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Q.

The fundamental frequency of a sonometer wire of length $l$ is $f_{0}$ . A bridge is now
introduced at a distance of $Δ l$ from the centre of the wire $(Δ l < < l)$ . The number of beats heard per second if both sides of the bridge are set to vibrate in their fundamental mode is

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a

$\frac{f_{0} Δ l}{l}$

b

$\frac{8 f_{0} Δ l}{l}$

c

$\frac{2 f_{0} Δ l}{l}$

d

$\frac{4 f_{0} Δ l}{l}$

answer is A.

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Detailed Solution

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$n = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = f_{0} ........ (1)$ $n_{1} = \frac{1}{2 l_{1}} \sqrt{\frac{T}{u}}$ , $n_{2} = \frac{1}{2 l_{2}} \sqrt{\frac{T}{u}}$
$l_{1} = \frac{l}{2} - Δ l$ , $l_{2} = \frac{l}{2} + Δ l$
$Δ n = n_{1} - n_{2}$ and $Δ l^{2}$ can be neglected.

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