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The ripples in a certain groove 10.8 $\mathrm{cm}…

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Problem 81 Easy Difficulty

Two strings on a musical instrument are tuned to play at 392 $\mathrm{Hz}(\mathrm{G})$ and 494 $\mathrm{Hz}(\mathrm{B}) .(a)$ What are the frequencies of the first two overtones for each string? (b) If the two strings have the same length and are under the same tension, what must be the ratio of their masses $\left(m_{\mathrm{G}} / m_{\mathrm{B}}\right) ?(c)$ If the strings, instead, have the same mass per unit length and are under the same tension, what is the ratio of their lengths $\left(\ell_{G} / \ell_{B}\right) ?(d)$ If their masses and lengths are the same, what must be the ratio of the tensions in the two strings?

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Chapter 15

Wave Motion

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Video Transcript

so the overtones are given by f N equals to end times F one, which is the fundamental pregnancy and n is toe three for and so on support. So all the anti jizz. So now for G? Um, it's F too, which is equal to twice off the fundamental frequency. 3 90 toe hurts and that's 7 84 hurts and, ah F three here is so that at me. I need a little bit off space here. So if T here is three times 3 92 herds, which is 1176 hurts, we can do the same thing for B. So for B F two is twice off for 94 hearts, which is equal to 9 88 herds on DA three. Here is three times 4 40 Herds, which is 14 82 hurts now for part B. If the two strings have the same length, they have the same wavelength, and the frequency difference is then, due to the difference in the whip speed caused by different masses of the spring. Sorry is master the strings and the mark. US enters the problem through the mass per unit length so you'll see in a minute. What I'm talking about. So we have f frequency JJ off G over frequency off here, which is velocity off G by reeling, divided by velocity off e over whaling. Now the mass enters here by mass per unit length calculations. So we have tension in the string divided by mass per unit length off G, divided by tension in the string eroded by Massport per unit length. And as we mentioned here that, um, the force on the strings are saying so from there we can, right, We can expand this mass per unit length expressions we have ft over MG divided by and divided by square root off FT. Divided by and a buy and and l are same for both. So from there we can get it off Ft and l. So all we have here is square root off AM a divided by mg. So if we take square on both sides, we have m a over M G. That's equal to F G over F A. Holds great. So that's 4 94 divided by 3 92 Who's squared, and that's equal to 1.5 tonight. Now, for part, see if the two strings have the same mass per unit length and the same tension than their waves, speed on both strings is the same. No, the frequency difference is due to the difference in we've linked only and, ah for the fundamental. The wavelength is basically twice the length off the stream, so f d over f B is equal. Do velocity over Lambda Gee, where, as we mentioned that, um, the wave speed or velocity is basically wave speeds. So we've speed a same onboard the strings. And a reason for that is they had the same mass per unit length and the same tension. So we have the over Lambda B. So from there we can write Lambda B over Lambda G. That's equal to twice off Lambda twice off L B times twice off L G. And this is because we know that four fundamental. The wavelength is basically twice the length off the street. So that's why we have to l B and two l g coming in the picture. So from there we can write l d over l G hopes equal do f B over f G because Stuart has matter here, which is 4 94 by 3 92 herds, and that's equal to 1.26 Now, finally, in party, if those two strings have the same length, they will have the same. So if they have the same length, they will have the same wavelength. And they also have the same mass per unit length, which is new basically. So the frequency differences then, due to the difference in we've speed caused by different tensions in the string. So the different thing here is Ft. So this is different and these are saying for both. So we have f B over F A. That's equal toe freebie over lander since Lambda I seem for both. So we have to be a over lambda, which gives us V B by V A. And from there we can write that in terms off tension and mass per unit length. So we have FT B bye mule, which is mastering it link with the same for both. So we have square root off f t a buy me, which gives us square root off f d be by f t eight on dhe. From there, if we take square on both sides. We have f T B by F d a. That's equal Do small left B by small f a a. And then there's a square. So now we can use Ah, the frequencies that's given So we have FT be by if looks f d a. Has equal do 3 92 That's F D by f averages for 94. Then there's a square. So that gives us 0.630 All right, Thanks for watching.

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Physics for Scientists and Engineers with Modern Physics

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Watch More Solved Questions in Chapter 15

Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Christina Krawiec

Marshall Styczinski

Meghan Miholics

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Christina Krawiec

Marshall Styczinski

Meghan Miholics

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Elyse Gonzalez

Christina Krawiec

Marshall Styczinski

Meghan Miholics

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics