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 (b) destructive interference occurs at point Q?

Accepted Answer

Hey everyone in this problem, we have a vehicle at rest. Okay, fitted with two sirens that are 2.2 m apart, the sirens are operated by the same circuitry admitting pure sine waves that are in phase A point S. Lies 1.1 m on one side, measured from the sirens for sound waves traveling straight from the siren. Two point S were asked to calculate the least frequency that causes destructive interference at point S. And so if you look at our diagram here, we have our two sirens, okay, that are 2.2 m apart and then this point S, which is 1.1 m from this last siren. Now we're talking about calculating a least frequency that causes destructive interference. Okay now we're talking about interference, destructive interference, an important value to calculate as the path difference. Okay, so we want to start by calculating the path difference and the path difference is going to be the difference in the distance that a wave has to travel from this. Further siren then from the closer sire. Okay, Alright, so let's start by finding the distance between each siren and the point. Okay and we're gonna call this siren on the left siren one and the one on the right siren too. So the distance D. one. Okay, this is gonna be the distance from siren one to our point. This. Okay, well what distance is that gonna be? Okay, we see we have 2.2 m to siren to and then another 1.1 m. So we get 2.2 m plus 1.1 m. Which gives us a distance of 3.3 m. Alright, we're gonna do the same for the second siren. K. So D two is gonna be the distance from siren to two. R .s In that distance were given in our Diagram as 1.1 m. So we have our two distances from each siren to the point s. Now let's calculate our path difference and our path difference again is going to be the magnitude of the difference between those two distances. Okay, so we're gonna call our path difference D. For this problem, we're just gonna take the magnitude of the absolute value of D one minus D. Two. Okay, you can do it D two minus D. One either way. It's fine because you're taking that absolute value. So we just want the magnitude. Okay, in this case we have the larger number first so we can drop the absolute value and we get 3.3 m -1.1 m. That gives us a path difference of 2.2 m. Alright, so we've calculated our path difference. We're trying to find the least frequency that causes destructive interference. Okay, what's the relationship between destructive interference and the path difference? Well, let's recall the destructive interference. I'm just gonna short form here interference. Okay, occurs when the path difference, which we've called the is a multiple of lambda over to the wavelength over two. So we have lambda over two. Three lambda over two. And I should say an odd multiple of five lambda over two dot dot dot. Okay, so we can write this as N. Lambda over two. Okay. Our path difference, we want to be an odd multiple of the wavelength. Okay. Where N is equal to one? 357 dot dot dot. Okay, I'm gonna put in brackets we have N is odd now. Why do we want this wavelength or the path difference to be a multiple of half of the wavelength? Well, if we do that, okay then when the sirens arrive at point S or the waves from the sirens arrive at point S. Okay, they're going to be out of phase. Okay. Because one is going to be a multiple of half a wavelength different than the other has to travel half a wavelength or a multiple of half a wavelength. Further case they're going to be out of phase. Okay, so those two waves, those two sounds are gonna kind of cancel each other out. And that's where we have destructive interference. Okay, so that's the conditions we want on our wavelength? Remember we're looking for a frequency. Okay, we're looking for the least frequency. So how can we relate the frequency to the wavelength? Will recall that the wavelength lambda is equal to the speed. V divided by the frequency F. Okay, this means that we can write our desired path link for destructive interference as D. Is equal to N. V. Divided by two F. Alright, so we've just replaced lambda here in this equation with V over F. We're trying to find a frequency. Okay, so let's isolate for the frequency F. And we're going to write F subscript N. Because the value is going to depend on end. We get N. V. Over two times of path lengthy. All right now we want to find the least frequency. Okay, So we want to find the smallest frequency that we can have in this situation. When is that going to occur? Well, N is in the numerator. So the smallest frequency is going to occur when N is the smallest And the smallest end we can have remember is one. So yeah, and that's equal to one. Now the least frequency is therefore going to be given by F. One which is going to be one times the speed, which in this case is the speed of sound. 343 m per second divided by two times our path difference, which was 2.2 m. Okay, the units of meter will cancel. And we're going to be left With the least frequency of 77.95 Hz. Okay. Alright. So our least frequency to have destructive interference is 77.95 hertz. And if we look at our answer choices. Okay, we can approximate and we're gonna see that we have answer F hertz. Thanks everyone for watching. I hope this video helped see you in the next one.

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