%28%20n_1%20-%20n_2%20%29%7D%20%5Cdots%20&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\cos \pi ( n_1 - n_2 ) t = 0 \\ \text{or} \pi ( n_1 - n_2 ) t = ( 2r + 1 ) \dfrac{\pi}{2} , r = 0,1,2,3, \dots \\ \text{The instant of minima are given by } \\ t = ( r + \dfrac{1}{2} ) ( \dfrac{1}{n_1 - n_2 )} \\ = \dfrac{1}{2 ( n_1 - n_2 )} , \dfrac{3}{2 ( n_1 - n_2 )} , \dfrac{5}{2 ( n_1 - n_2 )} \dots ' title='\cos \pi ( n_1 - n_2 ) t = 0 \\ \text{or} \pi ( n_1 - n_2 ) t = ( 2r + 1 ) \dfrac{\pi}{2} , r = 0,1,2,3, \dots \\ \text{The instant of minima are given by } \\ t = ( r + \dfrac{1}{2} ) ( \dfrac{1}{n_1 - n_2 )} \\ = \dfrac{1}{2 ( n_1 - n_2 )} , \dfrac{3}{2 ( n_1 - n_2 )} , \dfrac{5}{2 ( n_1 - n_2 )} \dots ' class='latex' /></p>
<p>The time interval between two consecutive minima   <img src='http://s.wordpress.com/latex.php?latex=%3D%20%5Cdfrac%7B1%7D%7Bn_1%20-%20n_2%7D%20%5C%5C%20%5Ctherefore%20%5Ctext%7BFrequency%20of%20minima%7D%20%5C%2C%20%3D%20%28%20n_1%20-%20n_2%29%20%5Ctext%7Bsec%7D%20%5E%7B-1%7D%20&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= \dfrac{1}{n_1 - n_2} \\ \therefore \text{Frequency of minima} \, = ( n_1 - n_2) \text{sec} ^{-1} ' title='= \dfrac{1}{n_1 - n_2} \\ \therefore \text{Frequency of minima} \, = ( n_1 - n_2) \text{sec} ^{-1} ' class='latex' /></p>
<p>The number of beats produced per second.</p>
<p>= <img src='http://s.wordpress.com/latex.php?latex=n_1%20%5Cto%20n_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n_1 \to n_2' title='n_1 \to n_2' class='latex' /></p>
<p><strong>Remark:</strong> When the prongs of a turning fork are loaded, the frequency of fork decreases and when they are filed; the frequency of fork increases.</p>
<p><strong>Stationary Waves:</strong></p>
<p>When two wave trains of same frequency and amplitude travel with the same velocity along the<br />
same straight line in opposite directions, they superimpose to, produce a new type of wave called stationary wave or standing wave.</p>
<p>The name stationary for such type of waves is justified because there is no flow of energy along the<br />
wave. Let the incident wave propagating along Y- axis be</p>
<p><img src='http://s.wordpress.com/latex.php?latex=y_1%20%3D%20a%20%5Csin%20%28%20%5Comega%20t%20-%20kx%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y_1 = a \sin ( \omega t - kx )' title='y_1 = a \sin ( \omega t - kx )' class='latex' /> …(1)</p>
<p>and the wave reflected from the boundary traveling along negative X-axis is</p>
<p><img src='http://s.wordpress.com/latex.php?latex=y_2%20%3D%20%5Cpm%20a%20%5Csin%20%28%20%5Comega%20t%20-%20kx%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y_2 = \pm a \sin ( \omega t - kx )' title='y_2 = \pm a \sin ( \omega t - kx )' class='latex' />&#8230;.(2)</p>
<p>The positive and negative signs are used: if the boundary is free or rigid respectively.</p>
<p><span style="text-decoration: underline;"> Case (i)</span>. If boundary is free: then equation (2) is,</p>
<p><img src='http://s.wordpress.com/latex.php?latex=y_2%20%3D%20a%20%5Csin%20%28%20%5Comega%20t%20%2B%20kx%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y_2 = a \sin ( \omega t + kx )' title='y_2 = a \sin ( \omega t + kx )' class='latex' />. &#8230;.(3)</p>
<p>The resultant displacement due to these incident and reflected waves is</p>
<p><img src='http://s.wordpress.com/latex.php?latex=y%20%3D%20y_1%20%2B%20y_2%20%5C%5C%20%3D%20a%20%5Csin%20%28%20%5Comega%20t%20-%20kx%20%29%20%2B%20a%20%5Csin%20%28%20%5Comega%20t%20%2B%20kx%20%29%20%5C%5C%20%3D%202a%20%5Csin%20%5Cdfrac%7B%5Comega%20t%20-%20kx%20%2B%20%5Comega%20t%20%2B%20kx%7D%7B2%7D%20%5Ccos%20%5Cdfrac%7B%5Comega%20t%20-%20kx%20-%20%5Comega%20t%20-%20kx%20%7D%7B2%7D%20%5C%5C%20%3D%202a%20%5Csin%20%5Comega%20t%20%5Ccos%20kx%20%3D%202a%20%5Ccos%20kx%20sin%20%5Comega%20t%20%5Cdots%20%284%29%20%5C%5C%20%3D%20A%20%5Csin%20%5Comega%20t&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = y_1 + y_2 \\ = a \sin ( \omega t - kx ) + a \sin ( \omega t + kx ) \\ = 2a \sin \dfrac{\omega t - kx + \omega t + kx}{2} \cos \dfrac{\omega t - kx - \omega t - kx }{2} \\ = 2a \sin \omega t \cos kx = 2a \cos kx sin \omega t \dots (4) \\ = A \sin \omega t' title='y = y_1 + y_2 \\ = a \sin ( \omega t - kx ) + a \sin ( \omega t + kx ) \\ = 2a \sin \dfrac{\omega t - kx + \omega t + kx}{2} \cos \dfrac{\omega t - kx - \omega t - kx }{2} \\ = 2a \sin \omega t \cos kx = 2a \cos kx sin \omega t \dots (4) \\ = A \sin \omega t' class='latex' /> ….(5)</p>
<p>Where <img src='http://s.wordpress.com/latex.php?latex=A%20%3D%202%20a%20%5Ccos%20kx%20%3D%202a%20%5Ccos%20%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D%20%5Cdots%20%286%29%20%5C%5C%20%5C%2C%20%28%20%5Ctext%7Bsince%7D%20k%20%3D%20%5Cdfrac%7B%5Comega%20%7D%7Bv%7D%20%3D%20%5Cdfrac%7B2%20%5Cpi%7D%7B%5Clambda%7D%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A = 2 a \cos kx = 2a \cos \dfrac{2 \pi x}{\lambda} \dots (6) \\ \, ( \text{since} k = \dfrac{\omega }{v} = \dfrac{2 \pi}{\lambda} )' title='A = 2 a \cos kx = 2a \cos \dfrac{2 \pi x}{\lambda} \dots (6) \\ \, ( \text{since} k = \dfrac{\omega }{v} = \dfrac{2 \pi}{\lambda} )' class='latex' /></p>
<p>is the amplitude of resultant wave.</p>
<p>At positions where <img src='http://s.wordpress.com/latex.php?latex=%5Ccos%20%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D%20%3D%20%5Cpm%201&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\cos \dfrac{2 \pi x}{\lambda} = \pm 1' title='\cos \dfrac{2 \pi x}{\lambda} = \pm 1' class='latex' /> , the displacement is maximum. Such points are called antinodes and are given by</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D%20%3D%200%20%2C%20%5Cpi%20%2C%202%20%5Cpi%20%2C%203%20%5Cpi%20%2C%20%5Cdots%20%3D%20r%20%5Cpi%20%28%20r%20%3D%200%2C1%2C2%2C%5Cdots%20%29%20%5C%5C%20%5Ctherefore%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20x%20%3D%20%5Cdfrac%7Br%20%5Clambda%7D%7B2%7D%20%3D%200%20%2C%20%5Cdfrac%7B%5Clambda%7D%7B2%7D%20%2C%20%5Clambda%20%2C%20%5Cdfrac%7B3%20%5Clambda%7D%7B2%7D%20%2C%202%20%5Clambda%20%5Cdots%20%28%207%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{2 \pi x}{\lambda} = 0 , \pi , 2 \pi , 3 \pi , \dots = r \pi ( r = 0,1,2,\dots ) \\ \therefore \, \, \, \, \, x = \dfrac{r \lambda}{2} = 0 , \dfrac{\lambda}{2} , \lambda , \dfrac{3 \lambda}{2} , 2 \lambda \dots ( 7 )' title='\dfrac{2 \pi x}{\lambda} = 0 , \pi , 2 \pi , 3 \pi , \dots = r \pi ( r = 0,1,2,\dots ) \\ \therefore \, \, \, \, \, x = \dfrac{r \lambda}{2} = 0 , \dfrac{\lambda}{2} , \lambda , \dfrac{3 \lambda}{2} , 2 \lambda \dots ( 7 )' class='latex' />
<p>The separation between two consecutive antinodes is <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Clambda%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{\lambda}{2}' title='\dfrac{\lambda}{2}' class='latex' /><br />
At positions where <img src='http://s.wordpress.com/latex.php?latex=%5Ccos%20%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\cos \dfrac{2 \pi x}{\lambda}' title='\cos \dfrac{2 \pi x}{\lambda}' class='latex' /> = 0, the displacement is always zero. Such points are called nodes and are given by</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D%20%3D%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%20%2C%20%5Cdfrac%7B3%20%5Cpi%7D%7B2%7D%20%2C%20%5Cdfrac%7B5%20%5Cpi%7D%7B2%7D%20%3D%20%28%202r%20%2B%201%20%29%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%20%28%20r%20%3D%200%2C1%2C2%2C%5Cdots%20%29%20%5C%5C%20%5Ctherefore%20%5C%2C%20%5C%2C%20x%20%3D%20%28%202%20r%20%2B%201%20%29%20%5Cdfrac%7B%5Clambda%7D%7B4%7D%20%3D%20%5Cdfrac%7B%5Clambda%7D%7B4%7D%20%2C%20%5Cdfrac%7B3%20%5Clambda%7D%7B4%7D%20%2C%20%5Cdfrac%7B5%20%5Clambda%7D%7B4%7D%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5Cdots%20%288%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{2 \pi x}{\lambda} = \dfrac{\pi}{2} , \dfrac{3 \pi}{2} , \dfrac{5 \pi}{2} = ( 2r + 1 ) \dfrac{\pi}{2} ( r = 0,1,2,\dots ) \\ \therefore \, \, x = ( 2 r + 1 ) \dfrac{\lambda}{4} = \dfrac{\lambda}{4} , \dfrac{3 \lambda}{4} , \dfrac{5 \lambda}{4} \, \, \, \, \dots (8)' title='\dfrac{2 \pi x}{\lambda} = \dfrac{\pi}{2} , \dfrac{3 \pi}{2} , \dfrac{5 \pi}{2} = ( 2r + 1 ) \dfrac{\pi}{2} ( r = 0,1,2,\dots ) \\ \therefore \, \, x = ( 2 r + 1 ) \dfrac{\lambda}{4} = \dfrac{\lambda}{4} , \dfrac{3 \lambda}{4} , \dfrac{5 \lambda}{4} \, \, \, \, \dots (8)' class='latex' />
<p>The separation between two consecutive nodes is  <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Clambda%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{\lambda}{2}' title='\dfrac{\lambda}{2}' class='latex' /><br />
From (7) and (8) it is obvious that at free boundary always an antinodes is farmed. Midway between the antinodes, there are nodes.<br />
The separation between a node and neighboring antinodes is <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Clambda%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{\lambda}{2}' title='\dfrac{\lambda}{2}' class='latex' /></p>
<p><span style="text-decoration: underline;">Case (ii) </span> If the boundary is rigid , then</p>
