The fundamental frequency of a sonometer wire of length $l$ is $n_0$ . A bridge is now introduced at a distance of $\Delta l ( < < l)$ from the centre of the wire. The lengths of wire on the two sides of the bridge are now vibrated in their fundamental modes. Then, the beat frequency nearly is
Diffcult
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Note that the beat frequency is $\left(\mathrm{n}_{1}-\mathrm{n}_{2}\right)$ and the corresponding vibrating lengths are $(\ell /2 - \Delta )$ and $(\ell /2 + \Delta \ell )$
Given $:{{\rm{n}}_0} = \frac{{\rm{V}}}{{2\ell }}$
$ \Rightarrow \frac{V}{\ell } = 2{n_0}$
$\mathrm{n}_{1}=\frac{\mathrm{V}}{2\left(\frac{\ell}{2}+\Delta \ell\right)}=\frac{\mathrm{V}}{\ell+2 \Delta \ell}$
$\mathrm{n}_{2}=\frac{\mathrm{V}}{2\left(\frac{\ell}{2}-\Delta \ell\right)}=\frac{\mathrm{V}}{\ell-2 \Delta \ell}$
Beat frequency $=\mathrm{n}_{2}-\mathrm{n}_{1}=\frac{\mathrm{V}}{\ell-2 \Delta \ell}-\frac{\mathrm{V}}{\ell+2 \Delta \ell}$
$=\frac{V}{\ell\left(1-\frac{2 \Delta \ell}{\ell}\right)}-\frac{V}{\ell\left(1+\frac{2 \Delta \ell}{\ell}\right)}$
$=\frac{V}{\ell}\left(1-2 \frac{\Delta \ell}{\ell}\right)^{-1}-\frac{V}{\ell}\left(1+2 \frac{\Delta \ell}{\ell}\right)^{-1}$
$=\frac{V}{\ell}\left(1+2 \frac{\Delta \ell}{\ell}\right)-\frac{V}{\ell}\left(1-2 \frac{\Delta \ell}{\ell}\right)\left[(1+x)^{n}=1+\operatorname{nx}\right]$
$=\frac{V}{\ell}+\frac{2 V}{\ell} \frac{\Delta \ell}{\ell}-\frac{V}{\ell}+\frac{2 V}{\ell} \frac{\Delta \ell}{\ell}$
$=2 \cdot 2 n_{0} \cdot \frac{\Delta \ell}{\ell}+2 \cdot 2 n_{0} \cdot \frac{\Delta \ell}{\ell}=8 n_{0}\left(\frac{\Delta \ell}{\ell}\right)$
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