Two loudspeakers, A and B (Fig. E16.35)

Question

, are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. What is the lowest frequency for which (a) constructive interference occurs at point Q

Accepted Answer

Hey everyone in this problem, we have two identical alarm speakers connected to the same circuit fitted to a building. They are separated by five m. Point P is located two m from alarm D. On the axis connecting the two speakers. If pure sine waves are emitted by both alarms are in phase and move straight to point P. Were asked to determine the least frequency capable of causing constructive interference at point P. Ok. All right now, when we're talking about constructive interference, let's recall that the path difference D. Is going to be important. Okay? So we want to go ahead and calculate that path difference and the path difference is the difference in the distance that the sound has to travel from one speaker over the other. Okay, so let's start by finding the distance between each speaker and the point P of interest. Okay? Alright. So we're gonna say D. D. Is going to be the distance from speaker D. To the point P. And we can see from our diagram. Okay or from the text, that point P is two m from alarm D. So this is gonna be equal to two m similarly D. C. This is gonna be the distance from speaker C. To the point or point of interest. Okay? Now we can see between point between alarm speaker, see an alarm speaker D. There's five m. Okay? And then from speaker D. Two point P it's two m. Okay, so the distance from speaker C. Two point P is going to be five m plus two m Which gives us seven m. Alright, so we've found the distance between each speaker and that point p Now we want to calculate the path difference. Okay, that's the value of interest to us and that's just going to be the difference in these distances. So D. Is equal to the path difference. Okay? We're gonna take the larger distance. Okay? So D. C. And subtract the shorter distance D. D. We get seven m minus two m which is equal to five m. And so this tells us that speaker see the sound waves from speaker C have to travel five m further than those from speaker D. To get to our point P. Alright. Now now that we know our path difference, what's the relationship between our path difference and constructive interference? Okay, well constructive interference is going to occur when the path difference. D. Is an integral multiple of the wavelength. Okay, so when D is zero lambda two lambda. Okay. Dot dot dot what does this mean? What this means? Is that when these two sounds reach point P they're going to be in phase. Okay because the difference, the extra distance at sea has to travel is a multiple of the wavelength so when they reach point P they'll be in phase. Okay and those two sounds will kind of add to each other, otherwise when they reach point P they're going to be out of phase and those two sounds kind of cancel each other out. Okay, so we want them to be an integer multiple of the wavelength. The path difference so that we have constructive interference because that's what we're looking for. Okay. Alright. So we've written this in terms of lambda. Okay. But we actually don't know information about lambda. Okay. And we want to find information about the frequency. So how can we relate the wavelength lambda to the frequency? F. Well, recall that lambda is equal to V over F. So, this means that we can write our path difference that we want D zero V over F. Two V over F. Okay. So we have an integer multiple of V over F now and we can simplify this to be and V over F. Where n is a positive integer 0123 dot dot dot. Okay. Alright. So this is an equation for D. And we're trying to find the frequencies F. So let's rearrange and write this in terms of F. So we get F. And we're going to call it subscript N because it depends on the value of N. Okay, It's going to be equal to N. The divided by D. Now, above where we've written this general information about constructive interference, we've said that we can have an equal zero. Okay, When N is zero we have D is equal to zero. Okay. So in this case we know that our path difference is not zero. So we know that D equals zero is not an option, which means that end cannot be zero. This means that N is going to start at one chris All right. So above when we've written D equals zero dot dot dot. That's in general. For our case we know D. Is not zero. So we can start at one. Okay. Alright. Well what is V? The speed. Okay, well we're talking about sound here. So we have the speed of sound which we can approximate to 343 m/s Divided by the path difference. D. We found which is five m and this gives us a frequency F N of n times 68. hertz. Okay. The units of meter cancels and we're left with hurts. Alright, so what was this problem asking us to find? Let's remind ourselves we're looking for the least frequency capable of causing constructive interference. Okay. We see that the least frequency is going to occur when we have the least end. Okay. We have N times 68.6. So when N is at its lowest, that's when the frequency will be at its lowest. The lowest value of N. We can have is one. So we get F one is equal to the least frequency, which is just going to be 68.6 hurts. And so that is our least frequency capable of causing constructive interference given this um system with our two speakers in our point p. Okay. And we see that that is going to be answered. G. Okay, 68.6 Hz. Thanks everyone for watching. I hope this video helped see you in the next one.

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