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Diarm Jddve rcdnrLntvibratino #uurcVeneralinu MpdCshallov: Trcunhwaleiseconda Tne ~Jvelenqth(henpplesDescribe thc motion of thc source $ and state its frequency 2 m...

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Diarm Jddve rcdnrLntvibratino #uurcVeneralinu MpdCshallov: Trcunhwaleiseconda Tne ~Jvelenqth(henpplesDescribe thc motion of thc source $ and state its frequency 2 mks With relerence t0 (he dagTaI alxve, wrile 'TWO slalernenls describirg the mnolion and frequency of particle. P of thc medium (watcr). nk$Calculate the velocity of the travelling wave.2 mks(iv}Slale T'WO di fTerences between travellirg and standing Waves_mksWilh Telerence l0 Lhe dagTaT alxve, Menlion ONE step thal could be

diarm Jddve rcdnrLnt vibratino #uurc Veneralinu MpdC shallov: Trcunh walei seconda Tne ~Jvelenqth (henpples Describe thc motion of thc source $ and state its frequency 2 mks With relerence t0 (he dagTaI alxve, wrile 'TWO slalernenls describirg the mnolion and frequency of particle. P of thc medium (watcr). nk$ Calculate the velocity of the travelling wave. 2 mks (iv} Slale T'WO di fTerences between travellirg and standing Waves_ mks Wilh Telerence l0 Lhe dagTaT alxve, Menlion ONE step thal could be laken t0 oblain a standing wavc from thc travclling ripples in thc trough mi



Answers

Traveling waves (for example, water waves or electromagnetic waves ) exhibit periodic motion in both time and position. In one dimension (for example, a wave on a string) wave motion is governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}}$$ where $u(x, t)$ is the height or displacement of the wave surface at position $x$ and time $t,$ and $c$ is the constant speed of the wave. Show that the following functions are solutions of the wave equation. $$u(x, t)=5 \cos (2(x+c t))+3 \sin (x-c t)$$

Hello students in this question we have to displacement equations. So excellent. It is equal to eight costs two pipe and VT minus X divided by lambda. And the second equation X. Two is a cause goodbye we plus delta. Sorry this is new. Ok New plus delta numerous player daytime t minus X divided by lambda. Plus delta lambda. Okay now we have to discuss the superposition of their displacement and show that the particular value of the X. The intensity will vary with. Okay, so natural displacement acts will be given by excellent. Plus X. two. Okay, this is the superposition accusation. So hence we can apply the formula of the cause A plus Cosby so we can write that cause A plus cost B is equal to the two course of A plus B. D. Were made to and because of a minus B divided by two. Okay so after substituting and solving this completely we will get that the net displacement occupation X will be given by two amplitude a. and costs of omega t minus K X. This omega is being replaced by two primary player by New and K. It is being replaced by two pi lambda. Okay, omega t minus K X. And minus of minus delta omega. He plus delta came or player by X divided by this too. And because of minus delta omega T plus delta X divided by two. Okay so this is the net displacement accusation. Okay now let us assume this value as five. So we can right here that this will also be five. So the situation will be replaced by X. With respect to time. T. Is equal to a cause omega t minus K. X minus five. And this cause fight. Okay, so this can all this can be rearranged to write to a cost five and cause of omega t minus K. X minus five. No, this is the result in displacement occupation. Okay. Mhm. Okay. No, we can solve for the intensity. So this is the value for the X. With respect to time T. We have five is equal to the minus of omega T. Plus delta X divided by two. Okay now under displacement decoration we have the amplitude of the wave supposed capital A is to a cost five. So this fight is variable with respect to time T. So hence we can write that the intensity I is equal to a constant C manipulated by amplitude squared. So constant simmer player B amplitude that is for a major player by causes square and five. From here is minus of delta omega T. Plus delta X divided by two. Okay, so it means we can say that this intensity is a function of time. So hence intensity depends upon good things upon time. Okay, so this is the answer for this problem. Okay, so thank you

Hello students in this question we have to displacement equations. So excellent. It is equal to eight costs two pipe and VT minus X divided by lambda. And the second equation X. Two is a cause goodbye we plus delta. Sorry this is new. Ok New plus delta numerous player daytime t minus X divided by lambda. Plus delta lambda. Okay now we have to discuss the superposition of their displacement and show that the particular value of the X. The intensity will vary with. Okay, so natural displacement acts will be given by excellent. Plus X. two. Okay, this is the superposition accusation. So hence we can apply the formula of the cause A plus Cosby so we can write that cause A plus cost B is equal to the two course of A plus B. D. Were made to and because of a minus B divided by two. Okay so after substituting and solving this completely we will get that the net displacement occupation X will be given by two amplitude a. and costs of omega t minus K X. This omega is being replaced by two primary player by New and K. It is being replaced by two pi lambda. Okay, omega t minus K X. And minus of minus delta omega. He plus delta came or player by X divided by this too. And because of minus delta omega T plus delta X divided by two. Okay so this is the net displacement accusation. Okay now let us assume this value as five. So we can right here that this will also be five. So the situation will be replaced by X. With respect to time. T. Is equal to a cause omega t minus K. X minus five. And this cause fight. Okay, so this can all this can be rearranged to write to a cost five and cause of omega t minus K. X minus five. No, this is the result in displacement occupation. Okay. Mhm. Okay. No, we can solve for the intensity. So this is the value for the X. With respect to time T. We have five is equal to the minus of omega T. Plus delta X divided by two. Okay now under displacement decoration we have the amplitude of the wave supposed capital A is to a cost five. So this fight is variable with respect to time T. So hence we can write that the intensity I is equal to a constant C manipulated by amplitude squared. So constant simmer player B amplitude that is for a major player by causes square and five. From here is minus of delta omega T. Plus delta X divided by two. Okay, so it means we can say that this intensity is a function of time. So hence intensity depends upon good things upon time. Okay, so this is the answer for this problem. Okay, so thank you

