A sonometer wire under a tension of $64 \mathrm{N}$ vibrating in its fundamental mode is in resonance with a vibrating tuning fork. The vibrating portion of the sonometer wire has a length of $10 \mathrm{cm}$ and a mass of $1 \mathrm{g}.$The vibrating tuning fork is now moved away from the vibrating wire with a constant speed and an observer standing near the sonometer hears one beat per second. Calculate the speed with which the tuning fork is moved. The speed of sound in air is $300 m{s}^{-1}.$

$\begin{array}{l}T=64\mathrm{N}\\ L=10\mathrm{cm}=0.1\mathrm{m}\\ m=\frac{1\times {10}^{-3}\mathrm{kg}}{0.1\mathrm{m}}={10}^{-2}{\mathrm{kgm}}^{-1}\end{array}$

The frequency of the sonometer wire vibrating in the fundamental mode is given by 

$v=\frac{1}{2L}\sqrt{\frac{T}{m}}=\frac{1}{2\times 0.1}\times \sqrt{\frac{64}{{10}^{-2}}}=400\mathrm{Hz}$

As the wise is in resonane with the tuning fork, the frequency of the fork $=$ frequency of the wire, i.e., $v = 400 Hz$ Let $u_s$ be the speed with which the fork is moved away from the observer standing close to sonometer. The apparent frequency of the sound of the fork is given by

$v^{\mathrm{\prime }}=\nu \times \frac{v}{\left(v+u_s\right)}$

where $v=$speed of sound $=300 m{s}^{-1}$

As 1 beat is heard per second, $v-v\text{'}=1$

or $v^{\mathrm{\prime }}=v-1=400-1=399\mathrm{Hz}$

Now $\frac{v}{v^{\mathrm{\prime }}}=\frac{\mathrm{v}+u_s}{\mathrm{v}}=1+\frac{u_s}{\mathrm{v}}$

or $\frac{v-v^{\mathrm{\prime }}}{v^{\mathrm{\prime }}}=\frac{u_s}{\mathrm{v}}\text{ or }{u}_{\mathrm{s}}=\mathrm{v}\times \frac{v-v^{\mathrm{\prime }}}{v^{\mathrm{\prime }}}$$=300\times \frac{(400-399)}{399}=0.752{\mathrm{ms}}^{-1}$