<img src='http://s.wordpress.com/latex.php?latex=y_2%20%3D%20-%20a%20%5Csin%20%28%20%5Comega%20t%20%2B%20k%20x%20%29%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5Cdots%20%28%209%20%29%20%5C%5C%20%5Ctherefore%5Ctext%7BThe%20resultant%20displacement%7D%20%5C%5C%20y%20%3D%20y_1%20%2B%20y_2%20%3D%20a%20%5Csin%20%28%20%5Comega%20t%20-%20k%20x%20%29%20-%20a%20%5Csin%20%28%20%5Comega%20t%20%2B%20k%20x%20%29%20%5C%5C%20%3D%202a%20%5Ccos%20%5Cdfrac%7B%5Comega%20t%20-%20k%20x%20%2B%20%5Comega%20t%20%2B%20k%20x%7D%7B2%7D%20%5C%5C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5Csin%20%28%20%5Cdfrac%7B%5Comega%20t%20-%20k%20x%20-%20%5Comega%20t%20-%20k%20x%7D%7B2%7D%20%5C%5C%20%3D%20-%202a%20%5Csin%20kx%20%5Ccos%20%5Comega%20t%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5Cdots%20%2810%29%20%5C%5C%20%3D%20A%20%5Ccos%20%5Comega%20t%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5Cdots%20%2811%29%20%5Ctherefore%20&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y_2 = - a \sin ( \omega t + k x ) \, \, \, \, \, \dots ( 9 ) \\ \therefore\text{The resultant displacement} \\ y = y_1 + y_2 = a \sin ( \omega t - k x ) - a \sin ( \omega t + k x ) \\ = 2a \cos \dfrac{\omega t - k x + \omega t + k x}{2} \\ \, \, \, \, \, \, \, \sin ( \dfrac{\omega t - k x - \omega t - k x}{2} \\ = - 2a \sin kx \cos \omega t \, \, \, \, \, \dots (10) \\ = A \cos \omega t \, \, \, \, \, \, \dots (11) \therefore ' title='y_2 = - a \sin ( \omega t + k x ) \, \, \, \, \, \dots ( 9 ) \\ \therefore\text{The resultant displacement} \\ y = y_1 + y_2 = a \sin ( \omega t - k x ) - a \sin ( \omega t + k x ) \\ = 2a \cos \dfrac{\omega t - k x + \omega t + k x}{2} \\ \, \, \, \, \, \, \, \sin ( \dfrac{\omega t - k x - \omega t - k x}{2} \\ = - 2a \sin kx \cos \omega t \, \, \, \, \, \dots (10) \\ = A \cos \omega t \, \, \, \, \, \, \dots (11) \therefore ' class='latex' />
<p>Amplitude of resultant disturbance,</p>
<img src='http://s.wordpress.com/latex.php?latex=A%20%3D%20-%202%20a%20%5Csin%20k%20x%20%3D%20-%202%20a%20%5Csin%20%5Cdfrac%7B2%20%5Cpi%7D%7Bx%7D%20%5C%2C%20%5C%2C%20%5C%2C%20%5Cdots%20%2812%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A = - 2 a \sin k x = - 2 a \sin \dfrac{2 \pi}{x} \, \, \, \dots (12)' title='A = - 2 a \sin k x = - 2 a \sin \dfrac{2 \pi}{x} \, \, \, \dots (12)' class='latex' />
<p>The positions of maximum displacement or antinodes are</p>
<p><img src='http://s.wordpress.com/latex.php?latex=%5Csin%20%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D%20%3D%20%5Cpm%201%20%5C%5C%20or%20%5C%2C%20%5C%2C%20%5C%2C%20%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D%20%3D%20%28%202r%20%2B%201%20%29%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%20%28%20r%20%3D%200%20%2C%201%20%2C%202%20%2C%203%20%2C%20%5Cdots%20%29%20%5C%5C%20%5Ctherefore%20%5C%2C%20%5C%2C%20%5C%2C%20x%20%3D%20%28%202r%20%2B%201%20%29%20%5Cdfrac%7B%5Clambda%7D%7B4%7D%20%3D%20%5Cdfrac%7B%5Clambda%7D%7B4%7D%20%2C%20%5Cdfrac%7B3%20%5Clambda%7D%7B4%7D%20%2C%20%5Cdfrac%7B5%20%5Clambda%7D%7B4%7D%20%5Cdots%20&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sin \dfrac{2 \pi x}{\lambda} = \pm 1 \\ or \, \, \, \dfrac{2 \pi x}{\lambda} = ( 2r + 1 ) \dfrac{\pi}{2} ( r = 0 , 1 , 2 , 3 , \dots ) \\ \therefore \, \, \, x = ( 2r + 1 ) \dfrac{\lambda}{4} = \dfrac{\lambda}{4} , \dfrac{3 \lambda}{4} , \dfrac{5 \lambda}{4} \dots ' title='\sin \dfrac{2 \pi x}{\lambda} = \pm 1 \\ or \, \, \, \dfrac{2 \pi x}{\lambda} = ( 2r + 1 ) \dfrac{\pi}{2} ( r = 0 , 1 , 2 , 3 , \dots ) \\ \therefore \, \, \, x = ( 2r + 1 ) \dfrac{\lambda}{4} = \dfrac{\lambda}{4} , \dfrac{3 \lambda}{4} , \dfrac{5 \lambda}{4} \dots ' class='latex' /><br />
The positions of zero displacement or nodes are</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Csin%20%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D%20%3D%200%20%5Ctext%7Bor%7D%20%5Cdfrac%7B2%20%5Cpi%20x%7D%7B%5Clambda%7D%20%3D%20r%20%5Cpi%20%28%20r%20%3D%200%20%2C%201%20%2C%202%20%2C%203%20%2C%20%5Cdots%20%29%20%5C%5C%20%5Ctherefore%20%5C%2C%20%5C%2C%20%5C%2C%20x%20%3D%20%5Cdfrac%7Br%20%5Clambda%7D%7B2%7D%20%3D%200%20%2C%20%5Cdfrac%7B%5Clambda%7D%7B2%7D%20%2C%20%5Cdfrac%7B3%20%5Clambda%7D%7B2%7D%20%2C%20%5Cdfrac%7B5%20%5Clambda%7D%7B2%7D%20%5Cdots%20&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sin \dfrac{2 \pi x}{\lambda} = 0 \text{or} \dfrac{2 \pi x}{\lambda} = r \pi ( r = 0 , 1 , 2 , 3 , \dots ) \\ \therefore \, \, \, x = \dfrac{r \lambda}{2} = 0 , \dfrac{\lambda}{2} , \dfrac{3 \lambda}{2} , \dfrac{5 \lambda}{2} \dots ' title='\sin \dfrac{2 \pi x}{\lambda} = 0 \text{or} \dfrac{2 \pi x}{\lambda} = r \pi ( r = 0 , 1 , 2 , 3 , \dots ) \\ \therefore \, \, \, x = \dfrac{r \lambda}{2} = 0 , \dfrac{\lambda}{2} , \dfrac{3 \lambda}{2} , \dfrac{5 \lambda}{2} \dots ' class='latex' />
<p>Thus a node is always formed at rigid boundary lowest.</p>
<p><strong>Fundamental tone, harmonics and overtones: </strong></p>
<p>The sound of frequency produced by a musical instrument is called the fundamental tone. The sounds of other frequencies produced by the musical instrument are called overtones. The overtones whose frequencies are integral multiplies of the fundamental frequency are called the harmonics. The fundamental tone is also called the first harmonic. If first harmonic is n, then the tones of frequencies 2n, 3n, 4n … are called the second, the third and the fourth harmonic respectively. If frequencies of sound emitted by an instrument are n, 1.5n, 2n, 2.5n, 3n etc, then the notes of frequencies 1.5n, 2n, 2.5n, 3n are overtones, while those of frequencies 2n, 3n are second and third harmonics respectively.</p>
<p><strong>Stationary Waves in Strings Fixed at Both Ends:</strong></p>
<p>For transverse vibrations in string, we have</p>
<p>Speed of waves v = n <img src='http://s.wordpress.com/latex.php?latex=%5Clambda%20%3D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7BM%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda = \sqrt{\dfrac{T}{M}}' title='\lambda = \sqrt{\dfrac{T}{M}}' class='latex' /></p>
<p>where T is tension in string and m is mass per unit length of string</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Ctherefore%20%5Ctext%7BFrequency%7D%20n%20%3D%20%5Cdfrac%7B1%7D%7B%5Clambda%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7Bm%7D%7D%20%5C%2C%20%5C%2C%20%5C%2C%20%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\therefore \text{Frequency} n = \dfrac{1}{\lambda} \sqrt{\dfrac{T}{m}} \, \, \, (1)' title='\therefore \text{Frequency} n = \dfrac{1}{\lambda} \sqrt{\dfrac{T}{m}} \, \, \, (1)' class='latex' />
<p>When the string is plucked in the middle, it vibrates in one loop with nodes at fixed ends and antinodes in the middle; so that length of string</p>
<div id="attachment_6694" class="wp-caption aligncenter" style="width: 249px"><img class="size-medium wp-image-6694" title="Tones" src="http://oscience.info/wp-content/uploads/Tones-239x300.jpg" alt="" width="239" height="300" /><p class="wp-caption-text">tones</p></div>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda%20_1%7D%7B2%7D%20or%20%5C%2C%20%5Clambda%20_1%20%3D%202l%20%5C%5C%20%5C%2C%20%5Ctherefore%20%5C%2C%20%5C%2C%20%5C%2C%20n%20%3D%20n_1%20%3D%20%5Cdfrac%7B1%7D%7B%5Clambda%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7BM%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B2l%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7BM%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda _1}{2} or \, \lambda _1 = 2l \\ \, \therefore \, \, \, n = n_1 = \dfrac{1}{\lambda} \sqrt{\dfrac{T}{M}} = \dfrac{1}{2l} \sqrt{\dfrac{T}{M}}' title='l = \dfrac{\lambda _1}{2} or \, \lambda _1 = 2l \\ \, \therefore \, \, \, n = n_1 = \dfrac{1}{\lambda} \sqrt{\dfrac{T}{M}} = \dfrac{1}{2l} \sqrt{\dfrac{T}{M}}' class='latex' />
<p>This tone is emitted is called the fundamental or first harmonic.</p>
<p>If the wire is plucked at one fourth of its length, the string vibrates in two loops, so that</p>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda%20_2%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda%20_2%7D%7B2%7D%20%3D%20%5Clambda%20_2%20%5C%5C%20%5Ctherefore%20%5Ctext%7Bfrequency%7D%20%5C%5C%20n_2%20%3D%20%5Cdfrac%7B1%7D%7B%5Clambda%20_2%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7Bm%7D%7D%20%3D%202%20.%20%5Cdfrac%7B1%7D%7B2l%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7Bm%7D%7D%20%3D%202n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda _2}{2} + \dfrac{\lambda _2}{2} = \lambda _2 \\ \therefore \text{frequency} \\ n_2 = \dfrac{1}{\lambda _2} \sqrt{\dfrac{T}{m}} = 2 . \dfrac{1}{2l} \sqrt{\dfrac{T}{m}} = 2n' title='l = \dfrac{\lambda _2}{2} + \dfrac{\lambda _2}{2} = \lambda _2 \\ \therefore \text{frequency} \\ n_2 = \dfrac{1}{\lambda _2} \sqrt{\dfrac{T}{m}} = 2 . \dfrac{1}{2l} \sqrt{\dfrac{T}{m}} = 2n' class='latex' />
<p>This tone is called the second harmonic or first overtone.<br />
In the string is plucked at one sixth of its length, the string vibrates in three loops, so that</p>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda%20_3%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda%20_3%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda%20_3%7D%7B2%7D%20%3D%20%5Cdfrac%7B3%20%5Clambda%20_3%7D%7B2%7D%20%5C%5C%20%5Ctherefore%20%5Ctext%7Bfrequency%7D%20%5C%5C%20n_3%20%3D%20%5Cdfrac%7B1%7D%7B%5Clambda%20_3%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7Bm%7D%7D%20%3D%203%20.%20%5Cdfrac%7B1%7D%7B2l%7D%20%5Csqrt%7BT%7D%7Bm%7D%20%3D%203n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} = \dfrac{3 \lambda _3}{2} \\ \therefore \text{frequency} \\ n_3 = \dfrac{1}{\lambda _3} \sqrt{\dfrac{T}{m}} = 3 . \dfrac{1}{2l} \sqrt{T}{m} = 3n' title='l = \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} = \dfrac{3 \lambda _3}{2} \\ \therefore \text{frequency} \\ n_3 = \dfrac{1}{\lambda _3} \sqrt{\dfrac{T}{m}} = 3 . \dfrac{1}{2l} \sqrt{T}{m} = 3n' class='latex' />
<p>This tone is called third harmonic or second overtone.<br />
In general when the string vibrates in p-loops, the frequency</p>
<img src='http://s.wordpress.com/latex.php?latex=n_p%20%3D%20%5Cdfrac%7B1%7D%7B%5Clambda%20_p%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7Bm%7D%7D%20%3D%20%5Cdfrac%7Bp%7D%7B2%20l%7D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7Bm%7D%7D%20%3D%20pn&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n_p = \dfrac{1}{\lambda _p} \sqrt{\dfrac{T}{m}} = \dfrac{p}{2 l} \sqrt{\dfrac{T}{m}} = pn' title='n_p = \dfrac{1}{\lambda _p} \sqrt{\dfrac{T}{m}} = \dfrac{p}{2 l} \sqrt{\dfrac{T}{m}} = pn' class='latex' />
