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Janlion1.837 PoneTha spood sound air i6 340 IVs; and the densily olair [s 1.2 K9/m-3 Whal micromaier and Ihe pressure variation aniplilude 348.39 Pa? none Ihese 0 5...

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Janlion1.837 PoneTha spood sound air i6 340 IVs; and the densily olair [s 1.2 K9/m-3 Whal micromaier and Ihe pressure variation aniplilude 348.39 Pa? none Ihese 0 537 0 895 0 716 0 358Irequanc; (In Hz) sound wave If tne diielcument amaltude

Janlion 1.837 Pone Tha spood sound air i6 340 IVs; and the densily olair [s 1.2 K9/m-3 Whal micromaier and Ihe pressure variation aniplilude 348.39 Pa? none Ihese 0 537 0 895 0 716 0 358 Irequanc; (In Hz) sound wave If tne diielcument amaltude



Answers

As a certain sound wave travels through the air, it produces pressure variations (above and below atmospheric pressure) given by $\Delta P=1.27 \sin (\pi x-340 \pi t)$ in SI units. Find (a) the amplitude of the pressure variations, (b) the frequency, (c) the wavelength in air, and (d) the speed of the sound wave.

In the party of this problem we have to find the amplitude pressure that is written as a um be maximum. The pressure variation is given by the equation as delta P is equal to 1.27 paschal sign of bikes -345 two. We call it equally number one. The general equating for the pressure variation is original. Delta P is equal to p maximum sign of K X minus omega T. So this will be our equation number two. Now using equation number one and number two we can white here that this p maximum is equals two 1.27 Basket. So this is a required answer. Now let's move to the body be of this problem. Okay in this case we have to find the frequency of the sound wave that is if campaigning equation number one with the question number two we can white the angle frequency as omega is equal to 340 by /s. So 345. Uh so we can write this, omega is omega is equals to two pi F. So this will be equals to 340 pie. So from here we can wire the safest physicals to 340 pi divided by two pi. So from here we will get the value for this frequency as their physicals to 170 yards. Mhm. So this is the answer to the party of this problem. In the party of this problem we have to find the equivalent of the here that is lambda. Once again comparing equation number one with the question number two, we can white. The weft number as case equals two. Bye per meter. We can define the wave number as case equals to two pi divided by lambda. So from here we can widely squatting for the lambda as lambda is equals to two pi divide by. Okay, So this can be written as a two by divide by pi formative. So from here we will get the value for this lambda As Lyme Disease Equals two two m. So this is the answer to the party of this problem. Yeah. In the part D of this problem we have to find the speed of the wave. That is we we can wipe the speed of the VFW as well as equals to F lambda. So we is equals two 170 Herds into two major. So from here we will get the value for this way as well as equals two 340 m/s. So this is the speed of the bath. Mhm.

