NEET Booster Week - Sound & Wave Motion

追加: 2026-05-23

[00:01:33]Hello students. Good afternoon. How are you?

[00:01:37]Our today's topic is wave motion.

[00:01:41]This chapter contributes to one or two questions in NEET exam.

[00:01:47]And this chapter you need to practice more numerical problems only in this chapter. Just like uh units and measurements. I told you know, units and you need to practice more problems. This chapter also, you need to practice more number of problems.

[00:02:04]Uh properties of progressive waves, properties of stationary wave, vibrations of air columns, vibrations of stretched and strings, beats.

[00:02:17]These five topics, problems, numerical problems, you need to practice. Come on, let us jump into problems.

[00:02:25]See the first problem. Two open organ pipes of fundamental frequencies N1 and N2 are joined in series.

[00:02:36]There are two open pipes.

[00:02:43]Their fundamental frequencies are N1 and N2.

[00:02:49]Their lengths L1, L2 you take.

[00:02:56]The fundamental mode of vibration In the fundamental mode of vibration, their frequencies are N1 and N2.

[00:03:08]Now, these two are joined in series.

[00:03:10]Means, these two open pipes of lengths L1, L2 join like this.

[00:03:17]Then, you'll have again open pipe only. Again open pipe of length L1 plus L2 you'll have. He's asking, "What is the fundamental frequency of this new air column in this new pipe?"

[00:03:33]So, what is this first pipe fundamental frequency N1?

[00:03:37]Velocity of sound in air by 2 L1.

[00:03:43]That means, L1 will be V by 2 N1.

[00:03:49]Similarly, this N2 will be v by 2 l 2.

[00:03:55]Means l 2 is v by 2 n 2.

[00:04:01]Now come to this join the two pipes joined in this case.

[00:04:07]The fundamental frequency must be v by 2 into l 1 + l 2. Now it's length.

[00:04:16]Then in place of l 1, I can write v by 2 n 1.

[00:04:27]And here v by 2 n 2, I can write v by 2 common here.

[00:04:33]That all v by 2, v by 2, v by 2, that gets cancelled. You will have 1 by 1 by n 1 + 1 by n 2.

[00:04:45]n 1 n 2 by n 1 + n 2.

[00:05:03]Okay, let us now move into next problem.

[00:05:22]See the second one. An open pipe of length 80 cm has second harmonic frequency equal to fundamental frequency of closed organ pipe. The length of the closed pipe is open pipe length they gave.

[00:05:39]Closed pipe length they're asking. And what is the condition given here? Second harmonic frequency of that open pipe is equal to fundamental frequency of closed pipe.

[00:05:55]For open pipe open pipe what is fundamental frequency V by 2L.

[00:06:06]Second harmonic means two times of V by 2L.

[00:06:11]That is second harmonic. Second harmonic means which is two times two fundamental frequency, no?

[00:06:19]This second harmonic frequency of open pipe is equal to fundamental frequency of closed pipe.

[00:06:28]Closed pipe fundamental frequency V by 4L. LCM writing closed pipe length.

[00:06:41]Here L not I'm writing open pipe. This second harmonic of open pipe is equal to fundamental of closed pipe. Equate them.

[00:06:51]V by 4LC is 2 into V by 2L not.

[00:07:00]And find LC.

[00:07:05]4LC equal to L not. LC is L not by 4.

[00:07:11]20 cm 80 by 4 20 cm.

[00:07:23]Now in this pipes another thing. Suppose there is an open pipe of length L.

[00:07:36]This open pipe fundamental frequency fundamental frequency Suppose 200 Hz given.

[00:07:52]This open pipe fundamental frequency 200 Hz given.

[00:07:57]Suppose the question is given like this.

[00:08:05]It is immersed in water.

[00:08:15]So that 1/3 of length is inside water.

[00:08:29]Then what is the new fundamental frequency?

[00:08:36]1/3 of length is immersed in water. What is the new fundamental frequency?

[00:08:45]is asked.

[00:08:47]If this open pipe of length L is immersed in water so that 1/3 of its length is inside water then how much length is outside?

[00:09:02]Means how much air column length is there? L minus L by 3.

[00:09:08]That is 2L by 3. Air column is there.

[00:09:13]And this air column is enclosed in which pipe? Now this is closed pipe, no?

[00:09:18]Earlier it is open pipe. Now it is closed pipe.