<p>This tone is called the p<sup>th</sup>. Thus in the case of string fixed at both ends, all harmonics even and odd are present.</p>
<p><strong>Melde’s Law:</strong></p>
<p>If N is frequency of tuning fork for a given length ‘l’ of a string vibrating in p-loops under 2; tension T, then Melde’s law states<br />
T.p<sup>2</sup> = constant or <img src='http://s.wordpress.com/latex.php?latex=p%20%5Csqrt%7BT%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p \sqrt{T}' title='p \sqrt{T}' class='latex' /> = constant</p>
<p><strong>Vibrations of Air columns in Organ Pipes </strong></p>
<p>The minimum frequency produced in a pipe is called fundamental and other notes are called overtones.</p>
<p><strong>Open organ pipe :</strong></p>
<p>An antinodes is always formed at the open end. Accordingly different notes produced in open pipe are shown in fig. In fundamental mode if <img src='http://s.wordpress.com/latex.php?latex=%5Clambda%20_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda _1' title='\lambda _1' class='latex' /> is wavelength, then</p>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda%20_1%7D%7B2%7D%20or%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5Clambda_1%20%3D%202l%20%5C%5C%20%5Ctherefore%20%5Ctext%7Bfundamental%20frequency%7D%20%5C%5C%20n%20%3D%20%5Cdfrac%7Bv%7D%7B%5Clambda_1%7D%20%3D%20%5Cdfrac%7Bv%7D%7B2l%7D%20%5C%2C%20%5C%2C%20%5Cdots%20%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda _1}{2} or \, \, \, \, \lambda_1 = 2l \\ \therefore \text{fundamental frequency} \\ n = \dfrac{v}{\lambda_1} = \dfrac{v}{2l} \, \, \dots (1)' title='l = \dfrac{\lambda _1}{2} or \, \, \, \, \lambda_1 = 2l \\ \therefore \text{fundamental frequency} \\ n = \dfrac{v}{\lambda_1} = \dfrac{v}{2l} \, \, \dots (1)' class='latex' />
<p><img class="aligncenter size-medium wp-image-6697" title="Fundamental frequency" src="http://oscience.info/wp-content/uploads/Fundamental-frequency-300x184.jpg" alt="" width="300" height="184" /></p>
<p>For first overtone if <img src='http://s.wordpress.com/latex.php?latex=%5Clambda%20_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda _2' title='\lambda _2' class='latex' /> is the wavelength , then</p>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda_2%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda_2%7D%7B2%7D%20%3D%20%5Clambda_2%20%5Cdots%20%282%29%20%5C%5C%20%5Ctext%7Bfrequency%20of%20first%20overtone%7D%20%2C%20%5C%5C%20n_2%20%3D%20%5Cdfrac%7Bv%7D%7B%5Clambda%20_2%7D%20%3D%20%5Cdfrac%7Bv%7D%7Bl%7D%20%3D%202%20%5Cdfrac%7Bv%7D%7B2l%7D%20%3D%202n%20%5Cdots%20%283%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda_2}{2} + \dfrac{\lambda_2}{2} = \lambda_2 \dots (2) \\ \text{frequency of first overtone} , \\ n_2 = \dfrac{v}{\lambda _2} = \dfrac{v}{l} = 2 \dfrac{v}{2l} = 2n \dots (3)' title='l = \dfrac{\lambda_2}{2} + \dfrac{\lambda_2}{2} = \lambda_2 \dots (2) \\ \text{frequency of first overtone} , \\ n_2 = \dfrac{v}{\lambda _2} = \dfrac{v}{l} = 2 \dfrac{v}{2l} = 2n \dots (3)' class='latex' />
<p>This frequency is double of fundamental frequency and is therefore called second harmonic.</p>
<p>For second overtone, if <img src='http://s.wordpress.com/latex.php?latex=%5Clambda%20_3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda _3' title='\lambda _3' class='latex' /> is the wavelength, then</p>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda%20_3%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda%20_3%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda%20_3%7D%7B2%7D%20%3D%20%5Cdfrac%7B3%20%5Clambda%20_3%7D%7B2%7D%20%5Ctext%7Bor%7D%20%5Clambda%20%3D%20%5Cdfrac%7B2l%7D%7B3%7D%20%5C%5C%20%5Ctherefore%20%5Ctext%7Bfrequency%20of%20second%20overtone%7D%20%5C%5C%20n_3%20%3D%20%5Cdfrac%7Bv%7D%7B%5Clambda%20_3%7D%20%3D%203%20.%20%5Cdfrac%7Bv%7D%7B2l%7D%20%3D%203n%20%5Cdots%20%284%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} = \dfrac{3 \lambda _3}{2} \text{or} \lambda = \dfrac{2l}{3} \\ \therefore \text{frequency of second overtone} \\ n_3 = \dfrac{v}{\lambda _3} = 3 . \dfrac{v}{2l} = 3n \dots (4)' title='l = \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} = \dfrac{3 \lambda _3}{2} \text{or} \lambda = \dfrac{2l}{3} \\ \therefore \text{frequency of second overtone} \\ n_3 = \dfrac{v}{\lambda _3} = 3 . \dfrac{v}{2l} = 3n \dots (4)' class='latex' />
<p>This is the third harmonic Frequencies</p>
<p>Frequencies n<sub>1 </sub>: n<sub>2</sub> : n<sub>3</sub> : … = 1 : 2: 3:…  i.e. in open organ pipe all harmonics even or odd are present.</p>
<p><strong>Stationary Waves and Harmonics in Closed Organ Pipe:</strong></p>
<p>A node is always formed at closed end and antinodes at open end. Accordingly different harmonies produced in closed organ pipe are shown in fig.</p>
<p><img class="aligncenter size-medium wp-image-6696" title="Closed organ pipe" src="http://oscience.info/wp-content/uploads/Closed-organ-pipe-300x172.jpg" alt="" width="300" height="172" /></p>
<p>In fundamental mode, if<img src='http://s.wordpress.com/latex.php?latex=%5Clambda%20_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda _1' title='\lambda _1' class='latex' /> is the wavelength of vibrations, then</p>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda%20_1%7D%7B4%7D%20%5Ctext%7Bor%7D%20%5Clambda_1%20%3D%204l%20%5C%5C%20%5Ctherefore%20%5Ctext%7Bfundamental%20frequency%7D%20n_1%20%3D%20%5Cdfrac%7Bv%7D%7B%5Clambda%20_1%7D%20%3D%20%5Cdfrac%7Bv%7D%7B4l%7D%20%5Cdots%20%285%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda _1}{4} \text{or} \lambda_1 = 4l \\ \therefore \text{fundamental frequency} n_1 = \dfrac{v}{\lambda _1} = \dfrac{v}{4l} \dots (5)' title='l = \dfrac{\lambda _1}{4} \text{or} \lambda_1 = 4l \\ \therefore \text{fundamental frequency} n_1 = \dfrac{v}{\lambda _1} = \dfrac{v}{4l} \dots (5)' class='latex' />
<p>&nbsp;</p>
<p>Frequency of first overtone, if <img src='http://s.wordpress.com/latex.php?latex=%5Clambda%20_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda _2' title='\lambda _2' class='latex' /> is the wavelength , then</p>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda_2%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda%20_2%7D%7B4%7D%20%3D%20%5Cdfrac%7B3%20%5Clambda%20_2%7D%7B4%7D%20or%20%5Clambda%20_2%20%3D%20%5Cdfrac%7B4l%7D%7B3%7D%20%5C%5C%20%5Ctherefore%7Bfrequency%20of%20first%20overtone%7D%20%2C%20%5C%5C%20n_2%20%3D%20%5Cdfrac%7Bv%7D%7B%5Clambda_2%7D%20%3D%203.%20%5Cdfrac%7Bv%7D%7B4l%7D%20%3D%203n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda_2}{2} + \dfrac{\lambda _2}{4} = \dfrac{3 \lambda _2}{4} or \lambda _2 = \dfrac{4l}{3} \\ \therefore{frequency of first overtone} , \\ n_2 = \dfrac{v}{\lambda_2} = 3. \dfrac{v}{4l} = 3n' title='l = \dfrac{\lambda_2}{2} + \dfrac{\lambda _2}{4} = \dfrac{3 \lambda _2}{4} or \lambda _2 = \dfrac{4l}{3} \\ \therefore{frequency of first overtone} , \\ n_2 = \dfrac{v}{\lambda_2} = 3. \dfrac{v}{4l} = 3n' class='latex' />
<p>This frequency is three times of fundamental. Therefore in closed pipe the first overtone is third<br />
harmonic.<em> </em><br />
For second overtone; if <img src='http://s.wordpress.com/latex.php?latex=%5Clambda%20_3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda _3' title='\lambda _3' class='latex' /> is the wavelength, then</p>
<img src='http://s.wordpress.com/latex.php?latex=l%20%3D%20%5Cdfrac%7B%5Clambda%20_3%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda%20_3%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Clambda%20_3%7D%7B4%7D%20%3D%20%5Cdfrac%7B5%20%5Clambda%20_3%7D%7B4%7D%20%5Ctext%7Bor%7D%20%5Clambda%20_3%20%3D%204l%20%5C%5C%20%5Ctext%7Bfrequency%20of%20second%20overtone%20%2C%7D%20%5C%5C%20n_2%20%3D%20%5Cdfrac%7Bv%7D%7B%5Clambda%20_3%7D%20%3D%205%20.%20%5Cdfrac%7Bv%7D%7B4l%7D%20%3D%205n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='l = \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{4} = \dfrac{5 \lambda _3}{4} \text{or} \lambda _3 = 4l \\ \text{frequency of second overtone ,} \\ n_2 = \dfrac{v}{\lambda _3} = 5 . \dfrac{v}{4l} = 5n' title='l = \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{2} + \dfrac{\lambda _3}{4} = \dfrac{5 \lambda _3}{4} \text{or} \lambda _3 = 4l \\ \text{frequency of second overtone ,} \\ n_2 = \dfrac{v}{\lambda _3} = 5 . \dfrac{v}{4l} = 5n' class='latex' />
<p>Thus in closed pipe</p>
<p>n<sub>1</sub> : n<sub>2 </sub>: n<sub>3</sub> : … = 1 : 3 : 5 : …</p>
<p>Hence in closed organ pipe only odd harmonics are present.</p>
<p><strong><span style="text-decoration: underline;">End Correction :<br />
</span></strong>So far we have considered that the antinodes is formed exactly at the open end of the pipe; but actually due to finite momentum of the particles the reflection takes place a little above the open end; that is why the antinodes is formed a little above the open end. For this a correction is applied being known as end correction.This is denoted by ‘c’ and its value to 0.6r. r being radius of the pipe. If l<sub>o</sub> is the length of pipe, then for closed pipe ‘l’ is replaced by l<sub>o</sub> + e while for open pipe ‘I’ is replaced by 1<sub>o</sub> + 2e .</p>
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		<title>Boolean Algebra</title>
		<link>http://oscience.info/mathematics/boolean-algebra-2/</link>
		<comments>http://oscience.info/mathematics/boolean-algebra-2/#comments</comments>
		<pubDate>Wed, 04 Apr 2012 19:00:27 +0000</pubDate>
		<dc:creator>ran</dc:creator>
				<category><![CDATA[Mathematics]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6686</guid>
