in-the-figure-below-two-speakers-mathrms_1-and-mathrms_2-emit-sound-waves-of-wavelength-2-m-in-phase

Question

In the figure below, two speakers, $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$, emit sound waves of wavelength 2 m, in phase with each other.
IMAGE IS NOT AVAILABLE TO COPY
Let $A_{p}$ be the amplitude of the resulting wave at Point $P,$ and $A_{Q}$ the amplitude of the resultant wave at Point $Q .$ How does $A_{p}$ compare to $A_{Q} ?$
(A) $A_{\mathrm{P}} < A_{\mathrm{Q}}$
(B) $A_{\mathrm{P}}=A_{\mathrm{Q}}$
(C) $A_{\mathrm{P}} > A_{\mathrm{Q}}$
(D) $A_{\mathrm{P}} < \mathrm{o}, A_{\mathrm{Q}}>0$

Linda Winkler · Accepted Answer

Yeah, in this example we are going to look at a situation in which waves are interfering in space. Um And although the example works with sound waves, the idea is the same, no matter what type of wave you have is that if waves are in phase when they get together at a point. Um So we&#x27;re looking at waves that have the same wavelength. Uh Sometimes those are called monochromatic late waves, especially if it&#x27;s light, but they have to have the same wavelength in order to get a coherent pattern. Um But what governs how they interfere is a path length Difference between two waves. So two waves that come together in space um Wave one, wave two, if they start out in phase and they stay in phase, you get what&#x27;s called constructive interference if they get out of phase um with uh huh half of a wave length difference between them, so that the crest of one wave adds to the trough of another, You get 180° out of phase. And that&#x27;s called destructive interference. And the condition for these two situations if they both start out together um from say the same source or two sources that have been carefully uh programmed to be in phase. Um The condition for constructive interference is a difference in path length in space equal to an integer number of full wavelengths. So and could be zero. Starting out at the source together. Could be 1, 2 et cetera. But they have to travel through space to get together again with an integer multiple of wavelengths difference in how they got there. Difference in path length for destructive interference, on the other hand, is an integer multiple of half wavelengths. So sometimes this is written as n plus one lambda over two. Where and again can be zero um two for etcetera. So if you you put in those even ends, um different ways to write it. But um if you put in even ends, uh then you wind up with uh say one half lambda path difference or three halfs lambda et cetera. Um So what matters is the difference in path link between two waves. So let&#x27;s take as an example, let&#x27;s pose that you have to sound generators will calm S. One and S. Two. And let&#x27;s say they have been carefully programmed to send out monochromatic sound waves. Both have λ Equal two m. I don&#x27;t know offhand whether those are something that you could hear, what you have to work that out, but they&#x27;re separated by three m. Down the middle is 1.5 m. But they&#x27;re full separation is three m. And you have um a sound detector that you can move around, but we&#x27;ll pretend that they also start out programmed in phase with each other at the source. So they have been fiddled with so that they&#x27;re putting out nice pretty sound waves do. Um I don&#x27;t know that&#x27;s the frequency. But anyway they&#x27;re putting out nice um Pretty sound waves that oscillate with a clean amplitude and period and wavelength. Um And we&#x27;ve got a detector that we could put at either point P. Or point Q. And point P. Is four m away. So here&#x27;s our detector four m away from source one. Um But it&#x27;s five m away that right triangle situation five m away from source to and we could ask ourselves what would we hear at point P. So that&#x27;s the question. Um And what we would have happening there is um Source one has traveled a path length of two full wavelengths. Two times two m is four m. Source too will have traveled 2.5 wavelengths Over that distance of five m, so five m divided by two m, gives us 2.5 way and this and notice that those are one half of a wavelength now out of phase. Um So they are one half lambda out of phase. And that means the path links have allowed them to shift into a situation more like the second one that I showed, and so they will actually cancel each other out with destructive interference. Um Meanwhile, if you moved your detector down to the central position here, um Each of those would have traveled basically five m, Well not quite five m, but um and I&#x27;m not even sure we need to figure that out with a pythagorean theorem. But what we see in um detector Q is that the two sources have traveled identical path lengths, identical paths through space, same length, um which means that they come together in constructive interference. And so you&#x27;d hear loud sound at point Q, and you would hear a quiet sound at point P. Or the amplitude of the resulting sound wave at Q should be much much greater than the amplitude at point P. Yeah.

