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The figure shows the output from a pressure monitor mounted at a point along the path takenby a sound wave ol J single frequency travelingat 343 MVs through air wi...

Question

The figure shows the output from a pressure monitor mounted at a point along the path takenby a sound wave ol J single frequency travelingat 343 MVs through air with a uniform density of 141kg/in: The vertical axis scaleis set by 4ps- 4.00 mPa Ifthe displacement furkction of the wave is written as slx t) Sm cos(kx Wt}, what are (a) S (b) k and (c) (? Theair is then cooled so that its density is 1.35 kz/m? and the speed of sound wave throughit is 321Ms The sound source again emits the sound wav

The figure shows the output from a pressure monitor mounted at a point along the path takenby a sound wave ol J single frequency travelingat 343 MVs through air with a uniform density of 141kg/in: The vertical axis scaleis set by 4ps- 4.00 mPa Ifthe displacement furkction of the wave is written as slx t) Sm cos(kx Wt}, what are (a) S (b) k and (c) (? Theair is then cooled so that its density is 1.35 kz/m? and the speed of sound wave throughit is 321Ms The sound source again emits the sound wave at the same frequency and same pressure amplitude: What noware (d) sm (e) k and (f) t? (mPul



Answers

Figure $17-32$ shows the output from a pressure monitor mounted at a point along the path taken by a sound wave of a single frequency traveling at 343 $\mathrm{m} / \mathrm{s}$ through air with a uniform density of $1.21 \mathrm{~kg} / \mathrm{m}^{3} .$ The vertical axis scale is set by $\Delta p_{s}=4.0 \mathrm{mPa}$. If the displacement function of the wave is $s(x, t)=s_{m} \cos (k x-\omega t),$ what are $(\mathrm{a}) s_{m},(\mathrm{~b}) k,$ and $(\mathrm{c}) \omega ?$ The air is then cooled so that its density is $1.35 \mathrm{~kg} / \mathrm{m}^{3}$ and the speed of a sound wave through it is $320 \mathrm{~m} / \mathrm{s}$. The sound source again emits the sound wave at the same frequency and same pressure amplitude. What now are (d) $s_{m},$ (e) $k,$ and (f) $\omega?$

For this problem on the topic of waves were shown in the figure the output from impression monitor That is observing a sound wave of a single frequency traveling at 343 m/s through a. It has a uniform density of 1.21 kg per cubic meter. And the vertical axis scale of the graph is set by delta P. S. Is for millie paschal's. Now, if the displacement function of the wave as a function of X and T is the displacement amplitude sm times the call sign of K x minus omega T. We want to find the displacement amplitude sm the values for K and the value for omega If it is then cool so that its density changes to 1.35 kg per cubic meter and the speed of a sound wave is now 320 m/s. We want to find the new values for S. M. K and omega. Now the period T we can see is too milliseconds, which we can write as 0.002 seconds and the pressure amplitude delta P. M. Is equal to eight millie paschal's. This is equivalent to 0.00 eight Paschal's or newtons per square meter. And the displacement amplitude, we know S. M. Is equal to the pressure amplitude. Datarpm over the times. Roll times omega, which we can right as data P. M divided by the times. Roll, multiplied by two Pi over tea. And so substituting the values above, we get this displacement amplitude To be 6.1 Times 10 to the -9 m. Next for part B, we want to find the angular wave number so the angular wave number K. Is equal to omega over V, which is two pi over the times T. We are given the speed of the wave To initially be 343 m/s. So we get the angular wave number To be 9.2 gradients perimeter, and the next in part C. We want to find the angular frequency. Omega angular frequency is two pi over the period T which is Approximately 3.1 Times 10 to the power three radiance per second. Now, for part D. Well, before we do party, we can summarize the results from above. As we have the wave function, the displacement wave function for the wave. Yes. As a function of X and T. Compute it and as 6.1 nanometers times the co sign of 9.2, permit to times x minus 3.1 times 10 To the Power three. The second times T. Now going on to part D. Using a similar reasoning, but with different values for the speed of sound and the density of air, we have the displacement amplitude Sm. Is the pressure amplitude out the PM of a V. Prime row prime mega. This is data p. M. Of V. Prime row prime times two pi over T. Which gives us the displacement amplitude. In the new Scenario to be 5.9 Times 10 to the -9 meters. So that's the new displacement amplitude. The new angular wave number K is equal to omega over the prime, Which is two pi over V. Prime times T. and using the prime in this case to be 320 m/s, We get K two B 9.8 radiance for me to and then lastly, for part F, the new angular frequency omega is equal to two pi over T. Which we see is the same as the old angular frequency 3.1 times 10 To the power three radiance per second. And so we can write the wave function under the new conditions, S. Of X and T. To be 5.9 nm times the co sign of 9.8 Permata times X -3.1 times 10 to the power three for a second times T.

