Small speakers A and B are driven in phase at 725 Hz by the same audio oscillator. Both speakers start out 4.50 m from the listener, but speaker A is slowly moved away (Fig. E16.34)
. (a) At what distance d will the sound from the speakers first produce destructive interference at the listener's location?

Question

Small speakers A and B are driven in phase at 725 Hz by the same audio oscillator. Both speakers start out 4.50 m from the listener, but speaker A is slowly moved away (Fig. E16.34)

Illustration showing two speakers A and B, listener, and distance x for wave interference.

. (a) At what distance d will the sound from the speakers first produce destructive interference at the listener's location?

Accepted Answer

Everyone in this problem. We have a radio with two detached speakers tuned to a radio station playing an alert tone at 1050 hertz. The speakers are initially placed at some 10500.3 point two m away from a listener. One speaker is moved backwards from the listener and were asked what distance X between the two speakers will create destructive interference for the second time at the listeners position. And so we look at our diagram, what we're looking to find is this value X. That distance between those two speakers. Alright, so we're talking about destructive interference here and we'll call the destructive interference is going to occur when the path difference DELTA X is an odd multiple of the wavelength of half, sorry, an odd multiple of half of the wavelength. And so we have the path difference is equal to lambda over two, 3 λ over two five lambda over two etcetera. Okay, so if we want to write this in general we have that. The path difference DELTA X is equal to n. Lambda over two Where N can be 135. Okay, these odd integers. And when we say delta X here, let me write that. This is the path difference. Why do we want that? Well, for destructive interference, if we have the difference in the distance between these two speakers being a multiple, an odd multiple of half the wavelength that when they get to the listener they're going to be out of phase and the waves will kind of cancel each other and there'll be very little sound that's actually hurt. Okay, that's what we have in destructive interference and that's why we want the path difference to be land over 23 lambda over two, et cetera. So that those waves are arriving out of phase now the second time for destructive interference. Okay, that's what we're trying to figure out. The second time we have destructive interference is going to be when n is equal to three the first time and is equal to one the second time and is equal to three. And so in order for this to occur, we want the path difference delta X. Two, B three lambda over two. Alright, so we have this equation that we want to satisfy, that has some information about the wavelength and information of a delta X. Now let's try to write delta X in terms of the value X that we're looking for. So what's that? D. one B. The distance from speaker one to the listener. And we're gonna say that this speaker on the left hand side is speaker one. And the speaker on the right hand side is speaker two. So the distance D one from the first speaker to the listener is going to be X plus 3.2 m. If we do the same for the distance from speaker to We're gonna have the D2 is just equal to 3.2 m. The distance from this right hand speaker to our listeners. So we have d. two is equal to 3.2 m. Now our path difference, DELTA X is going to be the difference between these distances. Okay, how much more does the sound have to travel from one speaker than it does from the other speaker. And so we get the absolute value of D one minus D. Two. Now in this case D one is larger, it's further away, the distance is larger, so this is going to be a positive value. So we don't have to worry about the absolute value, but it's a good thing to include um just in case. And so we have X plus 3.2 m minus 3.2 m. We get the absolute value of X. X is a distance, so it's going to be positive. And so we just get X. And so our path difference, delta X actually equals the distance we're looking for X. And so if we look at this equation, we had that we want to satisfy where the path difference is equal to three lambda over two. So that we have the second time of destructive interference, we can now write this as X is equal to three lambda over two. And so we have that X is equal to 3 λ over two. Now I have a way to solve for this value X. That we want, we just need to find the wavelength lambda. Now we aren't given the wavelength and the problem, but what we are given is the frequency. Now let's recall how can relate wavelength and frequency. Well the wavelength is going to be equal to the speed V. Over the frequency f. We know the speed V. Because we're talking about sound. The speed of sound is a standard value that we can look up in our textbook. And so this is gonna be three halves times 343 m per second, divided by the frequency we were given in the problem of 1050 hertz And we get a value of 0.49. Okay. The unit of per second and hurts will cancel out. And we are left with the unit of meter which is what we want for our distance X. And so the distance X that we were looking for is going to be . m. If you go back up to your answer traces, we see that that is going to be answer choice. E. Okay, so the distance X between the two speakers that we want in order to create destructive interference for the second time is going to be E 0.49 m. Thanks everyone for watching. I hope this video helped see you in the next one

Verified Solution

Key Concepts

Wave Interference

Path Difference

Frequency and Wavelength