		<description><![CDATA[Boolean Algebra Just like sets and logic, Boolean algebra is a modern concept. It is concerned with statements which are either true or false. It was George Boolen who developed this theory and its significance was realized when Claude Shanon introduced circuit algebra to deal with relay circuits in 1938. The digital computer which contains [...]]]></description>
			<content:encoded><![CDATA[<h2><strong>Boolean Algebra</strong></h2>
<p><strong><br />
</strong></p>
<p>Just like sets and logic, Boolean algebra is a modern concept. It is concerned with statements which are either true or false. It was George Boolen who developed this theory and its significance was realized when Claude Shanon introduced circuit algebra to deal with relay circuits in 1938. The digital computer which contains a large number of logic circuits in small space and works in switching them opened a wide field for this algebra and its recognition as significant part of modern mathematics.</p>
<p>&nbsp;</p>
<h3><strong>Definition</strong></h3>
<p>&nbsp;</p>
<p>A non-empty set B = (a, b, c …..) with two binary operations OR or JOIN denoted by +), AND or meet (denoted by <img src='http://s.wordpress.com/latex.php?latex=%5Cbullet&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bullet' title='\bullet' class='latex' />) and a unary operation <strong>complement</strong> (denoted by’) is a Boolean Algebra if the elements satisfy the following axioms:</p>
<p>B.I. <em>commutative law,</em></p>
<p>(i)a + b = b + a</p>
<p>(ii) a.b = b.a</p>
<p>&nbsp;</p>
<p>B II. <em>Distributive law,</em></p>
<p>(i)a.(b + c) =a.b + a.c</p>
<p>(ii) a+(b.c) = (a + b).(a + c)</p>
<p>&nbsp;</p>
<p>B III. <em>Existence of identity elements,</em></p>
<p>There exist elements, 0,<img src='http://s.wordpress.com/latex.php?latex=%20%5Cin%20B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt=' \in B' title=' \in B' class='latex' /> such that</p>
<p>a + 0 = a</p>
<p>a.1 =a</p>
<p>&nbsp;</p>
<p>B IV. <em>Existence of complements,</em></p>
<p>a + a’ = 1</p>
<p>a.a’ = 1</p>
<p>a.a’ = 0</p>
<p>we may note here that associative property has not been assumed in the above definition but will be derived from the above definition but will be derived from the above axioms. Also the distributive law (ii) is different from ordinary algebra applied to numbers and the reader is advised to familiarize himself fully with this property applied both was, the uniqueness of the identity elements and the complement shall be proved later on. There is great similarity among the set theory, logic and Boolean Algebra as shown below</p>
<table border="1" cellspacing="0" cellpadding="0">
<tbody>
<tr>
<td valign="top" width="106"></td>
<td valign="top" width="135">Set theory</td>
<td valign="top" width="113">Logic</td>
<td valign="top" width="106">Boolean Algebra</td>
</tr>
<tr>
<td valign="top" width="106">Union</td>
<td valign="top" width="135"><img src='http://s.wordpress.com/latex.php?latex=%5Cbigcup&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bigcup' title='\bigcup' class='latex' /></td>
<td valign="top" width="113">V</td>
<td valign="top" width="106">+</td>
</tr>
<tr>
<td valign="top" width="106">Intersection</td>
<td valign="top" width="135"><img src='http://s.wordpress.com/latex.php?latex=%5Cbigcap&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bigcap' title='\bigcap' class='latex' /></td>
<td valign="top" width="113">^</td>
<td valign="top" width="106"><img src='http://s.wordpress.com/latex.php?latex=%5Cbullet&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bullet' title='\bullet' class='latex' /></td>
</tr>
<tr>
<td valign="top" width="106">Complement</td>
<td valign="top" width="135">‘</td>
<td valign="top" width="113"><img src='http://s.wordpress.com/latex.php?latex=%5Csim&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sim' title='\sim' class='latex' /></td>
<td valign="top" width="106">‘</td>
</tr>
<tr>
<td valign="top" width="106">Universal set</td>
<td valign="top" width="135">S</td>
<td valign="top" width="113">T</td>
<td valign="top" width="106">1</td>
</tr>
<tr>
<td valign="top" width="106">Null set</td>
<td valign="top" width="135"><img src='http://s.wordpress.com/latex.php?latex=%5Cemptyset&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\emptyset' title='\emptyset' class='latex' /></td>
<td valign="top" width="113">F</td>
<td valign="top" width="106">0</td>
</tr>
<tr>
<td valign="top" width="106">Implication</td>
<td valign="top" width="135">Mapping</td>
<td valign="top" width="113"><img src='http://s.wordpress.com/latex.php?latex=p%20%5Cto&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p \to' title='p \to' class='latex' /></td>
<td valign="top" width="106">P’ + q</td>
</tr>
<tr>
<td valign="top" width="106">Equivalence</td>
<td valign="top" width="135">One-one mapping</td>
<td valign="top" width="113"><img src='http://s.wordpress.com/latex.php?latex=p%20%5Cleftrightarrow&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p \leftrightarrow' title='p \leftrightarrow' class='latex' /> q[/latex]</td>
<td valign="top" width="106">P’q’ + pq</td>
</tr>
</tbody>
</table>
<p>&nbsp;</p>
<p>&nbsp;</p>
<p><strong>Some properties</strong></p>
<p>&nbsp;</p>
<p>Now we shall prove some properties of Boolean algebra.</p>
<p>&nbsp;</p>
<p><strong>Theorem I :</strong> <em>the two identity elements are unique.</em></p>
<p><strong>Proof</strong>: suppose if possible there are two zero elements <img src='http://s.wordpress.com/latex.php?latex=0_1%20%5Ctext%7Band%7D%200_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0_1 \text{and} 0_2' title='0_1 \text{and} 0_2' class='latex' /> in B. THEN</p>
<img src='http://s.wordpress.com/latex.php?latex=0_1%20%2B%200_2%20%3D%200_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0_1 + 0_2 = 0_1' title='0_1 + 0_2 = 0_1' class='latex' />
<p>And <img src='http://s.wordpress.com/latex.php?latex=0_2%20%2B%200_1%20%3D%200_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0_2 + 0_1 = 0_2' title='0_2 + 0_1 = 0_2' class='latex' />  (B II)</p>
<p>But <img src='http://s.wordpress.com/latex.php?latex=0_1%20%2B%200_2%20%3D%200_2%20%2B%201&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0_1 + 0_2 = 0_2 + 1' title='0_1 + 0_2 = 0_2 + 1' class='latex' /> (B I)</p>
<p>Hence <img src='http://s.wordpress.com/latex.php?latex=0_1%20%3D%200_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0_1 = 0_2' title='0_1 = 0_2' class='latex' /></p>
<p>Similarly we can prove for the second identity 1.</p>
<p>&nbsp;</p>
<p><strong>Theorem II:</strong> a + a = a and a.a = a (indemptent laws)</p>
<p>&nbsp;</p>
<p><strong>Proof</strong>: a  = a + 0 = a+ (a. a’) (B IV)</p>
<p>= (a + a). (a +a’) (B II)</p>
<p>= (a + a).1</p>
<p>= a +a</p>
<p>Also a = a.1</p>
<p>= a.(a +a’)</p>
<p>= a.(a + a.a’</p>
<p>= a.a + 0 = a.a</p>
<p>&nbsp;</p>
<p><strong>Theorem III:</strong> form every element <img src='http://s.wordpress.com/latex.php?latex=a%20%5Cin%20B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a \in B' title='a \in B' class='latex' /></p>
<p>A + 1 = 1, a.0 = 0.</p>
<p>&nbsp;</p>
<p><strong>Proof</strong>: 1 = a + a’</p>
<p>= a + (a’.1) = (a + 1).(a + a’)</p>
<p>= (a + 1).1 = a+ 1</p>
<p>Also</p>
<p>0 = a.a’ (B IV)</p>
<p>= a.(a’ + 0)  (B III)</p>
<p>= a.a’ + a.0   (B II)</p>
<p>= 0 + A.0  (B IV)</p>
<p>= a.0</p>
<p>&nbsp;</p>
<p><strong>Theorem IV:</strong> (Absorption laws)</p>
<p>A + (a.b) = a and a.(a + b) = a.</p>
<p>&nbsp;</p>
<p><strong>Proof</strong>:</p>
<p>a =  a.1</p>
<p>= a.(1 + b)</p>
<p>= (a.1) + (a.b) (B II)</p>
<p>= a + a.b</p>
<p>Also</p>
<p>a. (a + b) a.a + a.b</p>
<p>= a + a.b  (Th. II)</p>
<p>= a</p>
<p>&nbsp;</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cleft%28%20%5Cbegin%7Barray%7D%7Bc%7D%20a%20%2B%20x%20%3D%20a%20%2B%20y%20%5C%5C%20a%5E%7B%5Cprime%7D%20%2B%20x%20%3D%20a%5E%7B%5Cprime%7D%20%2B%20y%20%5Cend%7Barray%7D%20%5Cright%29%5CRightarrow%20x%20%3D%20y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\left( \begin{array}{c} a + x = a + y \\ a^{\prime} + x = a^{\prime} + y \end{array} \right)\Rightarrow x = y' title='\left( \begin{array}{c} a + x = a + y \\ a^{\prime} + x = a^{\prime} + y \end{array} \right)\Rightarrow x = y' class='latex' />
<p>&nbsp;</p>
<p>And</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cleft%28%20%5Cbegin%7Barray%7D%7Bc%7D%20a.x%20%3D%20a.y%20%5C%5C%20a%5E%7B%5Cprime%7D.x%20%3D%20a%5E%7B%5Cprime%7D.y%5Cend%7Barray%7D%20%5Cright%29%5CRightarrow%20x%20%3D%20y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\left( \begin{array}{c} a.x = a.y \\ a^{\prime}.x = a^{\prime}.y\end{array} \right)\Rightarrow x = y' title='\left( \begin{array}{c} a.x = a.y \\ a^{\prime}.x = a^{\prime}.y\end{array} \right)\Rightarrow x = y' class='latex' />
<p>&nbsp;</p>
<p><strong>Proof</strong> : we prove the first implication,</p>
<p>(a + x).(a’ + x) = a.a’ + x  (B II)</p>
<p>= 0 + x = x</p>
<p>(a + y).(a’ + y) = a.a’ + y</p>
<p>= 0 + y = y</p>
<p><img src='http://s.wordpress.com/latex.php?latex=%5Ctherefore&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\therefore' title='\therefore' class='latex' /> (a + x).(a’ +x) = (a + y).(a’ + y)</p>
<img src='http://s.wordpress.com/latex.php?latex=%5CRightarrow%20x%20%3D%20y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Rightarrow x = y' title='\Rightarrow x = y' class='latex' />
<p>&nbsp;</p>
<p>Similarly a.x + a’.x = (a + a’).x</p>
<p>= 1.x = x</p>
<p>&nbsp;</p>
<p>And a.y + a’.y = (a + a’).y</p>
<p>= 1.y = y</p>
<p>Implying x = y</p>
<p>&nbsp;</p>
<p><strong>Theorem VI:</strong> <em>addition and multiplication are associative,</em></p>
<p>i.e., a + (b + c) = (a + b) + c and a.(b.c) = (a.b).c</p>
<p>&nbsp;</p>
<p><strong>proof</strong>: Let a + (b + c) x, (a + b) + c = y.</p>
<p>Then a.x = a.{a+ (b + c)}</p>
<p>=a’    ( Th. IV)</p>
<p>&nbsp;</p>
<p>And</p>
<p>a.y = a.{(a + b) + c}</p>
<p>= a.(a + b) + a.c</p>
<p>= a + a.c = a</p>
<p>&nbsp;</p>
