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)/(4) f_1`
C. `n = 5, f_2 = (3)/(4) f_1`
D. `n = 5, f_2 = (5)/(4) f_1`

Answer

D

Explanation

`lamda = l`
`:. f_(1) = (v)/(lamda) = (v)/(l)`

`lamda = (4l)/(n)`
`:. f_(2) = (v)/(lamda) = (nv)/(4 l)` …(ii)
Here `n` is an odd number
`f_(2) = (n)/(4) f_(1)`
From Eqs. (i) and (ii), we get
For the first resonance
`n = 5`,
`f_(2) = (5)/(4) f_(1)`
Given `f_(2) gt f_(1)`.