[00:09:23]So what is the fundamental frequency of this air column enclosed in closed pipe of length 2L by 3? That is question.

[00:09:33]This 200 Hz is this complete open pipe fundamental frequency.

[00:09:40]Open pipe fundamental frequency V by 2L.

[00:09:46]This N 200 given.

[00:09:51]V by 2L is 200.

[00:09:54]Now what is the frequency of that air column in the closed pipe of length 2L by 3 is asking.

[00:10:03]Closed pipe fundamental frequency V by 4L.

[00:10:07]In place of L I have to take 2L by 3.

[00:10:12]Then you will get 3 by 4 times of V by 2L.

[00:10:19]This V by 2L given as 200. 3 by 4 into 200. 150 hertz you will get. Like this also questions can be given on this.

[00:10:58]Let us move into next problem.

[00:11:08]In Melde's experiment when the tension in the string decreases by.009 kg weight the number of loops changes from 4 to 5.

[00:11:25]The initial tension is dash.

[00:11:30]Melde's experiment means one stretched string will be there.

[00:11:37]And weight hanger to create tension, we put weights here.

[00:11:42]Uh this is stretched string. Uh one edge of this stretched string uh is attached to one vibrating tuning fork of some frequency n.

[00:11:58]This is Melde's experiment. Means this wire is made to vibrate with a frequency n.

[00:12:06]Now what you are doing, you are changing the tension. Again, frequency is that only.

[00:12:12]Means in this Melde's experiment, frequency of that stretched string is maintained constant.

[00:12:23]Now in this problem, you are changing the tension. Then the number of loops formed here change. Here loops will form when stationary wave forms. The number of loops changed from four to five, but frequency constant.

[00:12:40]Now what is the frequency of a stretched string? Uh You can write like this. Now p by 2l into root t by mu. Mu linear density.

[00:12:54]T tension. P number of segments, number of loops formed is p.

[00:13:00]Here what you are changing?

[00:13:04]Tension. Because of that, p changed.

[00:13:07]Remaining things same only now.

[00:13:09]Frequency constant, length of the wire constant, linear density, wire material same, wire thickness same, same wire now. So linear density constant. N constant, l constant, mu constant.

[00:13:25]Variables are p and t.

[00:13:28]So, you write like this, P into root T is a constant here.

[00:13:39]P into root T is a constant.

[00:13:43]That means P inversely proportional to root T.

[00:13:49]If you decrease tension, number of loops increase.

[00:13:55]So, P1 by P2 is under root T2 by T1.

[00:14:04]He's asking, find the initial tension.

[00:14:08]Initial tension T1.

[00:14:13]And the new tension, tension decreased by 0.009 kg weight he gave.

[00:14:20]So, the new tension is T2 minus 0.009.

[00:14:29]P1 four loops, P2 five loops he gave. 4 by 5, 4 by 5. Now, square both sides. 16 by 25 is equal to squaring.

[00:14:45]4 by 5 squared 16 by 25 is equal to T2 minus 0.009 by Sorry.

[00:15:02]This is T1.

[00:15:11]T1 minus 0.009 by T1. Cross multiply that. Find T1.

[00:15:41]>> Okay, let us move to next problem.

[00:15:59]See this. Sound waves travel at 350 m/s through warm air and at 3,500 m/s through brass.

[00:16:14]The wavelength of the wave of 700 Hz frequency see as it enters brass from warm air decreases by increases by this is asking. Mean wavelength changed by how much is asking.

[00:16:38]In air 350 m/s is the velocity of wave.

[00:16:42]In brass 3,500 velocity of the wave.

[00:16:52]You know, out of all gases, velocity of sound is maximum in hydrogen gas.

[00:17:02]Out of all gases, velocity of sound is maximum in hydrogen gas.

[00:17:10]But out of all different media that is uh solids liquids, and gases. Out of all these three media, velocity of sound is maximum in solids, more in solids.

[00:17:35]Because of reason, which reason? Solids are highly elastic than these liquids and gases.

[00:17:43]Velocity of any progressive wave in any medium is root e by d. Formula is there, now? Where e is elastic modulus, d is density.

[00:17:53]Elastic property of solids is more.

[00:17:57]That's why velocity of sound is more in solids than in liquids and gases. So, what is showing that only here? Velocity of sound in air 350, but in brass 3,500.

[00:18:17]Whenever a wave travels from one medium to another medium, one medium to another medium, whenever wave travels, frequency of the wave remains constant. You know this.

