An open pipe is resonance is its `2^(nd)` harmonic with tuning fork of freqency `f_(1)`. Now it is closed at one end. If the frequency of the tuning fork is increases slowly from `f_(1)`, then agaain a resonance is obtained with `a_(1)` frequency `f_(2)`. if in this case the pipe vibrates `n^(th)` harmonics, then

Options

A. `n=3, f_(2) =3/4 f_1`
B. `n=3, f_(2) =5/4 f_1`
C. `n=5, f_(2) =5/4 f_1`
D. `n=5, f_(2) =3/4 f_1`

Answer

C

Explanation

`f_("open") = (2v)/(2l)` [Resonance frequency]
`f_("closed") = (nv)/(4L)` [Resonance frequency]
`f_2 = n/4 f_1` [ n is odd and `f_2 gt f_1` ]
Hence, n = 5
`f_2 = 5/4f_1`