{Use of Tech } Minimizing sound intensity Two sound speakers...

Question

{Use of Tech } Minimizing sound intensity Two sound speakers are 100 m apart and one speaker is three times as loud as the other speaker. At what point on a line segment between the speakers is the sound intensity the weakest? (Hint: Sound intensity is directly proportional to the sound level and inversely proportional to the square of the distance from the sound source.)

Accepted Answer

Welcome back, everyone. In a linear park, two water fountains are located 200 m apart. The water pressure at one fountain is twice as high as the other, resulting in twice the sound intensity of water splashing. At what distance from the quieter fountain will the combined sound intensity of the splashing water be the weakest? Rounder the answer to two decimal places. Hint the intensity of sound is proportional to the sound level and inversely proportional to the square of the distance from the source. A says it's 141.50 m. B 131.50. C 121.50, and the D 111.50. Now to help us solve, let's try to visualize our problem here. So we're saying that we have 2 water fountains, OK? So let's call these fountains A and B. And our problem told us that we're in a linear water park, so the distance between them is linear, OK? Or the, yeah, the distance between them is linear. And we know that we're trying to find a point between them where the combined sound intensity of the splashing water will be the weakest, OK? In other words, here, if we have, we know our total distance is 200 m, they're 200 m apart. So we're trying to find the distance to this point. So let's call the length from A to the point X and the remaining length 200 minus X. That is the fountain B. Now, we can let Fontaine A be the one with the higher water pressure, where it's twice as high as Fontaine B, OK? And if we use that, then recalling the sound intensity formula based on the hint we are given. We know that the intensity is going to be equal to I not or reference intensity divided by X2d It directly proportional to the sound level inversely proportional to the square of the distance from the source, which means then that the sound intensity from Fontaine A at distance X is going to be 2 I 0 divided by X2. Because since Fontaine has twice its own intensity, we have to put a factor of 2 in a not. And the sound intensity at Fountain B. At a distance of 200 minus X is going to be I not divided by 200 minus X squared. Thus, that means the combined sound intensity. OK, the combined sound intensity. At any point, OK. is going to be the sum of the intensities from both fountains. In other words, we have I total of X is going to be equal to IA plus IB. So that's gonna be 2 I not divided by X squared. Plus I not divided by 200 minus X2. OK. So this is gonna be our combined or total sound intensity. Now remember, OK, we're trying to find the combined sound intensity where the splashing water or we're trying to find a point where the combined sound intensity will be the weakest. So to do that, essentially, we need to minimize the combined intensity, OK? So let's minimize, OK? And remember to minimize, to minimize our function, we're gonna find the first derivative, OK? Set it to 0, get a critical point, and then test if that critical point is the minimum. OK? So minimizing our combined intensity, which is I total of X, let's start with the first part of that. Let's differentiate I with respect to X. Now here, OK. That means what we're really doing is finding the derivative with respect to x of 21 0 divided by X2, which we can write as 210 X to the negative 2 plus I not divided by 200 minus X2, which is I not multiplied by 200 minus X rays to the power of -2. Now, when we differentiate those, we know that the derivative of 2 I X 2 will be -4 I X -3, which you can write as -4 I not divided by X to the 3. And for our second term we can differentiate it using the chain rule, OK, of differentiation, which is going to make it be equal to 20 multiplied by 200 to minus X to -3 or we can rate it as being divided by 200 minus X to the 3rd, OK. And know if we were to simplify this by writing it as one expression. Then it would be negative for I not multiplied by 200 minus X to the 3rd. Plus to I not multiplied by 3. All divided by X to the 3rd multiplied by 200 minus X to the 3rd, OK, where we know that X is between 0 and 200. Now that we have the derivative, we can set it to 0 to solve for X. So at the derivative of our combined intensity with respect to x equal to 0, OK, then the derivative will be 0 and the numerator is 0. So we can set our numerator to 0, -4 I not multiplied by 200 minus. X cubed plus 2 I not X cubed to be equal to 0. Now, in this case, then if we divide through uh here by 2 I not, OK. Or divide through by negative to not, then we're going to get 2 multiplied by 200 minus X cubed minus X cubed to be equal to 0. Thus 2 multiplied by 200 minus X cubed is gonna be equal to x cubed. And if we find a cube root of both sides. Then we're going to get to. Sorry, 200 minus X multiplied by the cube root of 2 to be equal to X. Now, remember, we're solving for X, so we can expand this to help us solve, OK? So, when we expand, we're gonna get 200 curoot 2 minus Q root 2 multiplied by X to be equal to X. And now if we put the X terms on the same side and factor X, then we'll get 1 plus the cube root of 2 multiplied by X to be equal to 200 cube root 2. Thus, that means X is going to be equal to 200 cube root 2. Divided by 1 plus the cube root of 2. And that is gonna be approximately equal to 111.50 m. So now we know this is a critical point. Now, we just need to figure out if this is going to make the value a minimum, OK? So how can we test whether it's a minimum? Well, we can confirm, I mean if X is a minimum. R X corresponds, OK. X corresponds to a minimum. Using the first derivative test. OK, what does that test say? Well, basically, it just tells us, OK, that if the derivative changes, the first derivative changes sent from positive to negative before and after the critical point, then it corresponds to a minimum. If not, then it corresponds to a maximum. OK. So let's test that. And since our reference intensity is always positive and the denominator is always positive, the sign of the derivative will then be determined by the numerator, OK? So, and again, we know that we're working with uh our expression that we we we we we we have here, OK. So in this case, When X is less than our value, 200 cube root 2 divided by 1 plus cube root 2, we can choose a value of X. So let's let X be equal to 111. Now when X is 111. Then our first derivative, OK. So let me just call that here. I prime total of 111. It is going to be equal to -4 and not multiplied by 200 minus 111 cubed plus 2 and not multiplied by 111 cubed. I know when we work the salt, OK, then we should find that we get it to be equal to negative 84,614 I not, OK? Thus, since, remember we said Ino is positive, then This value is negative, OK? This initial value is negative. Now, after, OK, that is for X greater than 200 Q2 divided by a 1 + Q2, OK. Then let's let X be equal to 112. So essentially, we're just going to do the same thing again, OK? And the derivative. Properly here. And a derivative here when X is 112, and now when we do that, we will eventually work it out to get it to be equal to 83,968 I not, which is going to be a positive value. So no, based on how our derivative sign changes here, that means. That X indeed corresponds. 1 + 22 corresponds to, OK, a minimum. And since our distance is X, therefore, X, this value X, which is approximately 111.50 m, is going to be the correct answer. In other words, this is the distance at which or or uh or splashing the sound in the combined sound intensity of our splashing water will be the weakest. D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Sound Intensity

Inverse Square Law

Proportional Relationships

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