[00:18:36]Velocity changes, wavelength changes.

[00:18:40]Velocity If velocity increase, wavelength increases. If velocity decreases, wavelength also decrease. v proportional to lambda, frequency constant.

[00:18:53]So, what I'm writing? Frequency is v by lambda, now?

[00:18:56]So, wavelength of velocity in air by wavelength in air is equal to velocity of sound in brass by wavelength in brass, so I'm writing.

[00:19:09]What is asking here is wavelength in brass is how many times of wavelength in air.

[00:19:16]That is asked here.

[00:19:18]So, lambda B is cross multiplying this.

[00:19:26]VB by VA into lambda A.

[00:19:34]Find this VB.

[00:19:36]3,500 VA 350 into lambda air 10 times.

[00:19:49]Wavelength increases by a factor of 10.

[00:20:14]Okay. Next.

[00:20:27]The velocity of sound in air is doubled when the temperature is raised from 0 to some alpha degree centigrade. What is the value of alpha?

[00:20:39]819 Kelvin.

[00:20:42]This is a common problem.

[00:20:45]Velocity of sound in gases is root gamma P by D, where gamma is ratio of specific heats of gas, P pressure, D density.

[00:20:58]This can also be written as root gamma RT by M.

[00:21:05]M molecular weight, gamma already you know, T absolute temperature.

[00:21:12]Newton-Laplace equation. Newton suggested that when longitudinal wave travels through a gas, the pressure volume they are changing, okay.

[00:21:23]Temperature remains constant to the changes are isothermal.

[00:21:28]Like that he suggested, but and gave a relation root P by D. Newton gave root P by D.

[00:21:35]Which is not correct. That is corrected later by Laplace.

[00:21:39]And Laplace suggested the changes in a gas, the pressure volume they will change, you know, when longitudinal wave travels through gas. The change in pressure volume, they are not isothermal, they are adiabatic. Sudden changes are always adiabatic. So, adiabatic bulk modulus you should take.

[00:22:01]Actually, what is velocity of sound in fluids formula? Root K by D. Where K is bulk modulus. But gases will have isothermal bulk modulus and adiabatic bulk modulus. Newton suggested you should take isothermal bulk modulus, which is wrong.

[00:22:19]Uh Laplace corrected it. You should take adiabatic bulk modulus. Adiabatic bulk modulus is gamma P, no?

[00:22:27]That that formula is this.

[00:22:31]Now, he is giving here a problem based on velocity of sound and temperature relation.

[00:22:40]Same gas, no?

[00:22:43]Gamma M same. V proportional to root T.

[00:22:47]Very easy this problem. V2 by V1 is root T2 by T1.

[00:22:55]Root T2 by T1. 0° C means 273 Kelvin.

[00:23:04]This V2 by V1 is two.

[00:23:08]V2 two times of V1 given a double.

[00:23:16]Then squaring and cross multiplying you'll get T2 equal to 1092 Kelvin. From this if you subtract 273, you'll get centigrade temperature 8 9 10.

[00:23:39]And in this velocity of sound in gases based on this relation.

[00:23:47]Another question also this Suppose they gave velocity of sound in helium velocity of sound in helium at a temperature T degree centigrade is equal to velocity of sound in hydrogen suppose they give in hydrogen at 27 degree centigrade then T is dash.

[00:24:35]Helium temperature is asking.

[00:24:39]So velocity of sound in helium at some T degree centigrade velocity of sound in hydrogen at 27 degree centigrade are equal. What is the T is asking.

[00:24:51]This when you do it be careful.

[00:24:55]One gas monoatomic gas given, helium.

[00:24:59]Other gas, diatomic gas given. Gamma also different. Usually one mistake that what students do is gamma also constant they'll take and cancel that.

[00:25:11]Here one is monoatomic, the other is diatomic. Gamma different.

[00:25:16]So, root gamma of helium R temperature of helium by molecular weight of helium is velocity of sound in helium.

[00:25:37]Is equal to root gamma of hydrogen R universal gas constant to temperature of hydrogen by molecular weight of hydrogen. Equate them. Both sides root cancel now when you square it, then R R cancel R R cancel now.

[00:25:57]Then substitute.

[00:25:59]Gamma of helium, monoatomic gas, gamma ratio of specific heats 5 by 3.

[00:26:08]Temperature of helium you want to find helium temperature T by molecular weight of helium.

