Question

A 75.0-cm-long wire of mass 5.625 g is tied at both ends and adjusted to a tension of 35.0 N. When it is vibrating in its second overtone, find (a) the frequency and wavelength at which it is vibrating and (b) the frequency and wavelength of the sound waves it is producing.

Short answer

The frequency and wavelength of the string is 136.6Hz and 0.5m respectively and the frequency and wavelength of the string is 136.6Hz and 2.518m respectively

Step-by-step solution

STEP 1 Concept of the frequency for standing wave for the strings fixed at both ends,

Thefrequency for standing wave for the strings is given as$\mathbf{f}\mathbf{=}\mathbf{n}\frac{\mathbf{v}}{\mathbf{2}\mathbf{L}}$were,f is the frequency of nth harmonic, v is the velocity of the wave,$\mathbf{n}\mathbf{\to }\mathbf{nth}$harmonic (n — 1, 3, 5, ...), L is the length of the pipe.

Calculate the frequency and wavelength of the string

For the string

$$\begin{gathered} \mathrm{\mu }=\frac{\mathrm{m}}{\mathrm{L}}=\frac{5.625\times {10}^{-3}}{0.75}=7.5\times {10}^{-3}\mathrm{kg}/\mathrm{m} \\ \mathrm{v}=\sqrt{\frac{\mathrm{F}}{\mathrm{\mu }}}=\sqrt{\frac{35\mathrm{N}}{7.5\times {10}^{-3}\mathrm{kg}/\mathrm{m}}}=68.313\mathrm{m}/\mathrm{s} \end{gathered}$$

Second overtone is third harmonic n=3

$$\begin{gathered} {\mathrm{f}}_3=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{L}}=3\times \frac{65.32\mathrm{m}/\mathrm{s}}{2\times 0.75\mathrm{m}}=136.6\mathrm{Hz} \\ \mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}}=\frac{68.32}{136.6}=0.5\mathrm{m} \end{gathered}$$

Therefore, the frequency and wavelength of the string is 136.6Hz and 0.5m respectively

Step 3 Calculate the frequency and wavelength of the sound

The string is the source of vibration, so, the sound has the same frequency 136.6Hz

${\mathrm{\lambda }}_{\mathrm{sound}}=\frac{\mathrm{v}}{\mathrm{f}}=\frac{344}{136.6}=2.518\mathrm{m}$

Therefore, the frequency and wavelength of the string is 136.6Hz and 2.518m respectively