The fundamental frequency of a sonometer wire of length l is f₀. A bridge is now introduced at a distance $Δ l$ from the centre of the wire $(Δ l < < l)$ . The number of beats heard if both sides of the bridge are set into vibration in their fundamental modes are :

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$2 f_{o} \frac{Δ l}{l}$

$f_{o} \frac{Δ l}{l}$

$8 f_{o} \frac{Δ l}{l}$

$4 f_{o} \frac{Δ l}{l}$

answer is A.

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Detailed Solution

$\begin{array}{l} We know fundamental frequency \\ f_{o} = \frac{v}{2 l} \\ \Rightarrow v = 2 l f_{o} \\ Beat frequency, f_{1} - f_{2} = \frac{v}{2 (\frac{l}{2} - Δl)} - \frac{v}{2 (\frac{l}{2} + Δl)} \\ \Rightarrow f_{1} - f_{2} = 2 f_{o} l \{\frac{4 Δl}{l^{2} - 4 {Δl}^{2}}\} \approx \frac{8 f_{o} Δl}{l} \end{array}$