[00:26:17]Substitute molecular weight of helium.

[00:26:22]4 Gamma of hydrogen, diatomic gas, gamma 7 by 5.

[00:26:29]Temperature of oxygen 27° convert into Kelvin. 300 273 + 27 by molecular weight of hydrogen, 2.

[00:26:43]Simplify this, find T. Like this, don't forget gamma different. Suppose you both the gases are monoatomic or both are diatomic, then gamma cancel both sides.

[00:27:08]Okay, I'm moving into next problem.

[00:27:26]One more Okay, one more. Wait, wait, wait, wait.

[00:27:33]One more I want to tell you in this gases.

[00:27:39]Suppose a problem given like this.

[00:27:42]Velocity of sound in hydrogen at certain temperature and pressure velocity of sound in hydrogen at certain temperature and pressure is V naught.

[00:28:17]The velocity of sound in a mixture of hydrogen and oxygen mixed in 2:1 ratio by volume, their volumes ratio 2:1 is dash at same temperature and pressure?

[00:28:49]Okay, at same temperature pressure in the mixture of hydrogen and oxygen mixed in 2:1 by volume, what is velocity of sound they're asking?

[00:29:02]So, in this V under root gamma P by D if you take Here first gas hydrogen, next mixture of hydrogen and oxygen both diatomic only.

[00:29:23]That means in both cases gamma same, pressure same you gave. So, V inversely proportional to root of density you can write. That means velocity of sound in the mixture by velocity of sound in hydrogen is root of density of hydrogen by density of mixture you can write. So, you have to find the mixture density in terms of hydrogen density.

[00:29:56]So, I'm writing that mixture density.

[00:30:00]Mixture density any density what is density? Mass by volume now.

[00:30:07]Mass by volume. So, in this there is hydrogen and oxygen. Mass of hydrogen, mass of oxygen by volume of hydrogen plus volume of oxygen.

[00:30:26]The density of mixture, density of mixture is equal to mass of hydrogen is a of hydrogen into density of hydrogen.

[00:30:47]Volume of hydrogen in this ratio, how much? Two two is to one. Two given. Two I'm substituting.

[00:30:55]Okay?

[00:30:56]Two into density of hydrogen. This is volume of mass of hydrogen.

[00:31:03]Mass of oxygen is there. Again, volume of oxygen into density of oxygen I have to write.

[00:31:11]Oxygen density one given in the ratio.

[00:31:14]One into oxygen density. One into oxygen density.

[00:31:21]Observe oxygen density. Oxygen density, how many times of hydrogen density? 16 times of hydrogen density.

[00:31:33]By hydrogen volume plus oxygen volume two plus one three.

[00:31:39]So, 18 DH by three you are getting. So, mixture density six times of hydrogen density you are getting. So, here in place of this mixture density, six times of hydrogen density you substitute.

[00:31:56]You'll get velocity of sound in mixture is velocity of sound in oxygen by root six you'll get. So, answer is velocity of sound in mixture is velocity of sound in hydrogen V not by root six. This also important model in this.

[00:32:31]Next.

[00:32:35]Next next question six, the equation of a stationary wave along a stretched string given.

[00:32:47]The separation between adjacent nodes.

[00:32:50]All are in centimeter here given.

[00:32:53]Uh what is stationary wave equation given here? Y equal to five sine pi x by three cos 40 pi t given.

[00:33:10]What is a general form of the stationary wave equation? Y equal to 2a cos kx or sine kx cos omega t like this, no?

[00:33:26]So in place of k, k, what is there here?

[00:33:31]Pi by three is there.

[00:33:34]First you have to find lambda here. K pi by three is there.

[00:33:39]So k means 2 pi by lambda, no?

[00:33:44]It is pi by three given.

[00:33:47]Which implies lambda equal to 6 cm Now what is the distance between two adjacent nodes in a stationary wave?

[00:34:01]What is the distance between two adjacent nodes? Node to node, node to node, lambda by two, no? Lambda six I got. Lambda by two, three.

[00:34:14]Three.

[00:34:33]Next The path difference between two particles of sound wave is 50 cm.

[00:34:48]Path difference 50 cm given between two particles in the wave.

[00:34:56]And the phase difference between them 1.8 pi.

[00:35:02]1.8 pi.

[00:35:07]If the speed of sound V is 340 m per second.

