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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 $\Delta l$ from the center of the wire $(\Delta l < < l) .$ The number of beats heard if both sides of the bridges are set into vibration in their fundamental modes are

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Solution:

$f_{0}=\frac{1}{2 L} \sqrt{\frac{T}{\mu}},$
$f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
$\Rightarrow \Delta f=2 f \frac{\Delta l}{l^{\prime}}$
$\Rightarrow \Delta f=2 f_{0} \frac{(2 \Delta l)}{l / 2}$
$=8 \frac{\left(f_{0} \Delta l\right)}{l}$