Ze-Han Lee · Answer

So in this problem we have two speakers s one that&#x27;s too. And their separation is ah, three meters and the Woodlands that they admitted as well. I mean, the way that they emitted have women&#x27;s alumni quit two meters. So we have two interesting points. Ah, why is over here the otherwise here, That is a point p disappoint cube and distance off this up on p. So this this distance is four meters, four meters and ah, the punk. You is on the ihsaa on the central line and it is Ah, the horizontal distance is also four meters. So we want to know the ah attitude off the wave at one p and point you so because, uh, that this has been a p and this one&#x27;s four meters. So, uh, the wave shape emitted by asked one would be something like this. So this is Ah, this location zero. I say this, that&#x27;s one okay. And because peas full meters away and this is only two meters, all right, because this is a this is a whole period. So we&#x27;re doing that. We need to get another period, so this will be the point p as soon as you can see that. Ah, the wave. Ah, admitted by ass one point p is the largest has the largest vibration. Right? So what about asked to so separation between past two and a cute and one point piers Ah ah! Square root three meters squared plus four meters squid, which is five meters ray. So now we look at it as too. It will be something different. Excuse me. So, uh, let&#x27;s say we draw another one. So this is as two at origin and we have also so there&#x27;s one period to period. So there&#x27;s the two meters. This is full meters and there&#x27;s the five meters. All right, so because the separation pizza as two and a piece of five meters. So the wave mated by us two at point P is at the lowest point, So Ah, this is at the highest point. Is that Lois point saying that says this. Do we cancel each other? Okay, so find that it amplitude at ah point p zero. So a p zero, But it is different at upon Q because the separation between s one Q is screwed 1.5 square two plus four square, which is about full 0.27 meters. And if you look at the full 0.27 it&#x27;s a somehow look at it over here. Hurry. So no matter what it is, no matter where looking at its one on OS to ah, the vibrations, they&#x27;re both positive. So a queue in this says is positive, so a Q is larger than a P.

Donald Albin · Answer

all right, I&#x27;ve drawn a diagram of this situation. Noticed that I have two speakers that are four meters apart. That&#x27;s because they go from y equals negative two to y equals positive to, um there&#x27;s also three meters between the speakers and the points in question. Um, I drew our sub A which would be and I would I would rather use a capital A here the distance two point a and notice that that is. And you know what? I&#x27;m gonna also write down one and two, so I&#x27;m gonna call these points 1.2 the distance from 0.12 point a is D squared plus R one squared. Where are one is three OK, the distance from part 2.2 two point a is also D squared plus R one squared and so the distances are the same. So at point A. When we write down that the amplitude is to a co sign of 1/2 delta theta where Delta Theta is two pi dollar are over Lambda plus Delta Phi subzero. Um, the phase shift is zero and delta are is also zero. And the co sign of zero therefore is is one. And so the amplitude not point a so I don&#x27;t write that there. I&#x27;m gonna write amplitude appoint a maybe already over here. Yeah, is to a all right, the amplitude of point B. So if we&#x27;re gonna do point B, it is half a meter upwards. So that means that the Y value is going to be closer to point to, and it&#x27;s gonna be farther away from 0.0.1. So are from two looks to to be is going to be the square root Ah, de minus 1/2 squared plus R one squared R one b is going to be the square root of D plus 1/2 plus R one squared. So Delta R is going to be are to be actually going to be our one b minus are to be because our one B is gonna be lot greater than are to be okay. No, Del, if I it&#x27;s going to be two pi r one b minus are to be over lambda. Okay, so the amplitude is going to be too times a times the co sign of half of Delta Phi. We&#x27;ll have a dollar. Five is just gonna be pi over a Lambda Times R one B minus are to be okay so I can put this part in a graphing calculator or in a calculator and ah, okay, calculate when it comes up with, but I want to do something else here for C. All I have to dio is multiply Put a multiply by two in here two times and ah, two times So for C, I just need to do to times for D. I just need to do three times for E. Do we have an E for E? I just have to do four times away because let&#x27;s say for you four times 1/2 is to so if I just count this off 1/2 1 1.5 to so I can make this easier. But I just putting that factor into the calculator, and I&#x27;m gonna use desmond&#x27;s dot com. So I&#x27;m gonna tape in two times the co sign. Oh, hi, Overland us. I&#x27;m just gonna put in pie over lambda Lambda was too. Yep. Times R one b minus are to be, um, are one b is going to be this square root Oh, d plus 1/2. But D is four plus 1/2 but I want to take a number and times 1/2 okay. And then I want to square it. Plus R one squared are one is three. And now I want Teoh Tract off. There&#x27;s gonna copy Would have already typed and change it to a minus end times 1/2 And I&#x27;m going Teoh in Dismas dot com You cannot a slider for end when n equals one I get I&#x27;m gonna write. This is a sabih 0.6 to Okay, I&#x27;m gonna change my slider so that now it equals, um to a sub sea. I&#x27;m getting negative. Although I don&#x27;t think that is particularly meaningful. One point six time The amplitude. If I make a equals 23 I get 1.7 and if I said a equal to four, I get zero point 32 times the amplitude. Well, not sure what just happened there, but it&#x27;s good backup. Now I&#x27;m thinking that if I said a I mean any equal to zero in this equation. So I&#x27;m saying and times that I said N equals zero, it&#x27;s go back. I get to which is exactly what I wanted to get right there. So, um, I&#x27;m finished. Those are the answers. But let me just be very clear about what I did into Dismas dot com. I entered to co sign of pi over two times the square root of four plus in times 1/2 squared plus three squared minus the square root. And this is basically the same thing only with a minus in there for minus and times 1/2 squared plus three squared and equals. Put in a slider for nd equals 0123 four or five points. Yep, 01234 And that gave me the answers. So all I had to do was put an end times this end times 0.5 into both sped up.