<p>Also x = a’.{a + (b + c)}</p>
<p>= a’.a + a’.(b + c)</p>
<p>= 0 + a’ .(b + c)</p>
<p>= a’.(b + c)</p>
<p>&nbsp;</p>
<p>And</p>
<p>a’.y = a’.{(a + b) + c}</p>
<p>=a’. (a + b) + a’. c</p>
<p>= {a’.a + a’.b} + a’.c</p>
<p>= {0 + a’.b} + a’ .c</p>
<p>= a’. b + a’.c = a’.(b + c)</p>
<p>&nbsp;</p>
<p>Hence a.x = a. y and a’.x = a’.y</p>
<p><img src='http://s.wordpress.com/latex.php?latex=%5CRightarrow%20x%20%3D%20y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Rightarrow x = y' title='\Rightarrow x = y' class='latex' />   (Th. V)</p>
<p>Also let a.(b . c) = p and (a . b) . c = q</p>
<p>&nbsp;</p>
<p>Then  a + p = a + a(a.(b.c)}   (B II)</p>
<p>= (a + a). (a + b.c)</p>
<p>= a.(a + b.c)</p>
<p>= a    (Th. IV)</p>
<p>&nbsp;</p>
<p>Also a + q = a- {(a.c).c}</p>
<p>= (a + a.b).(a + c)   (B II)</p>
<p>= a.(+c)</p>
<p>= a . a + a.c</p>
<p>= x = a.c = c</p>
<p>&nbsp;</p>
<p>a’ + p = a’ + {a, (b, c)}</p>
<p>= (a’ + a).(a’ + b.c)</p>
<p>= 1.(a’ +b .c) + a’ + b.c</p>
<p>= (a’ + b). (a’ + c)</p>
<p>= {1 .(a’ + b).(a’ + c)}</p>
<p>= {(a + a’). (a’ + b)}. (a’ + c)</p>
<p>= (a’ +a.b). (a’ +c)</p>
<p>= a’ + (a.b).c   (B II)</p>
<p>= a’ + q</p>
<p>&nbsp;</p>
<p>Here a + p = a + q</p>
<p>And a’ + p = a’ + q</p>
<p><img src='http://s.wordpress.com/latex.php?latex=%5CRightarrow%20p%20%3D%20q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Rightarrow p = q' title='\Rightarrow p = q' class='latex' />   (Th. V)</p>
<p><strong>Note</strong>: <em>some authors include associative law in the definition of Boolean Algebra.</em></p>
<p>&nbsp;</p>
<p><strong>Theorem VII.</strong> The complement of <img src='http://s.wordpress.com/latex.php?latex=a%20%5Cin%20B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a \in B' title='a \in B' class='latex' /> is uique.</p>
<p>&nbsp;</p>
<p><strong>Proof</strong>: let there are two complements a’ and a’’.</p>
<p>Then a + a’ = 1, a.a’ = 0</p>
<p>a + a’’ = 1, a.a’’ = 0</p>
<p>now a’ = 1.a’ = (a + a’’).a’</p>
<p>= a.a’ + a’’.a’</p>
<p>= 0 + a’’ . a’ = a.a’’ + a’’ .a’</p>
<p>=a’’.(a + a’) = a’’.1</p>
<p>= a’’</p>
<p>Here a’ is unique.</p>
<p>&nbsp;</p>
<p><strong>Theorem VIII</strong> (a’)’ = a</p>
<p>&nbsp;</p>
<p><strong>Proof</strong> : (a’)’ = a.a’ + (a’)’</p>
<p>= {a + (a’)’}. {a’ + (a’)’} …(B II)</p>
<p>= {a + (a’)’}.1</p>
<p>= {a + (a’)’}.{a + a’}</p>
<p>= a + (a’).a’ = a+ 0</p>
<p>=a</p>
<p>The theorem is direct result of <strong>B IV</strong></p>
<p>&nbsp;</p>
<p><strong>Theorem IX:</strong> complements of identities</p>
<p>0’ = 1 and 1’ =0.</p>
<p>&nbsp;</p>
<p><strong>Proof</strong>: the proof follows from B III and B IV as</p>
<p>1 + 0 = 1 and 1.0 =0</p>
<p>&nbsp;</p>
<p><strong>Theorem X:</strong> de morgan’s law.</p>
<p>(a +b)’ = a’.b’</p>
<p>And (a.b)’ = a’ + b’</p>
<p>&nbsp;</p>
<p><strong>Proof</strong>:</p>
<p>(a’.b’).(a + b) = (a’.b’) . a + (a’.b’).b</p>
<p>= a’.(b’.a) + a’.(b’.b)  (Th. VI)</p>
<p>= a’.(a.b’) + a’.0</p>
<p>= (a’.a).b’ + 0 = 0.b’ + 0</p>
<p>= 0</p>
<p>Also (a’.b’) + (a + b) = (a’.b’ + a) + (a’ + b + b)</p>
<p>= (a + a’).(a + b’) + (a’ + b).(b’ + b)     (B II)</p>
<p>= 1.(a + b’) + (a’ + b).1</p>
<p>= (a + b’) + (b + a’)</p>
<p>= a + (b’ + b) + a’ = a  1 + a’</p>
<p>= a + a’ = 1</p>
<p>Hence a’.b’ is the complement of (a + b)</p>
<p>Also (a.b).(a’ + b’) = {(a.b).a’} + {(a.b).b’}</p>
<p>= {a’. (a.b)} + a. {(b.b’)}</p>
<p>= {(a’.a).b}  {a.0}</p>
<p>= 0.b + a.0 = 0 + 0 = 0</p>
<p>And  (a.b) + (a’ + b’) = (a’ + b’) + (a.b)</p>
<p>= {(a’ + b’) + a}.{(a’ + b’) + b}</p>
<p>= {a + (a’ + b’)}.{a’ + (a’ + (b’ + b)}</p>
<p>= {(a + a’) + b’}. {a’ + 1}</p>
<p>= (1 + b’). (a’ + 1) = 1.1 = 1</p>
<p>Hence a’ +b’ is the complement of a.b.</p>
<p>&nbsp;</p>
<h3><strong>Duality</strong></h3>
<p>&nbsp;</p>
<p>Any result that is obtained from the axioms of Boolean Algebra remains true if  + and <img src='http://s.wordpress.com/latex.php?latex=%5Cbullet&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bullet' title='\bullet' class='latex' /> are interchanged along with the interchange of 0 and 1 throughout the statement of the results.</p>
<p>&nbsp;</p>
<h3><strong>Partial ordering</strong></h3>
<p>&nbsp;</p>
<p>Just like inclusion of a set into another set as the case of subsets, we have the idea of partial ordering in Booleean algebra. Thus</p>
<p><img src='http://s.wordpress.com/latex.php?latex=a%20%5Cleq%20b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a \leq b' title='a \leq b' class='latex' /> if a.b = a,a,b <img src='http://s.wordpress.com/latex.php?latex=%5Cin%20B&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\in B' title='\in B' class='latex' /></p>
<p>Or a + b = b</p>
<p>It can also be stated as</p>
<p><img src='http://s.wordpress.com/latex.php?latex=a%20%5Cleq%20b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a \leq b' title='a \leq b' class='latex' /> if a.b’ =0</p>
<p>And <img src='http://s.wordpress.com/latex.php?latex=a%20%5Cleq&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a \leq' title='a \leq' class='latex' /> if <img src='http://s.wordpress.com/latex.php?latex=b%5E%7B%5Cprime%7D%20%5Cleq%20a%5E%7B%5Cprime%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b^{\prime} \leq a^{\prime}' title='b^{\prime} \leq a^{\prime}' class='latex' /></p>
<p>With the above notation we define as:</p>
<p>If ( B, + <img src='http://s.wordpress.com/latex.php?latex=%5Cbullet&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bullet' title='\bullet' class='latex' />, … 0, 1) is a Boolean algebra, then <img src='http://s.wordpress.com/latex.php?latex=%28B%2C%20%5Cleq%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(B, \leq)' title='(B, \leq)' class='latex' /> is a partially ordered set with greatest element 1 and least element 0. Moreover each pair {a, b} of elemtns has a least upper bound a + b and a greatest lower bound <img src='http://s.wordpress.com/latex.php?latex=a%20%5Cbullet%20b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a \bullet b' title='a \bullet b' class='latex' /></p>
<p>&nbsp;</p>
<h3><strong>Switching algebra</strong></h3>
<p>&nbsp;</p>
<p>The knowledge of Boolean algebra received recognition mainly due to its application in electrical circuits C.E. Shanon in his paper in 1938 was the first to introduce circuit algebra as method of dealing algebraically with relay circuits. The emergence of digital computers brought this system into significance as a part of pure mathematics applied to electrical engineering.</p>
<p>With the symbols in the beginning of the chapter, we consider the system [0,1] +, <img src='http://s.wordpress.com/latex.php?latex=%5Cbullet&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bullet' title='\bullet' class='latex' />]. The system is a Boolean with the composition tables.</p>
<p><img class="alignnone size-full wp-image-6688" title="composition tables" src="http://oscience.info/wp-content/uploads/Untitled44.png" alt="" width="400" height="83" /></p>
<p>&nbsp;</p>
<p>Also 0’ 1 and 1’ = 0. In an electrical circuit if a switch is closed (or on) we denote it by 1 and when it is open (or off), it will be denoted by 0.</p>
<p><img class="alignnone size-full wp-image-6689" title="Openswitch" src="http://oscience.info/wp-content/uploads/Untitled.png1_.png" alt="" width="300" height="159" /></p>
<p>Switches in parallel are denoted by + and in series by <img src='http://s.wordpress.com/latex.php?latex=%5Cbullet&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bullet' title='\bullet' class='latex' />. Thus</p>
<p><img class="alignnone size-full wp-image-6691" title="switches in parallel" src="http://oscience.info/wp-content/uploads/Untitled1.png2_1.png" alt="" width="400" height="136" /></p>
<p>The above figure gives two switches in parallel and the truth table is as under:</p>
<table border="1" cellspacing="0" cellpadding="0">
<tbody>
<tr>
<td valign="top" width="84">x</td>
<td valign="top" width="84">y</td>
<td valign="top" width="84">x+ y</td>
</tr>
<tr>
<td valign="top" width="84">0</p>
<p>0</p>
<p>1</p>
<p>1</td>
<td valign="top" width="84">0</p>
<p>1</p>
<p>1</p>
<p>1</td>
<td valign="top" width="84">0</p>
<p>0</p>
<p>0</p>
<p>1</td>
</tr>
</tbody>
</table>
<p>&nbsp;</p>
<p>Hence we see that the current does not flow only in the case when both the switches are open.</p>
<p>Switches in series are denoted by <img src='http://s.wordpress.com/latex.php?latex=%5Cbullet&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bullet' title='\bullet' class='latex' />. The figure of two switches in series is</p>
<p><img class="alignnone size-full wp-image-6692" title="switches in series" src="http://oscience.info/wp-content/uploads/Untitled.png3_.png" alt="" width="400" height="62" /></p>
<p>The truth table is as under:</p>
<table border="1" cellspacing="0" cellpadding="0">
<tbody>
<tr>
<td valign="top" width="84">x</td>
<td valign="top" width="84">y</td>
<td valign="top" width="84">x+ y</td>
</tr>
<tr>
<td valign="top" width="84">0</p>
<p>0</p>
<p>1</p>
<p>1</td>
<td valign="top" width="84">0</p>
<p>1</p>
<p>1</p>
<p>1</td>
<td valign="top" width="84">0</p>
<p>0</p>
<p>0</p>
<p>1</td>
</tr>
</tbody>
</table>
<p>&nbsp;</p>
<p>In this case, the current flows in the case when both the switches are closed.</p>
<p>Some other circuits are below shown:</p>
<p>Draw figures to show the following equivalence of switches:</p>
<p>(i)x.y = y.x</p>
<p>(ii) x+ y = y + x</p>
<p>(iii) x + y .z = (x  + y). (x + z)</p>
<p><img class="alignnone size-full wp-image-6693" title="equivalence of switches" src="http://oscience.info/wp-content/uploads/Untitled.png4_.png" alt="" width="400" height="249" /></p>
<p>(iv) x.(y + z) = x.y + x.z</p>
<p>(v) x + x’ = 1 and x.x’ = 0</p>
<p>(vi) x + (y +z) = (x + y) + z</p>
<p>(vii) x.(y.z) = (x.y).z</p>
<p>&nbsp;</p>