Hi, everyone. This is the problem. Based on variation off pressure in air due to propagation off salt here, pressure and petition graph is given using it. We have to plot pressure. And Kangra we have to plot by an extra. We have to clock by tear up. Yeah, Dean, we have to calculate maximum velocity on maximum acceleration. He parked the veracity off Vibration off the drum off the pull off the speaker that is to start solving it first part. From the first part, we can see maximum pressure is 40 person speed of sound in air is 3 44 m per second. And from the figure we can find the wavelength movie while 2 m So time period off oscillation will be lined up on with That is quite to upon 3. 44 that is 5.81 tend to departments for second. So PT graph we can blood time tend to the par minus four second. Yeah, pressure maximum is 40 Paschal and minorities minus 40. Possible? Yeah, yeah. Uh huh. E o So here it is. Toe 0.95 point age and 11.6. No, be part pressure as a function off X is given by minus week del y upon Dylex. So why, it will be my husband upon me being DX. See question, but okay. Area of photograph. And we have waste into height while 05 into 14. So, bun meter Paschal. Yeah. So it will be a year upon me. No. And bulk models off here is So it is to be seven. Find Gino Fort in tow. 10 to the power. No, minus six. So absolutely, you will get No. Yeah. More why you are taking Intento de Paul minus 6 m X in me too. Seven minus seven. Great. Do not do it my while while 2.3. While Sorry. 0.4 and quite fight. Mm hmm. Third part B in third part. We have to plot displacement as a function of time biting graph. Oh, yeah. Um, yeah. Time between. Came to the apartment is four seconds. Mhm. All right. 2.9 seconds. 5.81 817 and 11.6. What deeper maximum velocity is be mix upon me and toe dispute off in the media, that is 14 upon 1.42 came to the par five into 3. 44. The maximum particle velocity is 9.7 in tow. 10 to the power bi. This too meter per second. Yeah. Hi. No maximum acceleration. Delta P upon Delta X divided by a room s So it is to be equal toe 40 upon minus 40. Divided by mhm white one minus zero upon density off air. That is 1.2. So maximum acceleration You will get right 6 67 m per second squared. Yeah. No. Ybor Speaker must move According toa displacement graph. Speaker cold Our civic with frequency be upon lambda. That is 3 44 upon man, that is Uh huh. Yeah. And And dispute off Cool is yeah. Could suit accelerate. Okay, there's a text for watching it.

