The fundamental frequency of a sonometer wire | vedclass

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(B) The fundamental frequency of the original wire of length $l$ is $n_0 = \frac{v}{2l}$,where $v$ is the wave speed.When a bridge is introduced at a

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$8 n_0 \Delta l / l$

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(B) The fundamental frequency of the original wire of length $l$ is $n_0 = \frac{v}{2l}$,where $v$ is the wave speed.When a bridge is introduced at a distance $\Delta l$ from the center,the wire is divided into two segments of lengths $l_1 = \frac{l}{2} + \Delta l$ and $l_2 = \frac{l}{2} - \Delta l$.The fundamental frequencies of these two segments are:$n_1 = \frac{v}{2l_1} = \frac{v}{2(\frac{l}{2} + \Delta l)} = \frac{v}{l + 2\Delta l}$$n_2 = \frac{v}{2l_2} = \frac{v}{2(\frac{l}{2} - \Delta l)} = \frac{v}{l - 2\Delta l}$The beat frequency is $|n_2 - n_1| = \frac{v}{l - 2\Delta l} - \frac{v}{l + 2\Delta l}$.Using the binomial approximation $(1 \pm x)^{-1} \approx 1 \mp x$ for $x \ll 1$:$|n_2 - n_1| = \frac{v}{l} \left( (1 - \frac{2\Delta l}{l})^{-1} - (1 + \frac{2\Delta l}{l})^{-1} \right)$$\approx \frac{v}{l} \left( (1 + \frac{2\Delta l}{l}) - (1 - \frac{2\Delta l}{l}) \right)$$= \frac{v}{l} \left( \frac{4\Delta l}{l} \right) = \frac{4v}{l} \cdot \frac{\Delta l}{l}$.Since $n_0 = \frac{v}{2l}$,we have $\frac{v}{l} = 2n_0$.Substituting this,the beat frequency is $4(2n_0) \frac{\Delta l}{l} = 8 n_0 \frac{\Delta l}{l}$.

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