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

Options

A. 5
B. 7
C. 8
D. 3

Answer

B

Explanation

Since, `f=(1)/(2l)sqrt((T)/(mu))`
Number of beats `=f_(2)-f_(1)`
`=(1)/(2)sqrt((T)/(mu))[(1)/(l_(2))-(1)/(l_(1))]`
No. Of beats = `=(1)/(2)sqrt((20)/(1xx10^(-3)))[(1)/(49.1xx10^(-2))-(1)/(51.6xx10^(-2))]`