Hello, everyone. This is the problem based on concept off interference off to acquire it. Sun bets here it is given intensity off south at a distance. D 100 units and its maximum body with 900 in its Yeah, the arm safety is given. Yeah, In the first part, we have to find the frequency in second part. We have to find the intensity off constructive and destructive interference Not to see it part A Since the sound intensity at the changed from yeah, minimum value took maximum value. When the arm is sacred White 1.65 centimeter. Thus l will be skull toe lambda by toe. So lambda will be 1.65 in tow to that is 3.3 scent emitted. So frequency is speed up on wavelength Spirit is given 3 43 in the problem meter per second and Babel It is 3.3 in tow, 10 to the power minus two Have you will get 10.4 and took 10 to the power three hurts the ratio off maximum to minimum intensity substitute develop. They're so thanks for watching it. So you see parties also the That's all Thanks

Hello, everyone. In this video we will be looking at Sanyo Soto sound waves propagating through different media. The three media that we will be looking at is water, air and iron. The air will specifically be at zero degrees associates. Now what we're going to do is look at the angular frequency and intensity as being the same through all media. And we will be looking for their wavelengths, this placement, aptitude and pressure amplitude. Yeah, and finally we will change the initial angular frequency and the the intensity, the reference intensity value. So let's start off with writing what we're given now. We will need the properties of each media, so we'll have air. So for error we will have of velocity and let me go ahead and write air at zero degrees Celsius. So air at their degrees Celsius will give us a velocity of 330 31 so of velocity, the is equal to 331 m per second Up next we'll have the density. So the density of air is 1.29 meters per second and this book tends to use 1.1 point 21 so we could also incorporate that. But for this story of this problem, we will using 1.29 more accurate version. So for water, we'll be using freshwater sore at a regular degree of Oh, our standard temperature. So water at standard temperature, which is usually 20 degrees Celsius. We'll have a velocity and which sound travels through well. So the velocity which sound travels through water is 1493 like meters per second. And the density of fresh water is one times 10 to the three, one times 10 to the three to part three, and that, that is, I believe, kilograms per meter squared and that is for both air and water. So to me, right kilograms per meter squared kilograms for meter square. Now I will exclude the units when I'm actually solving my problems just so that we can save some room. And some time when reading out the problems now for iron will have a velocity which sound travels through iron, a velocity of 960,000, 150. Let me just regret that all more clearly. So you guys, let's see better 5900 and 50 meters per second. And for our density, we have seven point 86 times 10 to the power of three kilograms per meter squares. Now we are not given values for the reference intensity or the initial intensity and angular frequency. However, what we can do is we can just give. Since the intensities are equal for all medium, we can just right in our own value for intensity or frequency. So for we'll use the like the usual value for reference intensity, which is one times 10 to the negative, 12 meters per second. Oh, that is not meters per second. Sorry. So, um, what we have and we could do the same thing for angular frequency. So we know the angular frequency and good er frequency is equal to to buy F. And for our frequency, we can say that we start at a frequency of 300 hertz. It's okay for us to put our own value for frequency as long as they're all the same. Because the angular frequencies is also the same soul. If we include that for our value off anger frequencies. So we have the this right in 300 Hertz. Okay. To find our angular frequency of 1000 800 in 84. So 1884 0.95 we'll keep more vegetables just for accuracy. And with that, we'll have all of our constant values. So up next, we are going to start with our calculations. So we are asked to find or to compare the values of wavelength and this problem again. This problem does not give us an exact values. We do get values for problem for part D of this problem at which we can. So we're going. So for party, we're gonna compare wavelengths. Okay, So for per deal of this problem, we will be given values which we can then plug in to find actual solutions. But they essentially just want us to do these in terms of variables. But we're going to import actual values to get a better analysis of the differences between or a better comparison so we can have a reference. Essentially. So, for the wavelength of air, we know that the general equation is velocity over frequency. Yeah, And from here, what we can do is look so we'll have