<p>In the above, we have studied cases where the switches are either open or closed independent of each other. Now we shall see same cases where x represents the closed switch and x’ the open in the same circuit.</p>
<p>&nbsp;</p>
<p><strong>The Two Switch Problem:</strong> we have to design a circuit such that the light of a bulb can be turned on or off by the change in state of any of the two switches in the circuit. The figure is shown below:</p>
<p><img class="alignnone size-full wp-image-6695" title="two switch " src="http://oscience.info/wp-content/uploads/Untitled.png5_.png" alt="" width="400" height="111" /></p>
<p>There are tow switches <img src='http://s.wordpress.com/latex.php?latex=S_1%20%5Ctext%7Band%7D%20S_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='S_1 \text{and} S_2' title='S_1 \text{and} S_2' class='latex' /> different from ordinary switches as they have two way contacts. While <img src='http://s.wordpress.com/latex.php?latex=x_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1' title='x_1' class='latex' /> state closes B and opens EF, <img src='http://s.wordpress.com/latex.php?latex=x_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1' title='x_1' class='latex' /> closes EF and opens AB, similar is the case with <img src='http://s.wordpress.com/latex.php?latex=S_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='S_2' title='S_2' class='latex' /> so that the current flows in the state</p>
<img src='http://s.wordpress.com/latex.php?latex=f%28x_1%2C%20x_2%29%20%3D%20x_1x_2%5E%7B%5Cprime%7D%20%2B%20x_2x_1%5E%7B%5Cprime%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x_1, x_2) = x_1x_2^{\prime} + x_2x_1^{\prime}' title='f(x_1, x_2) = x_1x_2^{\prime} + x_2x_1^{\prime}' class='latex' />
<p>&nbsp;</p>
<p>If the circuit is on with either <img src='http://s.wordpress.com/latex.php?latex=x_1x_2%5E%7B%5Cprime%7D%20%5C%3B%20or%20%5C%3B%20x_2x_1%5E%7B%5Cprime%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1x_2^{\prime} \; or \; x_2x_1^{\prime}' title='x_1x_2^{\prime} \; or \; x_2x_1^{\prime}' class='latex' /> a change in any switch will make the circuit off. Again any change in the state of any of the switches will make the circuit on. We can write</p>
<img src='http://s.wordpress.com/latex.php?latex=f%28x_1%2C%20x_2%29%20%3D%20%28x_1%20%2B%20x_2%29%20%28x_1%5E%7B%5Cprime%7D%20%2B%20x_2%5E%7B%5Cprime%7D%20%5C%5C%20%3D%20%28x_1%2Fx_2%20%3D%20%28x_2%2Fx_1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x_1, x_2) = (x_1 + x_2) (x_1^{\prime} + x_2^{\prime} \\ = (x_1/x_2 = (x_2/x_1)' title='f(x_1, x_2) = (x_1 + x_2) (x_1^{\prime} + x_2^{\prime} \\ = (x_1/x_2 = (x_2/x_1)' class='latex' />
<p>&nbsp;</p>
<p>A three switch can be designed as</p>
<img src='http://s.wordpress.com/latex.php?latex=f%28x_1%2C%20x_2%2C%20x_3%29%20%3D%20x_1x_2%5E%7B%5Cprime%7D%20x_3%5E%7B%5Cprime%7D%20%2B%20x_1%5E%7B%5Cprime%7D%20x_2%20x_3%5E%7B%5Cprime%7D%20%2B%20x_1%5E%7B%5Cprime%7Dx_2%5E%7B%5Cprime%7Dx_3%20%2B%20x_1x_2x_3%20%5C%5C%20%3D%20x_1%2Fx_2%2Fx_3%20%5C%5C%20%3D%20%28x_1%2Fx_2%29x_3%5E%7B%5Cprime%7D%20%2B%20%28x_1%5E%7B%5Cprime%7D%2Fx_2%29x_3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x_1, x_2, x_3) = x_1x_2^{\prime} x_3^{\prime} + x_1^{\prime} x_2 x_3^{\prime} + x_1^{\prime}x_2^{\prime}x_3 + x_1x_2x_3 \\ = x_1/x_2/x_3 \\ = (x_1/x_2)x_3^{\prime} + (x_1^{\prime}/x_2)x_3' title='f(x_1, x_2, x_3) = x_1x_2^{\prime} x_3^{\prime} + x_1^{\prime} x_2 x_3^{\prime} + x_1^{\prime}x_2^{\prime}x_3 + x_1x_2x_3 \\ = x_1/x_2/x_3 \\ = (x_1/x_2)x_3^{\prime} + (x_1^{\prime}/x_2)x_3' class='latex' />
<p>&nbsp;</p>
<p>A n switch circuit can similarly be designed.<strong></strong></p>
<p>&nbsp;</p>
<h3><strong>Truth table</strong></h3>
<p>&nbsp;</p>
<p>The truth of the above Boolean relations can also be shown with the help of a truth table. If a variable x is denoted by 1, x’ is denoted by 0. For + sometimes <img src='http://s.wordpress.com/latex.php?latex=%5Cbigcup%20%5C%3B%20or%20%5C%3B%20%5Cvee&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bigcup \; or \; \vee' title='\bigcup \; or \; \vee' class='latex' />, and similarly for <img src='http://s.wordpress.com/latex.php?latex=%5Cbigcap%20%5C%3B%20or%20%5C%3B%20%5Cwedge&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\bigcap \; or \; \wedge' title='\bigcap \; or \; \wedge' class='latex' /> are used. Thus to prove the relation a(a + b) = a, we form the table</p>
<table border="1" cellspacing="0" cellpadding="0">
<tbody>
<tr>
<td valign="top" width="85">a</td>
<td valign="top" width="85">b</td>
<td valign="top" width="85">a + b</td>
<td valign="top" width="85">a.(a + b)</td>
</tr>
<tr>
<td valign="top" width="85">1</p>
<p>1</p>
<p>0</p>
<p>0</td>
<td valign="top" width="85">1</p>
<p>0</p>
<p>1</p>
<p>0</td>
<td valign="top" width="85">1</p>
<p>1</p>
<p>1</p>
<p>0</td>
<td valign="top" width="85">1</p>
<p>1</p>
<p>0</p>
<p>0</td>
</tr>
</tbody>
</table>
<p>&nbsp;</p>
<h3><strong>Duality: Another definition</strong></h3>
<p>&nbsp;</p>
<p>In every (correct) formula of Boolean algebra, we can interchange addition and multiplication. However when an equality fulfills the laws of Boolean Algebra involves the ‘special’ elements 0 and 1, then the interchange of Boolean addition and multiplication in this equality must be followed by the interchange of elements 0 and 1. For instance the validity of</p>
<p>(A + B) (A + 1) + (A + B) (B + 0) = A + B</p>
<p>Results in</p>
<p>(AB + A0). (AB +B1) = AB</p>
]]></content:encoded>
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		<item>
		<title>Wave motion</title>
		<link>http://oscience.info/physics/wave-motion/wave-motion/</link>
		<comments>http://oscience.info/physics/wave-motion/wave-motion/#comments</comments>
		<pubDate>Wed, 04 Apr 2012 16:54:03 +0000</pubDate>
		<dc:creator>sandeep</dc:creator>
				<category><![CDATA[Wave motion]]></category>
		<guid isPermaLink="false">http://oscience.info/?p=6675</guid>
		<description><![CDATA[Wave motion: Wave: Wave is defined as the continuous transfer of state from one part of the medium to the other with finite velocity without changing its form. There are two types of waves : (i) Mechanical waves: The waves which require a material medium for their propagation are called mechanical waves. Sound waves in [...]]]></description>
			<content:encoded><![CDATA[<h1>
<strong>Wave motion:</strong></h1>
<p class="size-full wp-image-6676" title="Wave motion">
<h2><strong>Wave: </strong></h2>
<p>Wave is defined as the continuous transfer of state from one part of the medium to the other with finite velocity without changing its form.<br />
There are two types of waves :</p>
<p>(i) <strong><span style="text-decoration: underline;">Mechanical waves:</span></strong></p>
<p>The waves which require a material medium for their propagation are called mechanical waves. Sound waves in air, waves in stretched string are examples of mechanical waves.</p>
<p>(<span style="text-decoration: underline;">ii) <strong>Electromagnetic waves:</strong></span></p>
<p>The waves which require no material medium for their propagation are called electromagnetic waves. The examples of electromagnetic waves are light waves, heat radiations, X—rays, <img src='http://s.wordpress.com/latex.php?latex=%5Cgamma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\gamma' title='\gamma' class='latex' />-rays etc.</p>
<p><strong>Characteristics of medium for Mechanical waves :</strong></p>
<p style="padding-left: 30px;">&gt;The medium must be elastic.</p>
<p style="padding-left: 30px;">&gt;The medium must have inertia.</p>
<p style="padding-left: 30px;">&gt;The damping must be very small</p>
<p>That is why bell is made of metal and not of wood.</p>
<p><strong>Types of Mechanical waves : </strong></p>
<p>There are two distinct types of mechanical waves:</p>
<p><strong><span style="text-decoration: underline;">Transverse waves</span></strong> are the waves in which particles of the medium execute simple harmonic motion about their mean positions at right angles to the direction of propagation of wave. The example is a wave spreading over the surface of water when a stone is dropped in a pond. All electromagnetic waves (e.g. light waves, heat waves etc.) are transverse. For, the propagation of transverse waves in medium, the medium must be rigid. The waves within solids are transverse waves.</p>
<p><strong><span style="text-decoration: underline;">Longitudinal waves </span></strong>are the waves in which the particles of the medium execute simple harmonic motion about their mean positions along the direction of propagation of wave. The example is sound waves in air.<br />
When a metallic rod is rubbed along the direction of length, the waves produced in the rod are longitudinal.</p>
<p><strong>Progressive waves:</strong></p>
<p>A progressive wave is one in which the disturbance in continuously transmitted along the direction of propagation of wave. These waves transmit energy from one place to another place.</p>
<p><strong>Amplitude (A):</strong></p>
<p>The amplitude is maximum displacement of the vibrating particle from its mean position.</p>
<p><strong>Period (T):</strong> The period is the time taken by the vibrating particle in one vibration.</p>
<p><strong>Wavelength (<img src='http://s.wordpress.com/latex.php?latex=%5Clambda&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda' title='\lambda' class='latex' />) : </strong> The wavelength is defined as the distance traversed by the disturbance in time during which the vibrating particle makes one complete vibration. The relation between these quantities are</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Clambda%20%3D%20v%20T%20%2C%20n%20%3D%20%5Cdfrac%7B1%7D%7BT%7D%20%2C%20v%20%3D%20n%20%5Clambda&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda = v T , n = \dfrac{1}{T} , v = n \lambda' title='\lambda = v T , n = \dfrac{1}{T} , v = n \lambda' class='latex' />
