- Home
- Standard 11
- Physics
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
$n_0\ \Delta l/l$
$8\ n_0\ \Delta l/l$
$2\ n_0\ \Delta l/l$
$n_0\ \Delta l/2l$
Solution
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 $=\mathrm{n}_{2}-\mathrm{n}_{1}=\frac{\mathrm{V}}{\ell-2 \Delta \ell}-\frac{\mathrm{V}}{\ell+2 \Delta \ell}$
$=\frac{V}{\ell\left(1-\frac{2 \Delta \ell}{\ell}\right)}-\frac{V}{\ell\left(1+\frac{2 \Delta \ell}{\ell}\right)}$
$=\frac{V}{\ell}\left(1-2 \frac{\Delta \ell}{\ell}\right)^{-1}-\frac{V}{\ell}\left(1+2 \frac{\Delta \ell}{\ell}\right)^{-1}$
$=\frac{V}{\ell}\left(1+2 \frac{\Delta \ell}{\ell}\right)-\frac{V}{\ell}\left(1-2 \frac{\Delta \ell}{\ell}\right)\left[(1+x)^{n}=1+\operatorname{nx}\right]$
$=\frac{V}{\ell}+\frac{2 V}{\ell} \frac{\Delta \ell}{\ell}-\frac{V}{\ell}+\frac{2 V}{\ell} \frac{\Delta \ell}{\ell}$
$=2 \cdot 2 n_{0} \cdot \frac{\Delta \ell}{\ell}+2 \cdot 2 n_{0} \cdot \frac{\Delta \ell}{\ell}=8 n_{0}\left(\frac{\Delta \ell}{\ell}\right)$