to input the value for the velocity of air. And we are. We assume that we're going to be working with frequency of 300 hertz. So now now we can just plug in and look for our see what wavelength we get for air at a frequency of 300 hertz. So the velocity of air at one degree associates is 331 meters per second, divided by 300 huts, and that gives us a wavelength of 1.1 meters. We write that and mark clearly so 1.1 m and that is the wavelength for air at two degrees Celsius. Now let's look at water. So the wavelength of alright H 20 for water. So we'll have to write. So we have to input our velocity of water over frequency. Yeah, and for that problem or for for this velocity of water velocity of sound through water is 1493. So 1493 divided by 300 hertz. Now this will give us a value of 4.97 m. So from here we can already see that the wavelength off water is much as about five times larger than the wavelength of air. So that's the first comparison we see. So we can say that air so far has the smallest wavelength. Now let's look at irons. We can compare all three, so the wavelength of iron is equal to velocity of iron over frequency. So this is the velocity of sound through iron. So for this, the velocity of sound through iron is 5000 950 and that would be divided by 300 hertz, which gives us a value off 19 m. Exactly so 19 m. So from this we see that iron is nine time, 19 times larger than air, about 19 or more, so 18 and it is about four times larger than the wavelength of water. So these are our answers for part A where we say that air gives us the smallest value. So smallest, take thank you. Let me get a racer and make this even more neater. So for the wavelength off air, we have 1.1 m, which is the smallest value. And for iron, we have the largest wavelength, largest wavelength. So that's what we see from comparing toe the relationship between wavelengths off different media we have that the more the heavier atoms ours. So I iron is much more or the wavelength of iron is much greater than that of H 20 of water and error of the wavelength of water and air. And this is dude. If we look at the relationship between wavelength and frequency and velocity, it's simply that the velocity I wish the at which sound travels to the medium is what gives us the sought. It is what influences the size of the wavelength. So now let's move on to perp e Well, where we'll be looking at the displacement aptitudes. So we'll be looking at part for part B will be comparing the displacement attitude comparing displacement. Yeah, altitudes, Yes. So we'll be comparing with displacement amplitude. We'll we'll have Well, we can look at the relationship being the S Max. So s Max being displacement amplitude? Thank you. This basement attitude is equal to the change in or the yeah pressure displacement. So the spaceman temperature is equal to the pressure displacement over row, the omega, and that will weaken, then rewrite pressure amplitude as being the square root of the reference intensity. Times 10 the sound level divided by 10. So 10 to the power of the sound level divided by 10. Let me write this more neater as well. Okay, Okay. So the reference intensity times 10 to the power of the sound level, divided by 10. Mhm. Yeah, times to Roe v all divided by role v omega. So we'll have to do this. So we have to solve this with respect to each media. So we'll start with air again. So we'll do the change in the change in. So the displacement amplitude is equal to pressure. Well, right, this form the equation. Since we have everything we need in here, and we're going to make another assumption of the sound level and decibels being so, the sound level depends on the reference intensity, which will That value will change later. But we can for now make an assumption that the displacement and we'll be looking at this basement amplitude at 70 decibels. And as long as that's the same for all medium, it's okay to do that. So, yes, Max will be equal to the square root of the reference out and this is of error so that this basement amplitude of error but the in parentheses will be equal to one times 10 to the negative 12 oh times 10 to the power of 70 decibels divided by 10 times to the density of air. Again, the density of air would be 1.29 times the velocity where which is 331. So 331 and all of that will be divided by the density of air times the velocity of of air times the angular frequency. So the density of air 1.29 times the velocity of air. 331 times omega angular frequency, which we said is 1000, 1884.2 point 95 So 1884.9 five and all of that will be equal to one point one for eight times 10 to the negative seven meters. Okay. Oh, so for the displacement amplitude of air, we have that the length is 1.1 times 10, 10 later, 7 m. Now we can look at water so for water, and I'll write this a bit lower so we'll have more room since this equation is pretty big or have the displacement amplitude. So the change in S max displacement amplitude of water is equal