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Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is

1. 7
2. 8
3. 3
4. 5

Correct answer: 8

Solution

The frequency of a vibrating string is given by \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length, \( T \) is the tension, and \( \mu \) is the mass per unit length. Since \( T \) and \( \mu \) are the same for both strings, the frequency is inversely proportional to the length. Calculating the frequencies for the two strings and finding their difference gives 8 beats per second.

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