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A light steel string of unit length produces a fundamental frequency of $f$ . The wire is clamped close to the middle so that the two segments of wire differ in length by a small amount $e$ . If the two segments vibrate in fundamental mode, then the number of beats produced is

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$f e$

$4 f e$

$f e / 4$

$f e / 2$

answer is B.

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Detailed Solution

$f = \frac{v}{2 L} = \frac{v}{2}$

The lengths of the segments are $L_{1} = \frac{1}{2} + \frac{e}{2}$ and $L_{2} = \frac{1}{2} - \frac{e}{2}$

$∆ f = f_{2} - f_{1} = \frac{v}{2 L_{1}} - \frac{v}{2 L_{2}} = f (\frac{1}{\frac{1}{2} - \frac{e}{2}} - \frac{1}{\frac{1}{2} + \frac{e}{2}}) ≃ 4 f e$

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