<p><strong>Speed of Sound Waves:</strong></p>
<p>Sound is a form of energy which produces the sensation of hearing and propagates through mechanical waves.</p>
<p><strong>Longitudinal waves :</strong></p>
<p>The speed of Longitudinal waves is a medium depends on elasticity (E) and density (d) of given by</p>
<p>Newton’s formula v = <img src='http://s.wordpress.com/latex.php?latex=%5Csqrt%7B%28%20%5Cdfrac%7BE%7D%7Bd%7D%20%29%20%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{( \dfrac{E}{d} ) }' title='\sqrt{( \dfrac{E}{d} ) }' class='latex' /> …..(1)</p>
<p><strong><em>For longitudinal waves in a solid </em></strong>elasticity E is replaced by Young’s modulus, Y, so that</p>
<p><img src='http://s.wordpress.com/latex.php?latex=v%20%3D%20%5Csqrt%7B%20%28%5Cdfrac%7BY%7D%7Bd%7D%20%29%20%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v = \sqrt{ (\dfrac{Y}{d} ) }' title='v = \sqrt{ (\dfrac{Y}{d} ) }' class='latex' /> ( = 5300 m/s for steel ) ….(2)</p>
<p><strong><em>For longitudinal waves in a liquid</em></strong><em>  </em>E is replaced by Bulk modulus K, so that</p>
<p><img src='http://s.wordpress.com/latex.php?latex=v%20%3D%20%5Csqrt%7B%20%28%5Cdfrac%7BK%7D%7Bd%7D%20%29%20%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v = \sqrt{ (\dfrac{K}{d} ) }' title='v = \sqrt{ (\dfrac{K}{d} ) }' class='latex' /> = 1450 m/s in water&#8230;..(3)</p>
<p><strong><em>For longitudinal waves in a gas,</em></strong><br />
Newton assumed that the changes in air are isothermal<br />
Isothermal elasticity E<sub>T </sub>= P</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Ctherefore%20v%20%3D%20%5Csqrt%7B%5Cdfrac%7BP%7D%7Bd%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\therefore v = \sqrt{\dfrac{P}{d}}' title='\therefore v = \sqrt{\dfrac{P}{d}}' class='latex' />
<p>For at NTP, P = 10<sup>5</sup> N/m<sup>2</sup> , d = 1.29 kg/m<sup>3</sup></p>
<p><img src='http://s.wordpress.com/latex.php?latex=%5Ctherefore%20v%20%3D%20%5Csqrt%7B%5Cdfrac%7B10%5E5%7D%7B1.29%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\therefore v = \sqrt{\dfrac{10^5}{1.29}}' title='\therefore v = \sqrt{\dfrac{10^5}{1.29}}' class='latex' /> = 280 m/s</p>
<p>which is incorrect.</p>
<p>Then Laplace assumed that changes in air are adiabatic</p>
<p>Adiabatic elasticity E<sub>S</sub> = <img src='http://s.wordpress.com/latex.php?latex=%5Cgamma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\gamma' title='\gamma' class='latex' />P</p>
<p>Where <img src='http://s.wordpress.com/latex.php?latex=%5Cgamma%20%3D%20%5Cdfrac%7BC_P%7D%7BC_V%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\gamma = \dfrac{C_P}{C_V}' title='\gamma = \dfrac{C_P}{C_V}' class='latex' /> and P is pressure, so that</p>
<p>v = <img src='http://s.wordpress.com/latex.php?latex=%5Csqrt%7B%5Cdfrac%7B%5Cgamma%20P%7D%7Bd%7D%7D%20%5C%5C%20%3D%20%5Csqrt%7B%5Cdfrac%7B%5Cgamma%20RT%7D%7BM%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{\dfrac{\gamma P}{d}} \\ = \sqrt{\dfrac{\gamma RT}{M}}' title='\sqrt{\dfrac{\gamma P}{d}} \\ = \sqrt{\dfrac{\gamma RT}{M}}' class='latex' /> = 332 m/s for air at 0<sup>o </sup>C) …(4)</p>
<p>where T is absolute temperature, M is molecular weight of gas and R = 8.3 joule/gm-mol K) is gas constant for 1 mole. Obliviously the velocity of sound is maximum in solids and minimum in gases. Clearly at a given temperature</p>
<img src='http://s.wordpress.com/latex.php?latex=v%20%5Cpropto%20%5Csqrt%7B%5Cdfrac%7B1%7D%7BM%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v \propto \sqrt{\dfrac{1}{M}}' title='v \propto \sqrt{\dfrac{1}{M}}' class='latex' />
<p><strong>Effect of various factors on velocity of sound in gases :</strong></p>
<p><strong><span style="text-decoration: underline;">(i) Effect of pressure :</span> </strong></p>
<p>The velocity of sound is independent of pressure<strong>.</strong></p>
<p><strong><span style="text-decoration: underline;">(ii) Effect of temperature: </span></strong></p>
<p>The velocity of sound increases with increase of temperature</p>
<p>As v = <img src='http://s.wordpress.com/latex.php?latex=%5Csqrt%7B%5Cdfrac%7B%5Cgamma%20RT%7D%7BM%7D%7D%20%3B%20%5Ctext%7Btherefore%7D%20v%20%5Cpropto%20%5Csqrt%7BT%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{\dfrac{\gamma RT}{M}} ; \text{therefore} v \propto \sqrt{T}' title='\sqrt{\dfrac{\gamma RT}{M}} ; \text{therefore} v \propto \sqrt{T}' class='latex' /></p>
<p>where T is absolute temperature.</p>
<p>If v<sub>0</sub> is velocity of sound at 0°C, then the velocity of sound at in all gases is given by</p>
<p><img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7Bv_t%7D%7Bv_0%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7Bt%20%2B%20273%7D%7B273%7D%7D%20%5C%5C%20%5Ctext%7Bor%7D%20%5C%2C%20v_t%20%3D%20v_0%20%2B%20%5Cdfrac%7Bv%3Csub%3E0%3C%2Fsub%3E%7D%7B546%7D%20t&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{v_t}{v_0} = \sqrt{\dfrac{t + 273}{273}} \\ \text{or} \, v_t = v_0 + \dfrac{v&lt;sub&gt;0&lt;/sub&gt;}{546} t' title='\dfrac{v_t}{v_0} = \sqrt{\dfrac{t + 273}{273}} \\ \text{or} \, v_t = v_0 + \dfrac{v&lt;sub&gt;0&lt;/sub&gt;}{546} t' class='latex' /> …(6)</p>
<p>For velocity of sound in air , v<sub>0</sub> = 332 m/s</p>
<p>or,      v<sub>t </sub>= v<sub>0</sub> + 0.61 t m/s</p>
<p><strong><span style="text-decoration: underline;">(iii) Effect of humidity: </span></strong></p>
<p>The velocity of sound in air increases with increase of humidity since density of air decreases with increase of humidity and <img src='http://s.wordpress.com/latex.php?latex=v%20%5Cpropto%20%5Cdfrac%7B1%7D%7B%5Csqrt%7Bd%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v \propto \dfrac{1}{\sqrt{d}}' title='v \propto \dfrac{1}{\sqrt{d}}' class='latex' /><br />
the velocity of sound is maximum in hydrogen (= 1270 mls) being four times than that in oxygen.</p>
<p><strong>(iv)Effect of nature of gas</strong></p>
<p>As <img src='http://s.wordpress.com/latex.php?latex=v%20%5Cpropto%20%28%201%20%2F%20%5Csqrt%7BM%7D%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v \propto ( 1 / \sqrt{M} )' title='v \propto ( 1 / \sqrt{M} )' class='latex' />the velocity of sound is maximum in hydrogen (= 1270 m/s) being four times than that in oxygen.</p>
<p><strong><span style="text-decoration: underline;">(v) Effect of velocity of wind: </span></strong></p>
<p>The velocity of sound is greater in the direction of wind and smaller in an opposite direction.</p>
<p><strong>(vi) Effect of frequency:</strong></p>
<p>The velocity of sound in air is independent of the frequency of propagating wave.</p>
<p><strong>The speed of transverse waves in a string is given by</strong></p>
<img src='http://s.wordpress.com/latex.php?latex=v%20%3D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7BM%7D%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7BT%7D%7Bx%20r%5E2%20d%7D%7D%20%5C%5C%20%3D%20%5Csqrt%7B%5Cdfrac%7BT%2Fxr%5E2%7D%7Bd%7D%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7B%5Ctext%7Bstress%7D%7D%7Bd%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v = \sqrt{\dfrac{T}{M}} = \sqrt{\dfrac{T}{x r^2 d}} \\ = \sqrt{\dfrac{T/xr^2}{d}} = \sqrt{\dfrac{\text{stress}}{d}}' title='v = \sqrt{\dfrac{T}{M}} = \sqrt{\dfrac{T}{x r^2 d}} \\ = \sqrt{\dfrac{T/xr^2}{d}} = \sqrt{\dfrac{\text{stress}}{d}}' class='latex' />
<p>Where</p>
<p>T = tension in string,<br />
m = mass per unit length of string,                                                                                                                                                      r  = radius of string,                                                                                                                                                         d = density of string</p>
<p><strong>Equation of General Harmonic Wave:</strong></p>
<p>The equation of a progressive wave is expressed in the following equivalent forms:</p>
<img src='http://s.wordpress.com/latex.php?latex=y%20%3D%20A%20%5Csin%20%5Comega%20%28%20t%20-%20%5Cdfrac%7Bx%7D%7Bv%7D%20%29%20%5C%5C%20%3D%20A%20%5Csin%20%28%5Comega%20t%20-%20kx%20%29%20%5C%5C%20%3D%20A%20%5Csin%202%20%5Cpi%20%28%20%5Cdfrac%7Bt%7D%7BT%7D%20-%20%5Cdfrac%7Bx%7D%7B%5Clambda%7D%20%5C%5C%20%5Ctext%7Bwhere%7D%20k%20%3D%20%5Cdfrac%7B%5Comega%7D%7Bv%7D%20%3D%20%5Cdfrac%7B2%20%5Cpi%7D%7B%5Clambda%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = A \sin \omega ( t - \dfrac{x}{v} ) \\ = A \sin (\omega t - kx ) \\ = A \sin 2 \pi ( \dfrac{t}{T} - \dfrac{x}{\lambda} \\ \text{where} k = \dfrac{\omega}{v} = \dfrac{2 \pi}{\lambda}' title='y = A \sin \omega ( t - \dfrac{x}{v} ) \\ = A \sin (\omega t - kx ) \\ = A \sin 2 \pi ( \dfrac{t}{T} - \dfrac{x}{\lambda} \\ \text{where} k = \dfrac{\omega}{v} = \dfrac{2 \pi}{\lambda}' class='latex' />
<p>is propagation constant and points along positive X- axis. Let if the wave propagates along negative axis, x is replaced by so that above equations take the forms</p>
<img src='http://s.wordpress.com/latex.php?latex=y%20%3D%20A%20%5Csin%20%5Comega%20%28%20t%20%2B%20%5Cdfrac%7Bx%7D%7Bv%7D%20%29%20%5C%5C%20%3D%20A%20%5Csin%20%28%5Comega%20t%20%2B%20k%20x%20%29%20%5C%5C%20%3D%20A%20%5Csin%202%20%5Cpi%20%28%20%5Cdfrac%7Bt%7D%7BT%7D%20%2B%20%5Cdfrac%7Bx%7D%7B%5Clambda%7D%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = A \sin \omega ( t + \dfrac{x}{v} ) \\ = A \sin (\omega t + k x ) \\ = A \sin 2 \pi ( \dfrac{t}{T} + \dfrac{x}{\lambda} )' title='y = A \sin \omega ( t + \dfrac{x}{v} ) \\ = A \sin (\omega t + k x ) \\ = A \sin 2 \pi ( \dfrac{t}{T} + \dfrac{x}{\lambda} )' class='latex' />
<p>In above equation there may be phase factor <img src='http://s.wordpress.com/latex.php?latex=%5Cphi&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\phi' title='\phi' class='latex' /> so that the general equation of wave propagating along positive takes the form</p>
