Two small stereo speakers A and B that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point P starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker B
(Fig. E35.1). For what values of x will the sound this person hears be (b) cancelled? Limit your solution to the cases where x … 1.50 m

Question

Two small stereo speakers A and B that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point P starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker B
(Fig. E35.1). For what values of x will the sound this person hears be (b) cancelled? Limit your solution to the cases where x … 1.50 m

Diagram showing two speakers S1 and S2, distance d, and point P for wave interference analysis.

(Fig. E35.1). For what values of x will the sound this person hears be (b) cancelled? Limit your solution to the cases where x … 1.50 m

Accepted Answer

Everyone in this problem, we have to radio speakers separated by two m. The speakers are playing a note of wave length 68.6 cm that spreads evenly in all directions. A drone carrying a microphone orbits one of the speakers at a radius of 2.2 m and were asked to determine all values of D less than or equal to 2.2 m measured from the other speaker for which the microphone will not detect any sound. Alright, we're given a diagram here in the problem. So we have speaker one okay separated by a distance D to the microphone that's detecting the sound Speaker two is separated by a distance of 2.2 m to that same microphone detecting the sound. And we're told that Speaker one and Speaker two are separated by two m. So this distance D is what we are looking to find some possible values for D. The answer choices were given where you have five choices A through E and each of them have a list of possible values for D. Those options are in meters. So let's think about what we're given in the problem. We're given a wavelength lambda of 68.6 centimeters. Now, our answer choices are in meters. So before we even get started, let's go ahead and convert this to meters. We have centimeters. We're going to take that 68.6 cm. We're going to multiply by one m per cm. The unit of centimeter will divide out and we're going to be left with meters and we get 0.686 m. Okay. So essentially we're dividing by 100 there. Now we want the values of D where this microphone detects no sound Okay. Now, the speakers are playing the same wavelength that wavelength we've written here in m, m. So in order to not detect the sound, what we want is destructive interference. Now, why do we want destructive interference? What we call the destructive interference occurs when the path difference in this case, we'll call it delta X is equal to land over two, 3 λ over two five lambda over two dot dot dot Okay. So the path difference is an integer, multiple and odd integer, multiple of half of the wavelength. And we can write this in general Delta X is equal to end time 2 λ over two for N A positive odd number 1357, etcetera. So what is this telling us is the difference in the distance from speaker, one to the microphone to the distance from speaker to, to the microphone is going to be this odd multiple of half the wavelength. That means that when those two sounds from the two different speakers get to the microphone, they're gonna be out of face because they're separated by half a wavelength. If they're out of phase, they're the same wavelength, then that means we're going to detect no sound, those two waves are going to cancel each other up. Okay. So that's why we want destructive interference so that we don't detect sound, we know that conditions to have destructive interference and we want to figure out the values of D. So let's see if we can get D into this equation in order to solve. Now, we know that D is related to the path length. So the path difference delta X, it's just going to be the difference from the distance from speaker one to the microphone minus the distance from speaker to, to the microphone. And we're just going to take the absolute value there and we're just worried about the magnitude of the difference. This is gonna be equal to the absolute value of D, which is a distance from speaker one to the microphone at point P -2.2 m, the distance from speaker to, to that microphone. Now, we're told in the problem, we want the D values less than or equal to 2.2. Okay. That means that D -2.2 is going to be negative. So we take the absolute value this is going to become 2.2 m minus D okay because we want the positive value and D is going to be less than 2. m. Alright. So we have our path difference now written in terms of D so we can write out the condition on the wavelength recall that we had delta X is equal to and Times λ over two Delta X. a path difference is 2.2 m -D. You ever end value we're going to leave it as and for now λ, the wavelength which we found to be 0. Sorry, 0.6, 86 m divided by two. Alright. If we simplify on the right hand side, we leave the left hand side alone, we have 2.2 m minus D is equal to N multiplied by 0.343 m. And we want to isolate for D we're trying to solve for D. So we get that D is equal to 2.2 m minus N times 0. m. And just to reiterate we have N values 1357, etcetera, those positive odd integers. Okay. So we have this expression for D we want to solve for D let's start with our first, the first end value we can have is N equals one. If M is equal to one, then we get that the distance D is equal to 2. m minus one times 0.343 m, which gives us a distance D of 1.857 meters. now, that's less than 2.2 m. And so that is going to be one of the values, that's part of the answer. Now, we can move to our next end value. It's going to be and is equal to three. If N is equal to three, the distance D is equal to 2.2 m minus two. Whoops, not 23 minus three. The end value times 0.343 m which gives us a distance D of 1. m. Again, that's less than 2.2 m. It satisfies the condition that we have no sound okay because we have destructive interference. So that is going to be one of the values for the solution. We can move to the next and value in this case and is equal to five D is equal to 2.2 m -5 times 0.3, 4, 3 m. This gives us a value of 0.4, Leaders. Again, less than 2.2 satisfies the condition. That's a solution. We're gonna keep going, we go to that next end value. It's going to be seven and is equal to seven to get the distance D is going to be 2.2 m minus seven times 0.343 m. This gives us a value Of -0.201 m. Now, this, when we look at it, it's less than 2.2 m, but it doesn't make sense. We know that D is a distance, we can't have a negative distance between two points. And so this is not a solution. And if we keep going up with and values, this is only going to get more and more negative. So any further choice of N is not going to lead to a solution. This is not a solution. So we have the three values that we found with N equals 13 and five, we go back up to our answer choices, we can compare what we found and we found that the value of D Less than 2.2 m for which the microphone will not detect any sound. Approximately 1.86 m, 1.17 m and 0.4, nine m that corresponds with answer choice. E thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Wave Interference

Path Difference

Wavelength