Okay, so we have sound. Wait, That ex traveling in air We know the velocity of something there We know the frequency of the way we know What is the pressure on in the displacement camp? Little off the wave. We make unit compressions leaders and we know that the models of air is this body busca Yes. So we wanted to find What is the pressure amplitude for these waves travelling. So for this, we have to remember the the equation off the pressure attitude is the book modelos one supply by the wave number Work applies by the amplitude, the pressure, nothing in them, the displacement completely. So it is pretty simple because we also know that the weight of number is two pi be worried by lambda or two pi time Is the frequency divided by Thanks. So, actually, this is going to be be signs A I am Stoop, I the frequency all of these divided by the velocity on the way. So I, uh, input the balance here. I'm going to have when going for two times 10 to the five. Rascal, uh, climbs the other two disease five times. 10 to the minus six meter, uh, to fight on the frequency that is 150 and all of the right. And by the velocity, that is 344. Hey, if you make the speculations, you're going to find that the answer is 1.94 SPL's is the answer of the first. For the second part, we want to find the intensity. Okay? And we know that, uh, have the same data and us before. And the only thing the changes is the crucial facility equation for the city is one health. The angular frequency, the Volt models, the wave number on and the I'm a little this work. Okay, so we know your value for the wave number, But we also know that they angular frequency is actually so flight in a fight with the equipments. So if we saw this this into this equation, we are going to find this is 1/2 of to buy does the frequency since the bulk models times to buy things, the frequency divided by the velocity. This is the value of K. I want to play by a square. Okay. So, uh, to to stencil to their out and this is going to be so Is this the two left by square? Okay, the frequency square, Uh, times that both modelers multiplied by the after did square all of these divided by the velocity. Okay, so we just input the data. These data on this is to I square, 150 Hertz square. Uh, see the book, Vardalos, That is 1.42 times 10 to the five. Let's go. This is not square. Uh, signs. The amplitude. There is five. I understand. For the final six. Leader squared, divided by the velocity 344 interspersing So, making this calculations done, get the square diversity for going to 58. And then So there, minus three. What's the sport meter? This is the second. And for the final answer, I have to find their sound intensity level. Okay. And I do this with the year questions. Does that be? It's actually 10. This is else isn't just for the units things the low? Very, uh, the intensity divided by there, Uh, this value this convention. Okay, we know that this is, uh, one. I'm stand to the minus of what's the word meter Okay, so if I just use the intensity there count before that is actually this one. I'm going to find that this is 10 lover. It'll, uh, or 0.58 times 10 to the minus three. Divided by this value is is a four. They want times 10 to the minus. 12 on this is going to be 96. Ah, point six. This is the last answer.

So the PDR capital to is given Toby two milliseconds or see your point CEO 02 seconds. Excuse me. On the amplitude, the B m is eight on. This is a very violent or cedar point year 08 movement only. Describe now from the cushion 17 point 15 We can write them to be equal to l b m. It is this amplitude over Mule Joe Times omega. So this is actually very little time. So, maker on Omega here is the angular velocity So we can write that as to buy over Times media And now the values core density on DDE velocity to be 3 43 Meet Oppa Seconds of this system velocity of song. So we blood these four into the secretion and find I stand that is the this place now amplitude off the displacement function off the wave to be 6.1 times 10 to the minus mine weakness. Now, after we had this can find the angular, brief number to be Mingo would be. And this is basically a toe to buy overs velocity times time period on again. We can plug all the values into the seclusion and find this to be 9.2 radiant. We'll need it. Then The angular frequency can we find all found out by calculating to buy over time, period on. And we used this value over here and find the angular velocity Toby nearly to be equal to 3.142 times into the bar. Three Gideon per second. And this is approximately equal to 3.1 times into the part three video for second now, still, no, we can somebody's that isms. As so we're given the displacement, Toby. Like this. So this is the expression. And now that we have all the values, the foot in door, this expression so we can do that and find the displacement. Victor said we got well, s m family, dude to the 6.1 man on video cause off Gay gay. We found that the nine point do made that invest because this radiant is has no dimension. Or you can say that you Nicolas Dimes X minus chief 0.1. Many hopes into the bar three seconds in verse. Okay, the second in worse should be inside because that's the Unix Theis. And then deep. So this is, uh, the displacement Victor? No, for party. We can use similar reasoning as you did for but ABC. But oh, now we need to use new values for your dash on dhe readers, which are the air density and velocity. So the velocity or the speed is now 20 meters a second. And the new density is 1.35 Katie for me, don't you? So now we can find it's him going the same way. So s then comes up to be. So we block the values using the same expressions of eggs they'll be ends on dhe read as Rodas Times to buy over time. Did it? Remember that the time period on dhe the amplitude off the pressure they'll be in. They are Constance here. So we use the same values is before and then tacular this to be 5.9 times 10 to the minus nine meter. On the same day we find Kato be 9.8 lady in Lomita because they're Deena's. Basically no units, no dimensions. You can write that 900 meter investors like with we did over you on dhe. Find only got to be. Since Omega is just depending on the time period, which is a constant here. So maybe I should be going through this value and spots. Um, now that we have all the components, we can again substitute these values in this a question on find AJ Displacement function to be 5.99 on Lido Dying scores off 9.8 during laws Dines X linus she point one time stand to the party. Second in words, which is Oh, my God. Angsty. So this is knew the specimen function and this was 41.

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.


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