<img src='http://s.wordpress.com/latex.php?latex=y%20%3D%20A%20%5Csin%20%5B%20%5Comega%20%28%20t%20-%20%5Cdfrac%7Bx%7D%7Bv%7D%20%29%20%2B%20%5Cphi%20%5D%20%5C%5C%20%3D%20A%20%5Csin%20%5B%20%28%20%5Comega%20t%20-%20kx%20%29%20%2B%20%5Cphi%20%5D%20%5C%5C%20%3D%20A%20%5Csin%20%5B%202%20%5Cpi%20%28%20%5Cdfrac%7Bt%7D%7BT%7D%20-%20%5Cdfrac%7Bx%7D%7B%5Clambda%7D%20%29%20%2B%20%5Cphi%20%5D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = A \sin [ \omega ( t - \dfrac{x}{v} ) + \phi ] \\ = A \sin [ ( \omega t - kx ) + \phi ] \\ = A \sin [ 2 \pi ( \dfrac{t}{T} - \dfrac{x}{\lambda} ) + \phi ]' title='y = A \sin [ \omega ( t - \dfrac{x}{v} ) + \phi ] \\ = A \sin [ ( \omega t - kx ) + \phi ] \\ = A \sin [ 2 \pi ( \dfrac{t}{T} - \dfrac{x}{\lambda} ) + \phi ]' class='latex' />
<p>The quantity in bracket {<img src='http://s.wordpress.com/latex.php?latex=%5Cphi&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\phi' title='\phi' class='latex' />} is called the phase of the wave.</p>
<p>A transverse wave propagates by means of crests and troughs and there is no change in pressure in a transverse wave. A longitudinal wave propagates by means of compressions and rarefactions and there is always a change of pressure along a longitudinal wave, given by</p>
<p>Excess pressure p = &#8211; E <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7B%5Ctriangle%20y%7D%7B%5Ctriangle%20x%7D%20%3D%20E%20%5Cdfrac%7Bu%7D%7Bv%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{\triangle y}{\triangle x} = E \dfrac{u}{v}' title='\dfrac{\triangle y}{\triangle x} = E \dfrac{u}{v}' class='latex' /></p>
<p>Pressure amplitude, <img src='http://s.wordpress.com/latex.php?latex=P_%7B%5Ctext%7Bmax%7D%7D%20%3D%20E%20%5Cdfrac%7BU_%7B%5Ctext%7Bmax%7D%7D%7D%7Bv%7D%20%3D%20%5Cdfrac%7BEA%20%5Comega%7D%7Bv%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='P_{\text{max}} = E \dfrac{U_{\text{max}}}{v} = \dfrac{EA \omega}{v}' title='P_{\text{max}} = E \dfrac{U_{\text{max}}}{v} = \dfrac{EA \omega}{v}' class='latex' /><br />
where u is particle-velocity and v wave-velocity.<br />
The pressure being maximum at compressions and minimum at rarefactions.</p>
<p>Intensity of a wave <img src='http://s.wordpress.com/latex.php?latex=I%20%5Cpropto%20A%5E2%20%5Cpropto%20P_%7B%5Ctext%7Bmax%7D%7D2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I \propto A^2 \propto P_{\text{max}}2' title='I \propto A^2 \propto P_{\text{max}}2' class='latex' /></p>
<p><strong> Differential equation of Harmonic wave motion is </strong><img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7Bd%5E2%20y%7D%7Bd%20x%5E2%7D%20%3D%20%5Cdfrac%7B1%7D%7Bv%5E2%7D%20.%20%5Cdfrac%7Bd%5E2%20y%7D%7Bdt%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{d^2 y}{d x^2} = \dfrac{1}{v^2} . \dfrac{d^2 y}{dt^2}' title='\dfrac{d^2 y}{d x^2} = \dfrac{1}{v^2} . \dfrac{d^2 y}{dt^2}' class='latex' /></p>
<p>v being velocity of wave.</p>
<p>Thus a harmonic wave, is of the form y = f ( x – vt ) or y = f ( x + vt ) and for this <img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7Bd%5E2%20y%7D%7Bd%20x%5E2%7D%20%5Ctext%7Band%7D%20%5Cdfrac%7Bd%5E2%20y%7D%7Bd%20t%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{d^2 y}{d x^2} \text{and} \dfrac{d^2 y}{d t^2}' title='\dfrac{d^2 y}{d x^2} \text{and} \dfrac{d^2 y}{d t^2}' class='latex' /> must exit and satisfy the differential equation</p>
<img src='http://s.wordpress.com/latex.php?latex=%5Cdfrac%7Bd%5E2%20y%7D%7Bd%20x%5E2%7D%20%3D%20%5Cdfrac%7B1%7D%7Bv%5E2%7D%20%5Cdfrac%7Bd%5E2%20y%7D%7Bd%20t%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{d^2 y}{d x^2} = \dfrac{1}{v^2} \dfrac{d^2 y}{d t^2}' title='\dfrac{d^2 y}{d x^2} = \dfrac{1}{v^2} \dfrac{d^2 y}{d t^2}' class='latex' />
<p><strong>The equation of propagation of a wave pulse is</strong></p>
<img src='http://s.wordpress.com/latex.php?latex=y%20%3D%20%5Cdfrac%7Ba%7D%7Bb%20%2B%20%28%20x%20%5Cpm%20vt%20%29%5E2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = \dfrac{a}{b + ( x \pm vt )^2}' title='y = \dfrac{a}{b + ( x \pm vt )^2}' class='latex' />
<p>where a and b are constant (-)  sign is for pulse propagating along (+) X-axis and positive sign for a pulse propagating along negative X-axis, v is velocity of pulse propagation.<strong> </strong></p>
<p><strong>Relation between path difference <img src='http://s.wordpress.com/latex.php?latex=%5Ctriangle&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\triangle' title='\triangle' class='latex' /> and phase difference <img src='http://s.wordpress.com/latex.php?latex=%5Cdelta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\delta' title='\delta' class='latex' /> is</strong></p>
<img src='http://s.wordpress.com/latex.php?latex=%5Ctriangle%20%3D%20%5Cdfrac%7B%5Clambda%7D%7B2%20%5Cpi%7D%20%5Cdelta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\triangle = \dfrac{\lambda}{2 \pi} \delta' title='\triangle = \dfrac{\lambda}{2 \pi} \delta' class='latex' />
<p><strong> Intensity of a wave</strong> is defined as the energy passing per second through unit area perpendicular to direction of propagation of wave and is given by <img src='http://s.wordpress.com/latex.php?latex=I%20%3D%202%20%5Cpi%20%5E2%20n%5E2%20A%5E2%20d.v%20%5Ctext%7BObviously%7D%20%5C%2C%20I%20%5Cpropto%20A%5E2%20%5Ctext%7Band%7D%20I%20%5Cpropto%20n%5E2%20%5Ctext%7Bpressure%20amplitude%7D%20%5Cpropto%20A%20%5Ctext%7Band%20intensity%7D%20I%20%5Cpropto%20A%5E2%20%5Ctherefore%20I%20%5Cpropto%20P_%7Bmax%7D2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I = 2 \pi ^2 n^2 A^2 d.v \text{Obviously} \, I \propto A^2 \text{and} I \propto n^2 \text{pressure amplitude} \propto A \text{and intensity} I \propto A^2 \therefore I \propto P_{max}2' title='I = 2 \pi ^2 n^2 A^2 d.v \text{Obviously} \, I \propto A^2 \text{and} I \propto n^2 \text{pressure amplitude} \propto A \text{and intensity} I \propto A^2 \therefore I \propto P_{max}2' class='latex' /></p>
<p><strong>Limits of audibility of sound</strong></p>
<p>The limit of audibility of sound is from 20 Hertz to 20,000 Hertz. The sound haying frequency less than 20 Hertz is called infra-sonic and that having frequency greater than 20,000 Hertz is called ultrasonic. The ultrasonic can be heard by dogs, owls etc. The ultrasonic are produced by</p>
<p>(i) piezoelectric effect</p>
<p>(ii) magnetostriction method and</p>
<p>(iii) Galton’s whistle.</p>
<p>Ultrasonic are used to determine depth of mines and sea, stimulate the plant growth, kill bacteria and smaller animals like rats, frogs and fishes.<br />
<strong>Reflection of Sound: Echoes :</strong></p>
<p>As sound propagates in the form of waves, it exhibits the phenomenon of reflection and refraction. When sound wave travelling in a medium strikes the surface separating the two media, a part of incident wave is reflected back into first medium obeying ordinary laws of reflection while the remainder is partly absorbed and partly refracted (or transmitted) into second medium.<br />
When sound wave is reflected a rarer medium or free boundary, there is no but the nature of sound wave is changed i.e. on reflection the compression is reflected back as rarefaction and vice versa.<br />
If incident wave is y = a sin (wt — kx), then the equation of reflected wave takes the form</p>
<p>y = <img src='http://s.wordpress.com/latex.php?latex=a%27%20%5Csin%20%28%20%5Comega%20t%20-%20kx%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a&#039; \sin ( \omega t - kx )' title='a&#039; \sin ( \omega t - kx )' class='latex' /></p>
<p>where a’ is new amplitude of reflected wave.<br />
As wave reverses direction, x has been replaced by (-x).</p>
<p>When sound wave is reflected from a denser medium or rigid boundary, the phase of wave is reversed but the nature does not change i.e. on reflection compression is reflected back as compression and rarefaction as rarefaction.<br />
The equation of reflected wave in this case takes the form</p>
<img src='http://s.wordpress.com/latex.php?latex=y%20%3D%20a%27%20%5Csin%20%28%20%5Comega%20t%20%2B%20k%20x%20%2B%20%5Cpi%20%29%20%3D%20-%20a%27%5Csin%20%28%5Comega%20t%20%2B%20kx%20%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y = a&#039; \sin ( \omega t + k x + \pi ) = - a&#039;\sin (\omega t + kx )' title='y = a&#039; \sin ( \omega t + k x + \pi ) = - a&#039;\sin (\omega t + kx )' class='latex' />
<p><strong>Echoes</strong></p>
<p>The common experience of sound reflection is echoes heard in large halls and in the neighborhood of hills. An echo is simply the repetition of speaker’s own voice caused by reflection at a distant surface e.g. a cliff, a row of buildings or any other extended surface.<br />
If t is the time interval between production of sound from source and its echo at the site of source, then the distance between source and reflector (s) is given by</p>
<p>2s = vt</p>
<p>where v is the velocity of sound.<br />
The persistence of hearing is 0.1 sec, therefore in order that an echo of short sound (e.g. shot of clapping) may be heard distinctly, the echo must come 0.1 sec later than the sound.</p>
<p><strong>Articulate sound:</strong></p>
<p>In case of articulate sound, it has been found that one can hear or pronounce distinctly not more than 5 syllables per second. Therefore for monosyllabic sound the minimum time interval between sound and its echo is 1/5 sec, for disyllabic and trisyllabic sounds it is 2/5 and 3/5 sec respectively and so on.<br />
Multiple echoes are produced when there are several reflecting surfaces. The speaking tubes,<br />
stethoscope, whispering gallery and sounding boards are based upon the reflection of the sound.</p>
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