Answer to Question #180682 in Mechanics | Relativity for Royce Karam

Question #180682

A musical note on a piano has a frequency of 40 Hz. If the tension in the 2-m string is 308 N, and one-half wavelength occupies the string, what is the mass of the wire?


1
Expert's answer
2021-04-13T06:28:29-0400

Explanations & Calculations


  • Since a half of the wavelength is contained in the length of the wire as it make a stationary wave we can write if the length of the wire is l\small l and the wavelength is λ\lambda,

λ2=lλ=2l\qquad\qquad \begin{aligned} \small \frac{\lambda}{2}&=\small l\to\lambda=2l \end{aligned}

  • As by now the wavelength is known we can apply the common equation v=fλ\small v=f\lambda for the stationary wave generated on the wire,

v=40×2l=40×2×2m=160ms1\qquad\qquad \begin{aligned} \small v&=\small 40\times 2l\\ &=\small 40\times 2\times 2m\\ &=\small 160\,ms^{-1} \end{aligned}

  • Then, according to the equation for the speed of a wave on a wire as,

v=TmTlM\qquad\qquad \begin{aligned} \small v&=\small \sqrt{\frac{T}{m'}}\to\sqrt{\frac{Tl}{M}} \end{aligned}

where m\small m' is the linear density of the wire whose total mass is M\small M and the length is l\small l.


  • Finally combining altogether yield the needed answer,

TlM=160M=Tl(160ms1)2308N×2m1602m2s2=0.02406kg=24.06g\qquad\qquad \begin{aligned} \small \sqrt{\frac{Tl}{M}}&=\small 160\\ \small M&=\small \frac{Tl}{(160\,ms^{-1})^2} \to \frac{308N\times 2m}{160^2 \,m^2s^{-2}}\\ &=\small 0.02406\,kg\\ &=\small \bold{24.06\,g} \end{aligned}


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