The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by 8 N , it is in resonance with the first overtone of the closed tube. The initial tension in the wire is

Options

A. `16N`
B. `8N`
C. `4N`
D. `1N`

Answer

D

Explanation

`T_(2)=T_(1)+8`
`n_(1)=(V)/(4l)`, `n_(2)=(3V)/(4l)`
`n prop sqrt(T)implies(n_(1))/(n_(2))=sqrt((T_(1))/(T_(2)))implies(1)/(3)=sqrt(T_(1))/(T_(1)+8)`