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A $\mathrm{A} 75.0 \mathrm{cm}$ wire of mass 5.625 $\mathrm{g}$ is tied at both ends andadjusted to a tension of 35.0 $\mathrm{N} .$ When it is vibrating in its sec-ond overtone, find (a) the frequency and wavelength at whichit is vibrating and (b) the frequency and wavelength of thesound waves it is producing.

a) Round off to three significant figures, the frequency of the second overtone of the soundwave in the wire is 137 $\mathrm{Hz}$The wavelength of the second overtone of the sound wave in the wire is $0.500 \mathrm{m}$b)The frequency of produced sound wave for the second overtone is $137 \mathrm{Hz}$The wavelength of the produced sound wave for the second overtone is $2.51 \mathrm{m}$

Physics 101 Mechanics

Chapter 12

Mechanical Waves and Sound

Periodic Motion

Mechanical Waves

Sound and Hearing

University of Washington

Simon Fraser University

University of Sheffield

Lectures

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A thin, 75.0 -cm wire has …

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A thin, 75.0-cm wire has a…

so in the length is equal to 70 75 centimeters or we can save 0.75 meters. We know that the masses equalling 5.6 to 5 grams, we can convert this into 0.0 56 to 5 kilograms. And we know that the tension there's going to be 35 Newtons without will actually say forced tension. Okay. And at this point, they want us to find the speed of propagation and the frequency of the second overtone and the ah, the frequency and the wavelength of the second overtone. So we know that the velocity is equal to the square root of the force. Tension divided by, um, you and mu is simply going to be the linear densities. The mu is simply the mass divided by the length. And so we can say that the force rather force the frequency of the harmonic will be equal to end times we the speed of sound and air divided by two times out. Rather not space on the other. The speed of the wave, the speed of propagation and so for party. When we're trying to find the speed of propagation, we're simply going to solve for V, so V will simply be equal to the square root of 35.0 Nunes the force tension divided by the linear density. So this will be 0.56 to 5 and then divided by 0.75 and this is equaling 68.3 meters per second. So now if you wanted to find the frequency of the second overtone or the third Harmon Nick, this will be three v divided by two times Tell and this will be three times 68 0.3, divided by two times 20.75 and we're getting 135 hertz for the second overtone or again the third harmonic for the wavelength of the third harmonic or the second overtone. This will be the frequently of velocity divided by the frequency. So 68.3, divided by 135 and we're getting that thie a wavelength of the second overtone is 0.506 meters. So that's your answer to party. And then for part B, they want us to find the I knew the wavelength. If the speed of propagation was the speed of sound in air. So we know that the frequency is 137 hertz. But now the is going to equal the speed of sounded air 344 meters per second. So wave length would be again the frequently philosophy about about the frequency. So it will be 344 divided by 137 and this is giving us 2.51 meters. And as you can see, the wavelength is increasing as the speed of the propagation speed increases. So we can say that the wave length is going to be directly proportional to the speed of the wave. That is the end of the solution. Thank you for watching.

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