to the square. Root is equal to the spirit roots of okay, one times 10 to the negative tall the reference intensity times 10 to the power of the sun level divided by 10. And the meat right to Simon, 70 for the sound level again. So 70 divided by 10 which is seven so 10 to the power of seven times to role row being the density of water. So to times the density of water which is so the density of water is 1493. So 1493 times the velocity of water of sound through water is 1000. Oops, I have switch these two terms sold up density of water is actually one times 10 to the three. So then sit so we'll have two times one times 10 to the power of three. So that so. That's two times the density of water times the velocity which is 1493. So all of that will be divided by the density of water times the velocity of water time, the angular frequency. So the density of water again is one times 10 to the power of three. And the velocity is 1400. So 1000 1400 in 93 mhm times the angular frequency which is 1000 800. So any other frequency is 1800 84 point 95 Plugging in for all these values we get to the we get that the spaceman amplitude of water is one point 9381.938 times 10 to the negative, nine to negative 9 m. Also box or answer for All right, for the displacement Look into the air. And from this we can see that the displacement amplitude of water is much, much lower than the displacement amplitude of air. Because we have this 10 to the power of negative to their that we see, it makes this an order of magnitude of 1000 or ah 100 less so Well, so pretty much. We are 1000 1000 decibels shorter than the displacement amplitude of air. So so far we have the air has the largest displacement amplitude. Now let's look at iron, the displacement, amplitude of iron. So the displacement amplitude of iron is equal to square root, the square root of the reference intensity. So one times 10 to native 12 times will write 10 to the seventh. So we know that this is our relationship of the pressure on YouTube, where we have 10 to the power of the sound level, divided by seven. So I'm just dividing Samity, divided by 10, which gives us 10 to the power of seven, as I mentioned appear. So we have 10 to the power of. So this part 10 to the power of 70 divided by 10 is just 10 to the power seven. So 10 to the power of seven times, two times the density of iron, which is 7.867 point 86 times 10 to the power of three. And so it will be the density times the velocity. So now we have to multiply this by the velocity of of sound through iron, which is 5000 5000 900. So 5900 and 50. So 5000 950 m per second. All of the all of this will be square rooted and divided by row The, um Rovio Mega which is the density of iron. So the density of iron again is 7.867 point 86 times 10 to the power of three times the velocity of arm, which is 5000 150 times the angular frequency which again is 1000 800 1884.95 1824 0.95 Yeah, so for this we get the that. The displacement aptitude of iron is three points 47 times 10 to the negative 10 m. So from this we see the iron actually has the smallest, this basement amplitude of all the media. Now, if we look again and what are what we're doing with the relationships of displacement, displacement, amplitude, we see that the displacement amplitude is the pressure amplitude divided by the density, velocity and angular frequency. So the media's with the highest densities and the highest velocity well, really make the will really have a shorter displacement amplitude. So if we look at our densities, we have that The largest density is iron, and the largest velocity is also iron. Which is why we have a very small value for the displacement aptitude of iron because it's being divided by thes these big numbers. So we're being so This is divided by this big number for the density and this big number for the velocity compared to let's say, air, which has a very small value for density and the very a small value for velocity relative to the other media. So that's the relationship we see there. Now let's move on to part C of this problem. So for part C, we are told to find the this pressure MPA too. So we're gonna be comparing the pressure amplitude. So compare pressure amplitude, and we kinda already did this for part B. As you can see for we find this basement amplitude, we found the pressure amplitude over the density times, velocity times omega. So we're going to be doing the same thing we did there and look at the pressure amplitude and its relationship of yeah to to roll the and we're gonna use this relationship because we it would. It would give us a similar values that we did for Part B. So in part B of this problem, we had this relationship. I used this relationship rather than but we what we could have done is just use this to Roe V. So we can say that the pressure amplitude is just described it of to Roe v The Value by Roe v. Omega. But this will become useful for part D of this problem where were given a different value for reference