Jayashree Behera · Answer

so in order to be out of phase or and thus result in destructive interference. If the super pose the bought the friends, let&#x27;s call that they&#x27;re lambda should be equal to land over to our some old integral multiple off land overto like tree Lambda overdose, which is on ord individual multiple. Similarly five Lambda Phi Tau, several Lambda I do and so on. I&#x27;m here. We see their path. Difference is l So they&#x27;re lander is actually and therefore the least possibility We must have l. Toby. The smallest value off this city&#x27;s the smallest passage, the smallest value for the bath difference that this land Abi Landover And then for the next possibility, we must have l Toby the next large large value. So that should be sorry the next smallest value after Landover too. And that this tree, Lando. And from here, we can find Q, which is L over lambda. I&#x27;m loving this value over here. They get cute, Toby three overto on solving that we can&#x27;t kill Toby 1.5

Dading Chen · Answer

Okay, So the question was asking us to create Uthappa re presentation for such question. The first hype over representation we need to create his way from re presentation. So from question, we know that tool source waves are two meter apart and this is the first representation I created, which is way from representation. And you see, here there&#x27;s a line which is are the you can see the points intersects when two crests met each other and this will produce and constructive interference along this line because it&#x27;s crass. Me crashed and at San tidal waves. So you only anything inside of cancel out. So yeah, so this is constructive interference on this lie and for question, be willing to create ah graphical representation which this pressure were just this in type graph. And here this is the first source and I create the graph of the first sores and you see, the pressure is on why access and the distance on the X axis and here and this is the wave graph for the first source. An aptitude for the first horse is a And for the second graph is a second graf. The graph for the second source and the Y axis pressure as well in the units, new tongue over me to square. And the access is just the position yourself. So this is the wave graph, but a second source those things that you face. So when any together you create, constructed, interferes and when you create a constructing universities, especially in this case, is a perfect strong interferences that perfectly in face, their amplitude double jeans, this a plus a plus a so the crashed and crashed from these two different sources. Edyta, which is a plus a and give us to a and this is the final graph when they are in face, when they are interested, was each other and the weapons do the same but and the MP to change while in my opinion, I think the first graph, which is the way from presentation who give us a better view as now, is a vital view to see the the relationship between these two resources. Thank you

Kai Chen · Answer

So, Mrs Question, we have intra office of two different, uh, wastes that has a face difference when it&#x27;s first rate out. The general form of these type of questions, if we have one way set, is why am sign? OK, X minus on me got t. And then we had another waves that is completely the same except for face depressed close fight. Then what we can do is that we can synch up sis ISS Cy X minus omega T plus flyover to minus for over two. It says its side K X minus omega T plus five over too, plus file over to why? Because, um then we can use and go addition formula for the sine function. So this one if we cause this are for and we cause this beta and if signori because his heart. But because this beta, this is equal to sign. Arfa Ko Sam Baker minus Cosette in Marfa Sign beta. Plus this one we have signed our first sign, baby, because I beta close call sign our first sign later. So these two cancel and we&#x27;re left with two times. Sign Arfa co sign Beta in this case call sign later is fly over to inside our fires. Kate, explain this. Omega T plus I work too. So if you put back the number we have two times by over two isn&#x27;t sign. Okay, X minus omega T plus five over two. So this is what happens when we add up to, um said you slightly Christmas with his face difference. In this case, we have a face difference off pie over too. So we have five equals tiny over, too. And we&#x27;re looking for is implicated over the result. So it was empty Tuna presented with silence. There should be on why I am over here for everything. So new amplitude here. Is that you for this new functions? This is a sign part is chaos minus omega T plus something shift. This is Jose two times call sign. If I over two times why am is the new amplitude? Okay, so here we want to find this new amplitude in terms of why em&#x27;s and that is just two times consigned priority. So there&#x27;s two times co signed by over two is pi over four, and this equal to two times co signed 45 degree is square to over two. So this is equal to square two, which is one point for one. So then your aptitude is 1.41 times original Aptitude white. Yeah.