Consider a sound wave in air that has displacement amplitude 0...

Question

Consider a sound wave in air that has displacement amplitude 0.0200 mm. Calculate the pressure amplitude for frequencies of (a) 150 Hz; (b) 1500 Hz; (c) 15,000 Hz. In each case compare the result to the pain threshold, which is 30 Pa

Accepted Answer

Hey everyone in this problem. We have an online to generator creating sign sound waves in the air with the displacement amplitude of 0.4 millimeters. And we have two frequencies F one is 200 hertz in F two is kilohertz. Okay. And we are asked to find the pressure amplitude P max at each frequency. Okay. And then were also asked to specify under these conditions, if the sound frequency is harmful to human hearing. Okay. We're told that the pressure variation that corresponds to harm is pascal's. Okay. Alright. So we're asked to calculate the pressure amplitude P max. Let's recall that we can write P max is equal to B. Okay, A A Where B is a bulk module asse. And A Is the amplitude. Okay, let's also recall that we can write K as two pi over lambda. Okay. That will relate directly to our wave. So let's go ahead and do that. So, we have P max is equal to b times two pi over lambda. The wavelength times a The amplitude. All right, so be the book module is well, this is going to be the book modulates of air. Okay, we're in air You can look this up in your textbook or in a chart that your professor provided but the bulk modules of air is going to be 1.4 Times 10 to the five Pascal's Okay. Our amplitude a well, this is given in the problem 0.04 mm. And if we want to put this into meters into our standard unit, we're gonna have times 10 to the negative three. We already have 10 to the negative two. And so this is gonna be four times 10 to the negative five m. All right, so we have our equation. We have B. We have a The last thing we need to do is find lambda and lambda is going to change based on the frequency. Whether we're at frequency one or frequency to Okay, so let's start with F one equals 200 hertz. Recall that we relate frequency and wavelength through the velocity? So we have F one is equal to the velocity over lambda one. The wavelength. Alright, well this means that λ one it's going to be equal to the velocity over the frequency. Mhm. Now in this case we're talking about the speed, we are talking about the speed of sound waves. Okay, so we have 300 43 m per second. Okay, so the speed of sound in air and our frequency hertz. And this is going to give us a wavelength of 1.715 m. Alright, so now we have B. We have a and we have our wavelength for the first frequency we're looking for. So we can find the pressure amplitude P max. So we get P max is equal to B 1. times 10 to the five pascal's. Okay, times two pi divided by lambda 1.715 m. Many times the amplitude. A four times 10 to the - m. The unit of meter will cancel, We're gonna be left with P max. It's equal to 20 0. pascal's Writing in scientific notation. We get 2.08 Times 10 to the one. Ask us. Okay. And so at the first frequency we have our P max here 2. times 10 to the one pascal's now were also asked to determine if this is harmful. We're told that the harm threshold is 30 pascal's okay, so we're below the harm threshold so this is going to be not harmful. Okay? Alright, so now we're going to do the same thing but we're gonna do it with the second frequency. So let's give ourselves some more space And we have F two is equal to Kilohertz. Okay. Which is equal to 20,000 hertz. Alright, so lambda to in this case just like when we did lambda one, lambda two is going to be equal to V over F two. Again, speed of sound air 343 m/s Divided by the frequency 20,000 Hz. And this is gonna give a wavelength λ two of 0. 1715 m. Now we have our wavelength lambda. So we can go ahead and find the pressure amplitude P max which is equal to again be the bulk modulates of Air 1. times 10 to the five pascal's times two pi divided by lambda, 0.1715 m times the amplitude four times 10 to the negative five m. Okay. The unit of meter will cancel, we're going to be left with the unit of pascal and we are going to get P max is equal to 0.96 pascal's writing this in scientific notation with three significant digits. 2.8 times 10 to the three pascal's. Okay. And now we just need to determine if this is harmful. Okay? The threshold is 30 pascal's were well above that at at 2000 pascal. Sorry. So this is going to be harmful. Alright, so let's go back up to our answer choices. We found ap max at the first frequency of 2.8 times 10 to the one pascal's that's not harmful to human hearing. Okay. And we found a P max at the second frequency of 2.8 times 10 to the three pascal's which is harmful to human hearing. So that's gonna be answer. C Thanks everyone for watching. I hope this video helped see you in the next one

Consider a sound wave in air that has displacement amplitude 0.0200 mm. Calculate the pressure amplitude for frequencies of (a) 150 Hz; (b) 1500 Hz; (c) 15,000 Hz. In each case compare the result to the pain threshold, which is 30 Pa

Key Concepts

Sound Wave Properties

Pressure Amplitude

Threshold of Pain

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