intensity. This I not so we'll be so we'll deal with that later. For now, we'll be well. Let's look at the pressure amplitude in the three different media. So let's look at air first as always. So the pressure amplitude of air at zero degrees Celsius is equal to the square root of two times the density of air. 1.29 times velocity. She's 331 now. For that we get the pressure. Amplitude of air is 29 point to to okay Newtons per meter squared. So now let's look at the pressure amplitude of water. So the pressure on the two of water will be equal to the square root off to times the density of water which is one times 10 to the power of three times the velocity of water. A sound through water, which is 1000 493. That gives us a value of 1000 700 on 28. Okay, let me write that eight A bit nicer. So 1728 Newtons per meter squared. Okay, so so far, we see that that pressure amplitude of water is a much, much greater than the pressure attitude of air. And this is again if we just look at the relationship of pressure, amplitude and dent with density and velocity. So if for larger values of density and velocity, we will get larger values for pressure altitude. So we can already guess that iron will have the greatest pressure amplitude because it has the greatest density and the greatest velocity and air will have the smallest other three media because it has the smallest as the smallest density and the smallest velocity through it. Our sound waves have the smallest velocity through air. So now let's this prove our hypothesis by showing or by finding the pressure amplitude of iron. Oh, so the pressure imperative of iron is equal to the square roots of to times the identity of 7.86 times 10 to the three to the 10 to the power of three times the velocity of iron, which is 5000 950 which is then equal to 9671. So 9000 671 0.3 Newtons per meter squared. And from here we have completed part C. And again we prove that iron has the largest pressure amplitude of them all because of its large, much larger density and much larger velocity at which sound travels through iron. So for part D of this problem, we are essentially wrapping up everything that we did. And we will be making all of our calculations. What we doing all of our calculations again with values. So let's say we'll analyze the song leaves analyze sound waves. Yeah, sounds sound Wave waves at angular frequency. Our initial angular frequency off 200 or 2000 rather for a value of 2000. Hi Her For eso the units of the units of England frequency is grabs per read per seconds. So radiance per second. Let me write that her second and for for our reference intensity will be using We'll be using one times 10 to the negative six watts per meter squared so far reference intensity will be using one times 10 to the negative six watts per meter square Oh oh, and again will be analyzing the wavelength displacement, amplitude and pressure amplitude. So let's start with part A and I will so with so that we don't get confused I'll put Roman numerals So Roman numeral one Not right subscript a just to remind you days part a will be finding the wavelength which is just the wavelength, is equal to velocity over frequency and will be Since we're given, we are given angular frequency. We know that angular frequency is equal to pi f so we can rearrange that and find f as and as omega over two pi which is then equal to so, um so omega over two pi would be to 1000. Hi. So 2000 hi divided by two pi so divided by two pi, the pies cancel. So is so. Since the pies cancel, we essentially have. Yeah, we will have the is a privilege to be 1000 hurts. And just to verify that we can I'll just show you that the pies will cancel like this and with it will be to 2000 divided by two, which you can just say is we leave. If you divide the tools and leave the one hundreds or leave the zeros, you'll get 1000. So the so 2000 divided by two is just 1000. So we're left with 1000 hurts. Now let's look at the wavelength of air. The wavelength of air will be 331. So 331 divided by our frequency that we found here to be, which is 1000 hertz and they gave me a value of 0.331 meters. So our wavelength for air is 0.331 m. Now let's look at water. Okay, the remember This is a rare at zero degrees Celsius. So for Aaron, we'll have that our velocity of air or velocity of sound through water. Sorry is 1493 divided by 1000. So if you put that into our calculators, we get that our wavelength for water is 1.493 which I will round two. Okay, well, until our around to 1.49 instead. 