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Suppose that depicts standing waves on a metal string under a tension of $88.2 \mathrm{~N}$. Its length is $50.0 \mathrm{~cm}$ and its mass is $0.500$ g. ( $a$ ) Compute $v$ for transverse waves on the string. (b) Determine the frequencies of its fundamental, first overtone, and second overtone. (a) $v=\sqrt{\frac{\text { Tension }}{\sqrt{\text { Mass per t unit length }}}}=\sqrt{\frac{88.2 \mathrm{~N}}{15.00 \times 10^{-4} \mathrm{~kg} /(0.500 \mathrm{~m})}}=297 \mathrm{~m} / \mathrm{s}$ (b) We recall that the length of the segment is $\lambda / 2$ and we use $\lambda=$ $\mathrm{u} / \mathrm{f}$. For the fundamental: $$ \lambda=1.00 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{1.00 \mathrm{~m}}=297 \mathrm{~Hz} $$ For the first overtone: $$ \lambda=0.500 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{0.500 \mathrm{~m}}=594 \mathrm{~Hz} $$ For the second overtone: $$ \lambda=0.333 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{0.333 \mathrm{~m}}=891 \mathrm{~Hz} $$

In this problem were given an equation representing the frequency at which gets a guitar string is by breeding. If you'd like more awful details, I would recommend looking at the problem in your textbook. We know that the length of the string is half a foot and that the ratio of the tension to the density is 220 feet per second were also given that the length is changing at a rate of negative four feet per second. Given those conditions were asked to find F Ramadi. So first, let's put in the information we knew it all requisition so ever be 1/2 times l and we knew that this ratio is equal to 200 20. So that simplifies all the equation for an to 110 invented by the length of this of the street and can remember This also means oh can be rewritten as 1 10 times alter the negative one so that when we find if prime, we have negative 1 10 or her l queen. So now that we have the equation for at fine when the string is half a foot, then the frequency is changing at a rate Oh, 1760 hurts for a second now. We were also told, given this reach, how can How long will it take for could pitch to be raised one octave? Or how long will it take to double the frequency? If in order to find that out first, I will find what efforts when the length is 1/4 exploding at 1/4 but 1/2 and that is 200 and 20 hertz ical. So to find a home long text during two double the frequency from Tu 22 for 40 her, we'll take, Ah, 220 Hertz, divided by our current route 1760 and that gives one of a second. So when the length is one, have put the frequency changes at a ridge of 1760 hertz per second on it would take 1/8 of a second to double the frequency, or is the pitch by one octave

So here we know that the length is the length is going to be equal to 63.5 centimeters or 0.635 meters. We know that the fundamental frequency is equal in 245 hertz. We know that the wavelength is equal to two times the length and that for part a, the speed of propagation will equal the wavelength times the frequency times of fundamental frequency. And this legal to two times the length times the fundamental frequency. And this will equal two times 20.635 times 245. And this is equaling 311 meters per second. Now for part B, we know that the frequency is going to be directly proportional to the speed of procreation. Therefore, the frequency is also going to be directly proportional to the square root of the period tee. So we can say that of the new fundamental frequency is going to be equal. Tio 245 times the square root of 1.1 And this equals 246 hertz. Now for port. See, they're asking us for the wavelength in air and this is simply going to be the velocity of the air divided by the frequency. And this will be 344 meters per second. This would be the speed of sound and air, divided by the frequency of 245 the fundamental frequency. And this is equaling 1.40 meters. Now here the frequency is going to be the same because it doesn't matter for the frequency. It doesn't matter about the medium. So the frequent. So the medium does not affect the frequency in this case. And the wavelength is changed from two times l 21.40 meters and commenting on this. We know that increasing the force tension ah is going to be is going to increase the wave speed and then this in turn increases the frequency. That is the end of the solution. Thank you for watching

In this question. The given pressure function is people's too. 0.003 sign 225 90 plus 0.003 divided by three science 665 80 plus 0.003 divided by five sign 1100 pity plus zero point 003 divided by seven sign 1540 piety. This motion position of three parts. We will start with party in this part we have to duplicate the graph as given in the question. So here is a craft. Now the graph of P over the film viewing window. Zero comma 0.0 T. Bye -0.005 Coma zero point It was 05 is given by So this is our graph, this is our X axis and this is our by access. Now with a look party in this part we have to describe the shape of the sound wave that that is produced. Now from this graph we can say that the gulf is periodic and the wave is Dell square tops and four times now the new party in this part we have to do the mine The times over the interval 0-0.03 when this will occur. So using the graphical calculator, it is observed that in the intervals Little Point Little Little 4 5 coma zero point People you know 91 coma zero point Do you know 136 00.0182 and zero point do you know double to seven coma 0.0273 over the interval 0: 0.03. The ear drum the most outward. Thank you.

So for this problem, we're taking a look at the pressure that sound makes according to the function that's given in the problem. So part is just to make a graft and you know, it's it's notice. It's kind of level, um, up it's the top and then goes down. Uh, the more frequencies we put in here and on frequencies, um, the more it will turn into a square wave. And if you go on in math, it's called a Fourier analysis. Um, where we're doing this. So, um, heart B is gonna ask us to calculate the derivative and interpret that. So first, we're just gonna need to go ahead and differentiate this thing. So, um, in each of these pieces were going to get to 20 pie co sign to 20 pi ti plus ah, third of the 6 60 So we have 1/3 of the 6 60 pi times co sign 6 60 pi ti plus, we get 1/5 of the 1100 co sign 1100 high tea and 1/7 of the co sign 15 40 pi ti, um and with that work's gonna go ahead and plug in our 0.2 Right here. And it comes out to be about 1.5 Um, and the units on that thing, we're going to get pounds per square foot, and that's per second because it's a rate. So the way we're gonna interpret that as we go back to our original graph and we're looking at the function Um right, let's get to our pencil here. So we get to our point right here the slope on this graph right here, um is a bounce 1.5 pounds per square foot right here. So that's how much it's changing at time. 1.54 point 002


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