[00:35:15]Find the frequency of wave. Frequency means first giving velocity frequency is asking means first lambda you should find wavelength.

[00:35:25]Phase difference between two particles and path difference relation. 2 pi by lambda into path difference now.

[00:35:37]Lambda is 2 pi by delta into X.

[00:35:42]1.8 1.8 means 9 by 5 now.

[00:35:53]1.8 pi means 9 pi by 5.

[00:35:58]Substitute.

[00:36:00]2 pi 9 pi by 5 X in cm there here X 50 cm you get 50 cm.

[00:36:11]50 cm means half meter now.

[00:36:15]Half meter.

[00:36:18]Find lambda value. 2 2 pi by cancel. 5 by 9 m lambda you will get. Now find frequency n v by lambda.

[00:36:34]V 340 given 340 by 5 by 9.

[00:36:40]Simplify that.

[00:36:55]Next A progressive wave of frequency 500 Hz is traveling with a velocity distance between two points same delta equal to 2 pi by lambda into x is asking x here.

[00:37:22]Path difference is asking phase difference given.

[00:37:27]Okay, path difference 60° means pi by 3.

[00:37:32]2 pi by lambda.

[00:37:37]What is the lambda? V by n.

[00:37:41]V 360 m per second only given SI unit.

[00:37:46]Frequency 500 given.

[00:37:49]So lambda 360 by 500 into x. Pi pi cancel.

[00:38:02]1000 1 by 3 equal to 1000 by 360 into x. Find x.

[00:38:17]0.12 you will get 0.12.

[00:38:28]>> Next.

[00:38:39]Two waves of amplitudes A1, A2 are superimposed.

[00:38:44]The ratio between maximum and minimum intensities of the resultant wave 9 is to 4.

[00:38:51]A2 A1 I max by I minimum What is I max by I minimum?

[00:39:09]formula Intensity proportional to square of amplitude.

[00:39:20]Intensity proportional to square of amplitude now intensity This time question you will get even in uh wave optics Young's experiment.

[00:39:37]There also you'll get.

[00:39:39]I max by I minimum root I1 plus root I2 by root I1 minus root I2 whole square This you'll get now.

[00:39:58]And I proportional to A square means root I proportional to A.

[00:40:05]So, this is A1 A1 plus A2 root I1 in place of root I1 A1 I can write root I2 A2 I can write by A1 minus A2 whole square.

[00:40:25]I max by I minimum 9 by 4 given now 9 by 4.

[00:40:31]So 9 by 4 is A1 plus A2 by A1 minus A2 A2 whole square. That is 3 by 2 equal to A1 plus A2 by A1 minus A2. Cross multiply find A1 by A2. A2 by A1 is asking. See, A2 by A1 is not asking not A1 by A2. Find that.

[00:41:28]Next.

[00:41:34]Next.

[00:41:39]For a wave equation Y equal to 8 sign so and so.

[00:41:44]Find the phase difference in between two particles separated by 2 cm.

[00:41:51]This 2 cm is path difference.

[00:41:57]Path difference.

[00:42:01]He's asking phase difference.

[00:42:07]Lambda you have to find.

[00:42:09]See, this 2 pi inside you take.

[00:42:13]This equation will become y equal to 8 sin 2 pi into.1.2 pi x 2 pi into 2 -4 pi t Now this 0.2 pi must be k So k 2 pi by lambda is 0.2 pi means 2 pi by 10. You will get lambda equal to 10.

[00:42:58]10 10 10 10 cm Here see unit cm given here.

[00:43:05]10 cm you will get.

[00:43:08]So lambda 10 cm this is 2 cm So what you will get 2 pi by 10 into 2 that is 2 pi by 5 radian Here is giving answers in degrees.

[00:43:30]2 by 5 into pi radian means 180 degrees now.

[00:43:36]360 by 5 72 I think.

[00:43:51]72 degrees Next.

[00:44:14]>> Yeah.

[00:44:18]A wire of length 2L is made by joining two wires A and B.

[00:44:24]This is one wire. This is another wire.

[00:44:27]Two different wires they are joined.

[00:44:32]Of same length. Length same for both.

[00:44:34]Length same. But different radii.

[00:44:37]One radius so this A radius so given as R, B radius given as 2R.

[00:44:45]And they are made of same material.

[00:44:48]Means their density same.

[00:44:52]It is vibrating at a frequency such that the joint of the two wires is a node.

