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(1I) Two piano strings are supposed to be vibrati…

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Problem 58 Hard Difficulty

(II) Two loudspeakers are placed 3.00 $\mathrm{m}$ apart, as shown in
Fig. $37 .$ They emit $494-\mathrm{Hz}$ sounds, in phase. A microphone is placed 3.20 $\mathrm{m}$ distant from a point midway between the two speakers, where an intensity maximum is recorded. (a) How far must the microphone be moved to the right to find the first intensity minimum?
(b) Suppose the speakers are reconnected so that the $494-H z$ sounds they emit are exactly out of phase. At what maximum are the intensity maximum and minimum now?

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 16

Sound

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Video Transcript

speakers placed three meters apart as shown and figure 37. They emit a 494 hertz sound. In phase, A microphone is placed 3.20 meter distance from between the two speakers where an intensity maximum is recorded. How far must the microphone be moved to the right to the first intensity minimum for part A and B supposed to. Speakers are reconnected to that 494 hearts set on, they admit, or exactly out of face. I want maximum Are the intensity maximum and minimum now? Okay, So I wrote down here what we were given. I wrote that the distance between the speakers and this is according to the figure that I also drew D is three meters. The frequency is 494 hurts the microphone a distance l is 3.20 meters. Also wrote that we have the speed of sound in the medium visa mess eyes 343 meters per second. Okay. And so also, if you look at the diagram, I drew the distance s one s two, which is the distance from the speakers which I wrote Drew and read to the microphone, which is a distance. X X is what we're trying to find to the right of the midpoint. So that microphone I represent is green is a distance X away from the midpoint. Okay, that's what we're gonna try to find. Well, um, the microphone must be moved to the right until the difference and distance from the two sources is exactly 1/2 wavelength. Okay, so I wrote down what s to an s one are equal to okay. Using Pythagorean theorem es tu is equal to this as two squared is equal to 1/2 d plus X squared us elsewhere. And then, um s one squared is equal to 1/2 D minus X squared. Plus elsewhere this the length of the two sides makeup the size of Earth. I've got news for that. Okay. And so we know then that the difference must be between these two. Must be 1/2 the wavelength. So let's go ahead and write that out. This is part s are gonna start working part right here. So s too minus s woman equals 1/2 lambda. Okay, well, Landa is equal to Visa Bess divided by frequency. So we're gonna be able to calculate land out when mean beaks. We know both of those values, and we know everything in s one of us to accept for X. Let's go ahead and, uh, plug in these values and then solve for X and exit since exited over trying to find. So we have here. Now, let's go and write this down here. Let's draw a line and the separates out working out a from everything else that we were given. So we have the square root Mrs s too, so as to is equal to the square root of 1/2 D plus x all quantity squared plus l squared. Great. Plus, I'm sorry. Minus s one. Which is the screw of 1/2 D minus X squared plus l squared. Extend the square root cause that includes all of that, that's gonna be equal to 1/2 believing. Okay, well, again, we're trying to solve for X, So let's go ahead and square both sides. Another money. Move the value, which represents s one to the right side of the equator. Besides, and then, uh, I'm sorry. I'm gonna move that value to the right side of the equation. So I'm gonna move all of this here to the right side of the equation, and then I'm gonna square both sides of the equation. So let's work that out on the next page. So that comes out to be 1/2 d plus x square. Listen elsewhere. Because the square root goes away. Who's three Draw that? It's gonna be equal to 1/4 Landis squared, Clint. 1/2 lambda get squared. Plus two times 1/2 lambda oh, times the square root of 1/2 D minus X square, plus l scream. Oh, could extend that route. Okay, Plus. And then now that square roots gonna be squared. So we're gonna have 1/2. Do you mind a sex close elsewhere with square root is gone. Okay, well, what do we notice? First thing we notice is there's a couple of things we can get rid of that cancel on both sides of the equation. That's going that's gone. Okay, so with those gone, another thing that we can do is, um Now, go ahead and square both sides once again gone. We can go ahead and square both sides once again, and, um, we will have to d. I'm sorry. Uh, we're not gonna square both sides, But now that we've cancelled that, we're going to go ahead and take the square of this value here. We're gonna take the square, that value, so we have to d x minus 1/4. Well, I'm just weird is equal to Lambda, right? Because this here, these two twos cancel that cancels with that. Right? And you have one for Lambda Times, the square root of one Cass D my squared plus l