1.49 meters again. Our wavelength for water is higher than air and we should see the same relationships. Even though we have these initial these conditions we know that generally are air. Our wavelength for air will be the smallest and iron will be the largest with water being in the middle. And we'll get the same relationships for the displacement attitudes where the displacement aptitude for air will be the largest. Let me write that largest and the displacement amplitude for iron will be the smallest. So it'll be the smallest and same thing will apply for our pressure attitudes where we'll have that the pressure aptitude for air will be the smallest and the pressure amplitude for iron will be the largest. So large. Just so. Those relationships we should see again in part D do you so for the wavelength of iron will have that The velocity at which sound travels through iron is 5950. Oh, let me write thes fives better. I'll just rewrite the whole thing. 5950. I hope so, yes, it seems that my pants still ain't that a bit bigger, so I'll just write larger So 5000 900 50 divided by our frequency, which is 1000 which will give us five point 95 meters. So 5.95 m And that is our largest wavelength, as we expected. So 5.95 m because people for our we've length of iron And now we can move on to part B where we look at the and I'll label that Roman numeral to with a subscript be at the bottom. Well, look at the max displacement amplitude. So again we'll be using S Max is equal to the change in the pressure amplitude. And so the change in P max divided by a row the Omega. So for this we can input our values for for part, we couldn't put our our values for parts for per be or purchase C and well So we're gonna actually skipped part C instead. And let me write this in blue. No, Uh, mhm, actually, I will. Yeah, actually, what? Yeah, we will write this in blue. I'll do part. Let's see here. We'll find the pressure amplitude, and we'll just plug that value. And so we'll find we'll do part C first to find pressure, aptitude. And then just plug that end for our value for part being. So, for part be, we'll have again this relationship of pressure impotent. So this is actually so this equation isn't exactly complete the actual Well, what I left out is the reference intensity. I'm not. And we left that out because it is the same. It is the same for all media. But for this problem, we are given the reference intensity before we assume that this one times 10 to the native 12. But we were just comparing the pressure amplitude, so I didn't need to include that since it was common among all media. This time we will include include that value. So the pressure, the max pressure amplitude is equal to the square root of I not to Roe v Yeah. So now I can write that the max. So the pressure amplitude for error is equal to the square root of one times 10 to the negative six. So one times 10 to the native, six times two times the density of error 1.29 times the velocity of air, which is 331 meters per second. Which gives us a value off 0.0 29 Newtons per meter squared, meters squared and all right, that meet am a bit neater. So Newton so 0.0 to 9, known as per meter squared and from previous problems, we should see that we should see that, uh, pressure amplitude for water should be the smallest. So now that's right. So now let's find the pressure amplitude for water. Mhm to the pressure institute for water is equal to the square root screw of one times 10 tint. Native six. Uh huh. So, the square root of one time sent to native six times, two times and I'm gonna run out of room here, So let me expand or move this. Yeah, move this over here a bit so we can have more room, so we'll have to times the density of water, which is one which is didn't see a water is one times 10 to the part of three and at times the velocity of water which velocity of sound through water, which is 1493. But and if you plug all this into a calculator, we get that the max, the displacement aptitude or the pressure amplitude of water is 1.7 28 Uh huh. So 1.7 to 8 meters Newtons Newman squared per meter or noon per meter squared. So one point 7 to 8 Miss per meter squared, please. And finally, our pressure amplitude for iron. Yeah, yeah, yeah, Get it? Yeah, Our pressure amplitude for iron is equal to the square root of the reference intensity. So the reference intensity is one times 10 to native six for this problem, one times 10 to negative, six times the density of iron seven point eight six times 10 to the 53 times are velocity of sound through iron, which is 5000, 5000, 950. So five 1000. Let me get rid of that comma. Yeah, 5000 915 so if we put all of this into a calculator, we get that are value for our pressure. Amplitude of iron is 9.67 New van's per meter square again, we see that iron has the largest pressure amplitude and air has the smallest. So now we can go over here and right our values for our displacement amplitude. So the displacement amplitude of air is our pressure aptitude divided by So for our pressure, amplitude of air. We got 0.0 to 9. So 0.0 to 9 did fight it by okay, density off the density of air. Just 1.29 times the velocity. Yeah, of air, which is which is 331 in times the times the angular frequency, which is 2000 2000 times pi. Uh huh. So you scroll this down, and if we write that into a calculator will get we'll get that are value for pressure. Amplitude would be to see 12 we'll get there are displacement. Aptitude