[00:44:58]Here node is there. Here here node is there. Maybe here some loops form like this.

[00:45:07]Like this. Here node is there. Joint is a node.

[00:45:13]If the number of antinodes in wire A is P and in B Q, then P is to Q. Number of antinodes. This is antinode now.

[00:45:21]Antinode antinode. Number of antinodes is nothing but number of loops.

[00:45:27]Number of loops ratio is asking.

[00:45:30]So what is the frequency formula? Number of loops segments by 2L into root T by mu.

[00:45:42]This is under root This is P by 2L into under root T by this mass per unit length linear density can be written as area of cross section into density, you know, mass per unit volume, this density row.

[00:46:07]Pi r squared row you can write, now.

[00:46:09]Write like that because radius different there. That we write it in terms of radius.

[00:46:16]Now, see.

[00:46:17]Whenever wave travels from one medium to another medium, frequency constant. So, frequency constant here.

[00:46:26]For both the wires, frequency same. And both are under same tension. Tension same here.

[00:46:34]And length same. So, frequency, tension, length, these three are same for both the wires here. And density also. They are of same material. Density also same.

[00:46:51]So, what are different? Number of loops and radius.

[00:46:57]That is different. Means P by root 1 by r squared. All the remaining things constant, now.

[00:47:05]Root r squared is r. Means P by r. This value is same for both.

[00:47:14]So, equate them. Number of loops in first wire by number of loops in second wire, P1 by P2 must be equal to R1 by R2.

[00:47:29]R1 by R2, r by 2r, 1 by 2, 1:2.

[00:47:46]Okay. Next, go into next problem.

[00:48:20]>> Uh one more point I want to discuss about strings.

[00:48:24]See this question you see logically it is Suppose they give a stretched string is vibrating with some frequency. Suppose 100 Hz.

[00:48:44]100 Hz with some points with some points as anti-nodes. Some points are anti-nodes.

[00:49:02]Frequency 100 Hz, some points on the wire are anti-nodes.

[00:49:08]Now, what do you will ask is what is the next frequency for which the same points again become anti-nodes? Same point. Whichever is anti-nodes now, the same points should be anti-nodes again.

[00:49:22]What is the next frequency at which the same points again become anti-nodes?

[00:49:29]Like this, very important logical problem.

[00:49:32]For example uh that 100 Hz frequency is fundamental frequency, let us assume. Just assumption.

[00:49:40]Okay, he has not given. Some points as anti-nodes 100 Hz he gave. I'm assuming that is fundamental frequency. One single loop will form in fundamental frequency. Then this middle point is antinode, now? So, what is the next two frequency for which again this middle point only antinode is asking that.

[00:50:02]What is next to second harmonic if you take this the same wire of same length if it vibrates in the form of two loops, that is second harmonic.

[00:50:14]This frequency will be 200 Hz, now? Two times two fundamental frequency.

[00:50:20]In this case, this middle point is not antinode, it is node here.

[00:50:28]Next to third harmonic if you go, the same wire under same tension and same length, everything same.

[00:50:36]Vibrates in the form of three loops of stationary wave, now? Three loops like this.

[00:50:44]Third harmonic.

[00:50:46]In this case, frequency is three times two this, now?

[00:50:49]That is 300 Hz, now? Here you see this middle point this middle point which is antinode in first case again becoming antinode.

[00:51:01]Means, at what What is the next two frequency at which the same points again become antinodes? Means, definitely that next two frequency must be The next two frequency must be three times more than given frequency only.

[00:51:30]If this is given as 50 Hz, next frequency 150 Hz. If this is given as 200 Hz, next two frequency at which same points again become antinodes is 600.

[00:51:42]Three times only it is.

[00:51:48]Okay, I'm going to 12.

[00:52:00]Two waves with same amplitude, same frequency.

[00:52:05]First wave amplitude A, second wave amplitude also A.

[00:52:11]Superimposing.

[00:52:12]The ratio of resultant intensity when they arrive in phase to the resultant intensity when they arrive 90° out of phase. Okay. What is uh uh resultant amplitude, resultant intensity?

[00:52:33]That relation, resultant amplitude. What is the resultant amplitude? R I will use R symbol.

[00:52:47]A1 squared plus A2 squared plus 2 A1 A2 cos delta, where delta is the phase difference, no?

[00:53:00]If they arrive in same phase, if they are interfering at same phase, means phase difference zero, cos zero one, no?

