Question

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?

Answer 1

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 $\small l$ and the wavelength is $\lambda$,

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

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

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

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

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

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

• Finally combining altogether yield the needed answer,

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