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Each of the two strings of length 51. 6 cm and 49.1 cm. are equally tensioned by 20 N force. Mass per unit length of both the strings is the same and equal to 1 g/m. When both the strings vibrate simultaneously in fundamental mode the number of beats is

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Detailed Solution

$I_{1} = 0 .516 m, l_{2} = 0 .491 m, T = 20 N$

Mass per unit length, $μ = 0.001 kg / m$

Frequency, $v - \frac{1}{2 l} \sqrt{\frac{T}{μ}}$

$v_{1} = \frac{1}{2 \times 0.516} \sqrt{\frac{20}{0.01}}$

$v_{2} = \frac{1}{2 \times 0.491} \sqrt{\frac{20}{0.01}}$

$∴ Number of beats = v_{1} - v_{2} = 7$

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