A loud factory machine produces sound having a displacement ampli...

Question

A loud factory machine produces sound having a displacement amplitude of 1.00 mm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is 1.42 * 105 Pa. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit? Is this frequency audible to the workers?

Accepted Answer

Hey, everyone in this problem, we're told that high frequency sounds are common causes of ear damage, especially in Children. The maximum sound pressure, the child's ear can tolerate is about 20 pascals. A to speaker produces pure sign sound waves with a maximum displacement of 0.2 millimeters. If the bulk modulus of air is B equals 1. times 10 to the five pascals, we're asked to find two things. The first thing we're asked to do is to find the maximum tolerable sound frequency of the produced sound wave to prevent hearing loss. And the second thing we're asked is whether the child can hear at that frequency, we're given four answer choices. Option A and B share the same answer for frequency F is equal to 192.2 Hertz. In option A, we have the yes, that frequency is audible to the Children. And in option B, we have no, it's not. Option C and D also share the same frequency of 3844.4 Hertz. Option C has yes, the frequency is audible to the Children. And option D has no, it's not. So we wanna find this maximum tolerable sound frequency to prevent hearing loss. Now, to prevent hearing loss, we're given a maximum sound pressure. So let's think about this maximum sound pressure. Now, recall that we can write this maximum pressure P max is equal to B multiplied by K multiplied by A where B is the bulk modulus K is the wave number and A is the displacement amplitude. So now we have this equation with our maximum pressure, but we want to relate this to the frequency. So how can we do that? Well, recall that the wave number K and it can be written as two pi divided by the wavelength of lambda. So we can write our equation as P max is equal to B multiplied by two pi divided by lambda multiplied by A. Now we have our equation with this information of a wavelength and we know that we can relate the wavelength to the frequency. OK. So we called it, the frequency is equal to the speed V divided by the wavelength lambda which tells us that we can write one divided by lambda. OK. Like we have in our equation as a frequency F divided by the speed V. So now we can write our pressure equation as P max is equal to B multiplied by two pi F divided by B multiplied by A. Now we know the maximum pressure that we can have. OK. We're told that the he um child's ear can tolerate 25 scouts. We're told the bulk modulus of air, we wanna find the frequency F OK. We know that we're in error. So we know the speed of sound and air and we're also told this maximum displacement which gives us the amplitude. So now we have everything we need, we can plug our values into this equation and solve for the frequency F. So let me write out our variables on the right hand side here. So we have that P max the maximum pressure this child's ear can tolerate is 20 pascals. We have that the bulk modulus of air is 1. times 10 to the exponent five pascals. We have that the speed of sound and air is 343 m per second approximately. And the maximum displacement we're given which is our amplitude is 0.2 millimeters. OK. So amplitude A is going to be equal to 0.2 millimeters which we can write as two times 10 to the exponent negative three millimeters. And we move that decimal 30.1 to 3 positions to the right hand side to get two. And so we have 10 to the exponent negative three. Now this is in millimeters, we wanna convert to our standard unit. So what we wanna do to go from millimeter to meter with our units, we multiply by 10 to the exponent negative three. OK. The prefix milla means 10 to the exponent negative three. And so we get an amplitude of two times 10 to the exponent negative six m. OK? 10 to the exponent negative three times 10 to the exponent negative three. They have the same base, we can add those exponents and we get an exponent of negative six. OK. So now we have all of our variables written out. We know what equation we wanna use. We just need to substitute everything in and solve graph. So we have 20 pascals is equal to 1.42 times 10 to the exponent five pascals multiplied by two pi F divided by 343 m per second, multiplied by two times 10 to the negative six m. Now the unit of meter will divide it from the numerator and denominator. On the right hand side, on the left hand side, we still have our 20 past gals. On the right hand side. If we simplify, we have 1.42 times 10 to the exponent five multiplied by two pi multiplied by two times 10 to the exponent negative six divided by 343. And we get that, that is equal to 0. 52024. Now our units here are Pasco per second. OK? And this is multiplied by the frequency act. So now to find the frequency F, we are going to divide oops. And that's not per second. Sorry, that's pascal multiplied by second. OK? Because we have pascal divided by, per second. OK. So when we divide, we get that, our frequency is equal to 20 pascals divided by 0.52024 pascal seconds. And so we get a frequency of 3844.38. OK? For our units, the unit of pascal will divide out and we're gonna be left with the unit per second, which is equivalent to a Hertz. And so we find that the frequency is 3844.38 Hertz. OK. So that is part one of this question. Part two is Ken the child here at that frequency. Now, we're called up for humans. Humans can hear between 20 and 20,000 Hertz. Now 3844 falls in that range. And so yes, they will be able to hear it and that answers both of those questions. So if we go back to our answer choices, OK. The frequency is rounded to one decimal place. So we found a frequency of th 3844.4 Hertz and that it was audible to the Children and this corresponds with answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

Watch next

Standing Sound Waves practice set