An open pipe is in resonance in `2nd` harmonic with frequency `f_(1)`. Now one end of the tube is closed and frequency is increased to `f_(2)` such that the resonance again ocuurs in `nth` harmonic. Choose the correct option

Options

A. `n=3,f_(2)=(3)/(4)f_(1)`
B. `n=3,f_(2)=(5)/(3)f_(1)`
C. `n=5,f_(2)=(3)/(4)f_(1)`
D. `n=5,f_(2)=5/4 f_(1)`

Answer

D

Explanation

`therefore f_1=(2)(c)/(2l)=c/l`…(i)
`lambda =( 4 1)/n`….(ii)
`therefore f_2=(v)/(lambda)=(nc)/(4l)`
Given `f_2 gt f_1`
From Eqs. (i) and (ii), we get :
For the first resonance
`n=5, f_2=5/4 f_1`