[00:53:08]Then resultant amplitude will become A1 plus A2, that is 2 A. A plus A, 2 A.

[00:53:17]Resultant amplitude.

[00:53:19]Resultant intensity, intensity is some constant into amplitude squared, no?

[00:53:25]So, some constant K into 4 A squared.

[00:53:31]That is resultant intensity.

[00:53:36]If they are at 90° phase, when delta is 90°, then cos 90 0 then resultant amplitude how much you will get root a square plus a square that is root 2 a then the resultant intensity k into resultant amplitude square root 2 a square means 2 k a square. This is 4 k a square that is 2 k a square 2 is to 1 2 is to 1 next beats beats beats beats are produced by two sound sources of same amplitude and of nearly equal frequency nearly nearly equal not equal slightly different frequency when sound waves of two slightly different frequencies undergo interference then only you know beats are obtained nearly equal means slightly different their frequencies the maximum intensity of beats will be dash that of one source means one source intensity i other source intensity i the maximum intensity what is maximum intensity root i1 plus root i2 whole square now.

[00:55:42]Root I 1 is root I root I root I root I root I root I 2 root I whole square that is 4 I means maximum intensity is four times of intensity of any one source four times.

[00:56:09]Okay, next next.

[00:56:19]The equation of traveling wave traveling wave progressive wave equation given the ratio of maximum particle velocity to the velocity of wave propagation maximum particle velocity particles make simple harmonic motion in wave.

[00:56:45]Maximum particle velocity A omega and if wave equation is Y equal to A sin omega T plus or minus K X if this is wave equation omega 2 pi n K 2 pi by lambda omega by K equal to if you divide these two you will get n lambda that is velocity of wave.

[00:57:26]So wave velocity omega by K He's asking the ratio of maximum particle velocity to wave velocity.

[00:57:40]These two ratio is asking.

[00:57:44]Means A omega by omega by K is asking. Omega omega cancel. KA A 60 K 6 is there now here KX. In place of K 6 is there.

[00:58:07]This is 60. This is 6.

[00:58:11]And distance unit microns given micrometer.

[00:58:17]Means this amplitude is A. Amplitude is 60 into 10 power minus 6 meter.

[00:58:28]Amplitude micron given now there.

[00:58:33]A K is a 6.

[00:58:37]K is a constant now it is propagation constant of the wave.

[00:58:41]6 it is. So substitute them. Find KA.

[00:59:00]Find find find 16 into 60 60 60 into 10 power minus 6 into K value 6 into 10 power minus 6 micron given now.

[00:59:21]Finally this 5 into 36 into 10 power 11 something like that you are getting.

[00:59:28]Okay. Next uh wave equation two waves undergo interference. This is one wave.

[00:59:49]This is another wave. These two undergoing interference and producing beats. How many beats?

[00:59:58]Number of beats per second.

[01:00:01]And the intensity ratio between waxing and waning mean maximum minimum intensity ratio.

[01:00:10]We have to find what is the beat frequency.

[01:00:20]N1 difference N2 now.

[01:00:24]The difference of the frequencies of the two waves now number of beats per second.

[01:00:30]Find it.

[01:00:31]Omega 1 710 pi 2 pi N1 is 710 pi.

[01:00:47]Pi pi cancel.

[01:00:50]710 by 2 355 hertz you will get.

[01:00:59]Like that the other one is second one 2 pi N2 is 702 pi. Pi pi cancel.

[01:01:10]N2 702 by 2 351 you will get.

[01:01:19]So one wave frequency 355 Other is 351.

[01:01:24]Difference of these two is four.

[01:01:27]So, number of beats four four four.

[01:01:31]A C two options are there.

[01:01:39]Next intensity ratio intensity ratio maximum intensity I max and I minimum I max I minimum their ratio.

[01:01:56]What is maximum amplitude?

[01:01:59]Maximum amplitude A max If these two waves produce constructive interference maximum amplitude you'll get now that is A1 plus A2 4 plus 3 seven 49 is there C option C.

[01:02:22]A minimum minimum amplitude when they undergo destructive interference amplitude minimum now A1 minus A2 4 minus 3 one So, I max by I minimum is A max by A minimum whole square 49 is to one.

[01:02:46]Third option.

[01:03:05]Next Two waves given two waves if the frequency and amplitude of the resultant wave remain equal.

[01:03:28]These two waves have same amplitude.