for air is one 0.8 times 10 negative. Eight meters. Look. Yeah, and again from part be of this entire problem. We learned that this Max. This basement aptitude will be the largest because air has the smallest air has the smallest density and velocity at which sound travels through air. So now let's look at the displacement amplitude of water. This placement and put to the water is is one point. So, actually, for our our pressure amplitude, we did to get that the pressure amplitude of water is 1.7 28 newton per meter squared, divided by now. For this, we have to write the density of water. So the density of water, which is one times 10 to the three. Okay, so the density of water times see density of water times the density of water times the velocity at which water or sound travels through water. Okay, so that is, Let's see. Here we have 1493 1400 and 93 times the angler frequency and the English frequency is 2000 pie. Mhm. So, playing all of this into a calculator we have, let's see, 2000 high we have that are value for displacement. Amplitude. The displacement episode of water is 1.8 four times 10 to the negative 10 and this is in meters. So we have a very, very small spaceman aptitude for water. We're gonna even have a smaller one for iron. And we do see that we have for error for the displacement amplitude of air. We have 10 to the power of native eight and this spaceman amplitude of water. We have 10 to the part of negative 10. And for part B of this problem, we got that same difference of our magnitude difference of 10 to the power native to between air and water. So previously we got 10 to the power of negative seven for air and 10 to the negative nine for water. So we see that magnitude of 10 to the are negative to difference. So now let me solve for the displacement amplitude, this placement aptitude of iron. So s Max of of iron. For this we got so for a pressure amplitude we got nine 0.67 divided by so pressure Amplitude divided by yes, our density our identity of iron which is 7.86 times tend to the are of three the payment and that will be multiplied by the velocity velocity of 5950. So that's the velocity of sound. Traveling through iron times are angular frequency, which is 2000 pie. Okay. And this gives us a value of okay, give me a second to type this into a kind of regulator. So we have 9.67 divided by divided by 7.86 to the power of or 10 to the power of three times 5950 times 2000 and pie 2000 times pi we get that are pressure, amplitude or displacement. Amplitude is equal to three point 29 times 10 to the power of negative one. So again we have the displacement. Amplitude of iron is the smallest displacement amplitude of all media. And if you go back to part be of this problem we get, we see that again. We have this magnitude difference between water and iron to be 10 to the power of negative line and tend to the power negative 10. So that's the difference off magnitude of 10 to the negative one power which you see here as well. So for our answer for part d, we have 10 to the power of native 10 and 10 to the power of negative 11. So we see this difference in magnitude to remain. And so essentially we did the same thing. For part, we did part A, B and C in party with giving value. So let's do a quick review of the differences between these two or between the three media we have that we have. We have a wavelength of. So the wavelength for error we found was much smaller than it was the smallest of the three media. And that iron was the largest. And this relationship is just due to the velocity at which sound travels through the respective media. So air, having the smallest velocity gave it the smallest wavelength and iron. Having the fastest, largest velocity gave us the gave it the largest wavelength for part B. We compared the displacement amplitude where we saw that the displacement amplitude really depended on the value off the density and velocity for each media. Where had the smallest values which gave it the largest displacement attitude and iron having the largest identity and the largest velocity gave it the smallest value for this placement amplitude and for pressure amplitude. We got that. The relationship really relied on the density and velocity in which the the velocity are the pressure. Amplitude of air had the smallest value because of the small, because of how small the density and velocity of of airway waas with respect to the same leave and for the pressure, amplitude of iron Way got the largest value because of how big the density and the velocity waas of iron. And for part B, we repeated part a through C with using the values of omega, not the initial angular frequency given and the initial intensity given to us. And we did so we did the exact same thing and saw the exact same relationship, even with the different values given to us and that is it for this problem, I hope you enjoy it and got through the whole thing.


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