Question

A 5.00 m, 0.732 kg wire is used to support two uniform 235 N posts of equal length (Fig. P15.55), Assume that the wire is essentially horizontal and that the speed of sound is 344 m/s. A strong wind is blowing, causing the wire to vibrate in its 5^th overtone. What are the frequency and wavelength of the sound this wire produces?

Short answer

Thus, the frequency is $2064Hz$ and wavelength is $0.166m$.

Step-by-step solution

Given in the question

Speed of sound $v=344\mathrm{m}/\mathrm{s}$.

The mass of wire $m=0.732\mathrm{k}\mathrm{g}$.

Length of wire $L=5.00m$

Tension on the wire $F=235\mathrm{N}$.

Use formula of frequency and wavelength

The speed is$v=\lambda f=\sqrt{\frac{F}{\mu }}$.

Thus, the formula for frequency and wavelength is given by:

$f=\left(pthovertone+1\right)\times velocityofsoundinair$

Here, $F$ is tension, $\mu$ is linear density, $\lambda$ is wavelength and $f$ is frequency.

Calculate the frequency and wavelength

According to the question,

The frequency is calculated as follows:

$\begin{array}{c}f=\left(5+1\right)\times 344\\ =6\times 344\\ =2064Hz\end{array}$

Thus, the wavelength is calculated as follows:

$\begin{array}{c}v=\lambda f\\ 344=\lambda \times 2064Hz\\ \lambda =\frac{344}{2064}\\ =\frac{1}{6}\\ =0.166m\end{array}$

Hence, the frequency is $2064Hz$ and wavelength is $0.166m$.