Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone arranged as shown in

Fig. E16.39. For what frequencies does their sound at the speakers produce (a) constructive interference

Video transcript

Hey everyone in this problem we have a 75" led TV that has left and right stereo speakers separated by 1.5 m. The speakers which are run by the same sound card inside the tv are beeping the same notification tone. The microphone on a recording smartphone is placed perpendicular and 0.8 m directly in front of one of the speakers. And were asked to determine all frequencies between 20 hertz and 1.5 kilohertz that cause constructive interference at the location of the microphone. Alright, so when we're talking about constructive interference, okay, we know that there's a condition on the path difference. D Okay, constructive interference is going to occur when the path difference is a integer multiple of the wavelength. Okay so we're gonna get back to that but let's start we need to know the path difference. Okay so we need to know the distance between the left speaker and the microphone and the right speaker and the microphone. Okay so let's start with the left speaker. Okay so we're gonna say D. L. Is gonna be the distance from the left speaker to the microphone which is in that smartphone. Okay Now the distance between the left speaker and this smartphone is just that vertical distance. 0.8 m. Okay so D. L. is going to be 0.8 m. Okay? And then D. R. Well this is gonna be the distance from the right speaker to the microphone. Now gr it's a little bit um less straightforward than dl um but it's nothing that we can't handle. Okay we're just gonna have to do some triangle math. So we have our left speaker here. We have our right speaker here. We have our mike here. Okay, Now we know the distance between this left speaker and the mic is 0. m. We know the distance between the two speakers is 1.5 m. And the distance r from the right speaker to the mic is going to be this hypotenuse. Okay, so we have a right angle triangle, we're trying to find the hypotenuse. So let's use pythagorean theorem. D R squared is going to be equal to 1.5 m squared plus 0.8 m squared. Okay, this means that D. R squared is going to be equal to 2.25 m squared plus 0.64 m squared. Ok, add we take the square root, we get that D. R. It's plus or minus 1.7 m. Okay? And we're just going to take the magnitude. We just want to know that distance. Okay, so we got 1.7 m. Alright, so our distance from the left speaker is .8 m from the right speaker is 1.7 m and we want to know the difference between those two distances. Okay, how much further does one speaker or does one? Yeah, this one speaker have to admit it's sound than the other. Okay, so the path difference. We're going to call D and we're just going to take the speaker that's further away the right speaker. Okay, And subtract the distance of the closer speaker. D r minus D L. We at 1.7 m -0.8 m, which gives us a path difference of 0.9 m. Okay, Why is this path difference important? Okay. And I mentioned this earlier and we're coming back to it now. Well, we want constructive interference. This means that we want the speakers the waves to be in phase so that the sound kind of adds on top of each other. Okay? Instead of it being out of phase with destructive interference where the sounds kind of cancel each other out. So with constructive interference, this is going to occur. Okay, When the difference in those distances, So that path difference D is an integer multiple of the wavelength. Okay, So it's either zero lambda two lambda dot dot dot. Okay. If that's the case, then those waves are in phase and we get constructive interference like we want. Okay. Alright, So we have these conditions on the distance, the path difference D that we found. Okay, now we don't have information about lambda, but let's recall that lambda is related to the speed and the frequency through V over F. So, if we replace lambda with V over F, this means that we are distance D. We want to be either zero V over F. Two V over F dot dot dot. Okay. We want it to be an integer multiple of V over F. All right. So what does this tell us? Okay, well, in general we can write this as N times V over F. Okay. Where N is an integer? 012 dot dot dot. Okay. A positive integer. All right. So if we're looking at possible frequencies then are possible frequencies for constructive interference are going to be given by F. We're gonna say subscript N is equal to N. V over D. D is equal to envy over F. And if we rearrange to isolate F we get F is equal to envy over D. Alright, well, we know the speed because we're talking about sound. So we have the speed of sound. Okay. Is 343 m per second. Approximately Our path difference D We calculated to be 0.9 m. Mhm And this is going to give us possible frequencies F n of n Times 381. repeated. Okay. The unit of meter will divide out and we're left with hurts. Alright. So the possible frequencies where we have constructive interference are given by this equation here where we have N is equal to 012 dot. Now an equal zero doesn't necessarily make sense in this case because we know that our path difference D is not equal to zero. Okay, So we have end starting at one. Okay. And we want to find all of the frequencies where this occurs between 20 hertz and 1.5 kilohertz. Okay. And now when we're talking 1.5 kilohertz this is going to be 1. kilohertz. Okay. Times 1000 hertz per kilohertz. Which gives us 1500 hertz. Okay, so we're looking for all of the frequencies between 20 hertz and 1500 hertz that satisfy the equation we've written down here. Okay. All right so let's start with N equals one. Okay that gives us F one is just equal to 381.11 hurts. Okay this is in our range. Okay. Alright. The range here we want 20 hertz To 1500 Hz. Okay, just to remind ourselves so this is in our range. Okay so we're going to include this one now if we go to F. Two. Okay this is gonna be two times 381 hertz. Which is going to give us 762 . Hz. Now 762 is between 20 and 1500. So this is also in our range. So we're going to include that one. We go to our next end value F three. We get three times 381.11 Hz. Which gives us a value of 1,143. Hz. And those should be repeated. Okay this is also between 20 and 1500 hertz. So we include that one. Okay We go to our next and value of four, we got four times 381.11 hertz. Which gives us an f. four value of 1,524.44 repeated hurts. Okay now this one is bigger than hertz. Okay so this one we're not going to include. Okay we're gonna cross that out. We're not going to include that. It's not in the range we were looking for. And you'll notice that because we have end times 381 every time we have a bigger end our frequency is going to get bigger. Okay? So if F four is too big than every frequency after that, it's also going to be too big. Okay so we have the following three frequencies that can cause constructive interference in the range we were looking for. So we have 381 hurts. Approximately. 762 hertz approximately. And 1143 hertz approximately. And that's going to give us answer. C That's it for this one. I hope this video helped. Thanks everyone for watching. See you in the next one.