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. 7
B. 8
C. 3
D. 5

Answer

A

Explanation

`v_(1) = (1)/(2l_(1)) sqrt((T)/(m)), v_(2) = (1)/((2l_(2))) sqrt((T)/(m))`
`m = 1 g//m = 10^(-3) kg//m`
`v_(1) - v_(2) = (1)/(2) sqrt((T)/(m)) ((1)/(l_(1)) - (1)/(l_(2)))`
`= (1)/(2) sqrt((20)/(10^(-3))) [(l_(2) - l_(1))/(l_(1) l_(2))]`
`= (1)/(2) sqrt((20)/(10^(-3))) [(51.6 - 49.1)/(51.6 xx 49.1)] xx 10^(2)`
`= (sqrt(2) xx 0.1)/(2) xx 10^(2) ~~ 7`