The fundamental frequency of a sonometer wire | vedclass

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Question

Note that the beat frequency is $\left(\mathrm{n}_{1}-\mathrm{n}_{2}\right)$ and the corresponding vibrating lengths are $(\ell /2 - \Delta )$ and $(\

Vedclass · Accepted Answer

$8\ n_0\ \Delta l/l$

Vedclass · Answer

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 $=\mathrm{n}_{2}-\mathrm{n}_{1}=\frac{\mathrm{V}}{\ell-2 \Delta \ell}-\frac{\mathrm{V}}{\ell+2 \Delta \ell}$

$=\frac{V}{\ell\left(1-\frac{2 \Delta \ell}{\ell}\right)}-\frac{V}{\ell\left(1+\frac{2 \Delta \ell}{\ell}\right)}$

$=\frac{V}{\ell}\left(1-2 \frac{\Delta \ell}{\ell}\right)^{-1}-\frac{V}{\ell}\left(1+2 \frac{\Delta \ell}{\ell}\right)^{-1}$

$=\frac{V}{\ell}\left(1+2 \frac{\Delta \ell}{\ell}\right)-\frac{V}{\ell}\left(1-2 \frac{\Delta \ell}{\ell}\right)\left[(1+x)^{n}=1+\operatorname{nx}\right]$

$=\frac{V}{\ell}+\frac{2 V}{\ell} \frac{\Delta \ell}{\ell}-\frac{V}{\ell}+\frac{2 V}{\ell} \frac{\Delta \ell}{\ell}$

$=2 \cdot 2 n_{0} \cdot \frac{\Delta \ell}{\ell}+2 \cdot 2 n_{0} \cdot \frac{\Delta \ell}{\ell}=8 n_{0}\left(\frac{\Delta \ell}{\ell}\right)$

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

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