[01:03:32]The resultant wave amplitude also same is giving.

[01:03:38]When two vectors of same magnitude act at an angle 120 degrees, then only the resultant is also equal to either of them.

[01:03:52]One vector is of magnitude A. Another vector is also of magnitude A. Second vector magnitude also A.

[01:04:02]If these two vectors act at 120 degrees, then the resultant will be also equal to A only. A few days back this case I told you this.

[01:04:15]This is that case only. First wave amplitude A, second also A, the resultant also A means their phase difference is 120.

[01:04:24]Resultant amplitude root A1 squared plus A2 squared plus 2 A1 A2 cos delta.

[01:04:37]This is A.

[01:04:38]This is A. A2 also A. A into A cos 120 minus 1/2.

[01:04:49]Two two cancel you'll get resultant amplitude A only you'll get.

[01:04:53]If it is 120, so their phase difference is 120 degrees.

[01:04:59]2 pi by 3.

[01:05:17]>> Next.

[01:05:21]And this very easy equation of progressive wave given velocity of wave is asking omega omega by K is velocity of wave.

[01:05:32]Uh This is in the progressive wave formula in a sign in brackets you have KX minus omega T omega 400.

[01:05:46]K is 20.

[01:05:49]X and Y meters so directly meters only 20 meter per second easy.

[01:05:57]Next.

[01:06:06]Two sound waves of wavelengths 5 and 6 meter produce 30 beats in 3 seconds.

[01:06:13]Find the velocity.

[01:06:16]Wavelength of the wave lambda V by N N is V by lambda.

[01:06:26]Lambda less means frequency more. First wave frequency more N1 more.

[01:06:35]Here N1 is more.

[01:06:39]So number of beats N1 minus N2 equal to this is number of beats per unit time beat frequency.

[01:06:50]30 beats per 3 seconds means for 1 second 30 by 3.

[01:06:55]10.

[01:06:57]V by lambda 1 lambda 1 5.

[01:07:01]N2 V by lambda 2 6.

[01:07:05]Is 10. From this find we 6 V minus 5 V.

[01:07:15]6 V minus 5 V V 5 into 6 30 V equal to 300 And in this beats I want to tell you one more I want to discuss in this beats. So Suppose they make you a tuning fork and a string are in unison.

[01:08:29]If the tension in string tension in string increased by 4% then they produce five or six some beats they'll give.

[01:08:50]Six I'm giving.

[01:08:55]The frequency of tuning fork is like this they can give beats.

[01:09:07]So, this how to solve this?

[01:09:11]They are in unison initially, mean their frequencies are equal. Initially, tuning fork stretched string have equal frequency.

[01:09:24]So, there is a tuning fork, its frequency n.

[01:09:28]There is a stretched string under a tension t.

[01:09:33]Its frequency also n only, no? Unison means resonance, unison means frequency is equal.

[01:09:41]Now, what they are doing?

[01:09:44]Tuning fork frequency same as it is.

[01:09:48]Tension in this string increased.

[01:09:51]When tension increases, frequency increases in the string.

[01:09:56]Frequency increased by 4%. So, what is it? So, frequency will become n plus delta n.

[01:10:04]Frequency will increase now, n plus delta n after tension increased by 4%.

[01:10:13]So, this n n plus delta n different, no?

[01:10:18]Now, you'll get beats. Their difference, what is n n plus delta n different?

[01:10:23]Delta n.

[01:10:25]That difference only, no? Number of beats, that is given six. Delta n six.

[01:10:32]Now, n is asking. Now, what you do is for a stretched string, n proportional to root t.

[01:10:39]That is t to the power of half.

[01:10:43]Which implies delta n by n into 100 is equal to half into delta t by t into 100.

[01:10:56]This delta t by T into 100 percentage change in tension four given.

[01:11:03]So, substitute four there.

[01:11:06]Delta N by N into 100 is four.

[01:11:14]This delta N is six.

[01:11:18]Sorry, four into half I have to write now. This half half into four.

[01:11:23]This delta N is six. Substitute six here. Find N. Tuning fork frequency you'll get.

[01:11:31]300 you'll get.

[01:11:34]Like this this is also important model in beats problem. Beats stretched string combination.

[01:11:48]Okay.

[01:11:49]Time over.

[01:11:51]All of you students practice more problems in this chapter units and measurements. So, this chapter there are problems you need to practice. All the best.

[01:13:14]>> Mhm.