Question

Each of the 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 the same and equal to `1gm^(-1)`. When both the strings vibrate simultaneoulsy the number of beats is
A. `5`
B. `7`
C. `8`
D. `3`

Answer 1

Correct Answer - B
The number of beats will be the difference of frequencies of the two springs
Frequency of first string `f_1=(1)/(2l_1)sqrt((T)/(m))`
`=(1)/(2xx51.6xx10^-2)sqrt((20)/(10^-3))=137.03Hz`
Similarly frequency of second string
`=(1)/(2xx49.1xx10^-2)sqrt((20)/(10^-3))=144.01`
Number of beats `=f_2-f_1=144-137=7beats`