Yeah, I am proven. Here it is given in question. Off propagation. Off speed. Off service B in a region off uniformed depth. Yeah, she squares car two sigma by room 25 i lambda plus G lambda by to fight and off toe by edge of corn Lambda Mhm. You have to fight. Yeah. Dimension less group. See the solution, lambda upon and and upon end and see upon me not oh dr! In stillness them not not Stop glammed up on it. Register at opponent. Sister, see upon me not each are diamonds unless putting this value in give any question he was See we not square is card to Sigma Tau pipe A room named a star l learn duster held upon to pipe, you know, to buy register upon lamb tester Ed. Thank you. Okay, So seize stirred square cabinet And as Sigma upon who a we not x squared. Who? Private London. Not G L upon we not squared. Lambda stir upon to quite paying off to fight edges. Start upon them Gerston So dimensional! This group is G L upon. Do you notice? Script Which is okay. Yeah, Reciprocal! Off the square off the I'm going. I'm doing so Number and sigma upon room and we notice square, Which is the inverse off, however number. Yeah. Yeah, that's all. Thanks for

So rather displacement. Often molecule at the position yeah, position of X and T. To uh um time is given two. Uh It is like the general forum uh is I'm blue tube quite saying K. X minus omega T. Plus some farm to fit fine. And we know that molecule which which has the original position of right X one. Um 2.0. Mhm meters has at some instant time has a displacement. Has a maximum displacement off six. So S. Mhm for X one. Mhm ate some time. T. T. No. That's cool. Colleague is Mhm six. Six point oh The 6.0. Is the maximum displacement. This which means that this is S. M. Okay. This also means that while this is a maximum this function becomes magazine where co sign because maximum uh or one. So maximum value focusing is one. So this this happens when so co sign has this shape. Right right wave like shape. So maximum at one and something like this. And it goes on. Mm hmm. Yeah. So This line is one And the -1. Mhm fine. And it also have has a negative part. So so but let's imagine we are here and let's choose this point to be zero. So Uh the argument of course in is zero. We can choose to pi uh for pipe for simplicity uh Let's choose it. zero. So the condition will be omega three node fi the argument inside the co sign issue. So we have Mhm Yes, we have this condition and then we know that molecule which has uh the original position. It's To go on the 0. 7 m. Uh It is this uh it has a displacement. So 2.0 so At X. two and the same time at the same time moment is 2.0 and we know that all the molecules uh have the intermediate. This Uh displacements so between six and between two. This means that If this is zero we are ah here. Okay. For the second molecule we are somewhere here where the displacement is uh one Third why 1/3? Because this is causing this is not the function of displacement. And we will see just just in a moment why this supposed to be one third when we write the questions. So let's plug these uh numbers. So S. Is too So at 2.0 and then we write this uh General equation. So sm maximum is 6.0 because saying K. X two minus will make a peanut plus fee. Okay. Yeah. Mhm. And this means that mm co sign okay X. To end um Let's look at this term minus omega t. No plus fee from this condition minus omega T. No plus 55 is minus K. X. One. If we move this to the other side. So the second term is minus K. X. One. So we can write minus K X. One is one. So so let's open up a new tub. So we have on third for saying thank you. Okay X. Two. Okay explain. And yeah let's universe of course saying X two minus K X one is causing our cosign. What Earth which is one. What I'm going 23 X 2 -11 is zero point 03 meters. And we can calculate, okay Which is 123 divided by X 2 -11 which is point oh seven. And lee sees 17 point by them 85 meter 21. And we know that. Okay. The way the director is two pi over the wavelengths and wavelength is uh let me write it here, wavelength he's speed of sound, our frequencies and we plug this in here. We get to by Earth frequency oversee. And from here we can find we can find frequency F. Yes, C. K. Or two pi. And we have found Kay and C. Is 343 2nd. We have found K. And the frequency of the camps. Yeah. Nine. And we 59 points 97 hurts. So let me right here. Finally last year nine and 59 0.97 hurts. Mhm


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