squared. Okay. And the reason that you're only left with the two D X is because if you take the square of 1/2 D squared plus X squared and you take the square of 1/2 a T minus X squared, the only thing that is left is the D x on the one side or the negative DX on one side of the DX on the other. So you at that DX over and you get a two d X. Everything else will cancel because 1/2 D square will be on both sides of the equation. X squared will be on both sides of the equation. So now that we have here. We can go ahead and simplify this even further by. Go ahead. And now we're gonna square both sides. So if we square both sides, we have four dx squared. The reason we're spring. Besides, we want to get rid of that square root for G squared X squared minus four dx times 1/4 Landis squared. Okay. Plus 1/16 lam just the fourth. Then the other side of the equation was gonna times 1/2. That's where it's gone When half t minus X weird plus elsewhere box. I didn't know. That's good, right? That's that's illegible. Go box ended. Okay, well, cancels here. Well, these force cancel. Okay, so we can go ahead and rewrite this again as four dx squared. Excuse me for D squared X squared minus D X squared. I'm sorry. Minus d x lambda squared because the force cancel. Plus 1/16 claimed a squared or I landed to the fourth is equal to all right now, let's go ahead and carry out the square inside here. So this value here, let's go ahead and carry this square out. We're gonna have, um, 1/4 and we're also multiply that lamb. This great through, so I'm gonna 1/4 eastward limbs weird. Minus D X land a square plus x squared. Lambda squared. Okay. Plus lame. This weird elsewhere. Okay, well, we have some more that cancels here. This cancels with this, so we further simplified it a little bit. All right, Now, let's go ahead and move the X over to the same side of the equation. So if we do that, we're gonna have X squared D squared and by force. Let's not forget that four out front minus X squared. Blame the square. Okay? And that's on the one side of the equation. Well, let's go ahead and even make this easier. Let's pull that X squared out for so that way X squared is all by itself. So it's full that excrement out front. And that gets rid of the two experts in the middle or, uh, in the parentheses that's gone. That's gone. Okay. And then on the right, sir. And we're left with 1/4 de swirled Lambda Square, plus lame squared. Elsewhere, minus 1/16 landed a source. Okay, well, now we can pull the Lambda squared. We can pull the Lambert. Square it out So if you do that now, that's common in all of them. So go land this crate out front and then this is gone. This is gone, and this becomes a squared. Okay, well, then we could also isolate the X by dividing both sides by four D squared, minus lambda squared and then taking the square root. So let's go to a new page. We'll go ahead and do that and we find that X is equal to Lambda Times, the square root of 1/4 D squared Plus elsewhere, minus 1 16 Blameless squared. Okay. Oh, standing square root cause that includes all that. And then in the denominator, we have four d squared, minus lame. The square. Oh, okay. And if we remember that lambda is equal to the speed of sound divided by the frequency and we plug in all the values Girl, go back and show you. So lam does the speed of sound times the frequency which is written here. So you plug that in and then you plug in all the values for the speed of sound. The frequency the length l in the distance D in you will find that. Let's go back to the other. Page 84 You'll find the X is equal to 0.411 meters box that in another solution apart, eh? Now, for part B sold to see which this is part B for part B, we are asked you back to the question. Suppose the speaker's air reconnected so that the 49 hertz sounds they emit are exactly out of phase. Of what maximum are the intensities maximum and minimum now. Okay. Well, um, according to part A when the speakers are exactly out of phase, the maximum minimum will be interchanged. And thus the intensity maximum are 4.11 meters. Still left or right. Mid point which we found a part Ay. So here. Let's write that it's me. Um when eyes Max, this is that X is equal to plus or minus 4.11 I'm sorry. Point for women 0.411 meters to the left or right of the mid point. Okay. Well, then, when eyes minimum then could be act the midpoint Well, at the midpoint we have defined. This is the point of X is equal to zero, so X equals zero or this is X equals zero or the midpoint. Write this up or committed weaken box this in to part B.

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Physics for Scientists and Engineers with Modern Physics

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Watch More Solved Questions in Chapter 16

Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Farnaz Mohseni

Zachary Mitchell

Aspen Fenzl

Meghan Miholics

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Christina Krawiec

Andy Chen

Marshall Styczinski

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Farnaz Mohseni

Zachary Mitchell

Aspen Fenzl

Meghan Miholics

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics