A thin metallic wire of mass per unit length 0.02kg/m is clamped at three points A, B and C and a block of weight 50 N is suspended from the wire as shown. The length AC = 25 cm and the length CB = 26 cm. If both segments (AC and CB) of the wire are made to vibrate in their fundamental modes, the number of beats heard per second (to the nearest integer) is ________.

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answer is 4.

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Detailed Solution

Speed of sound in the wire, $c = \sqrt{\frac{50}{0.02}} = 50 m / s$
So, the beat frequency, $f_{B} = \frac{50}{2 (0.25)} - \frac{50}{2 (0.26)} \approx 4$

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