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

Answer

C

Explanation

For an open pipe , fundamental frequency `(N) = V/(2L) `
` :. ` For th second harmonic the resonant frequency
` f_(1) = 2N = (2N)/(2L) = V/L `
and for a closed pipe, the fundamental frequency
` = V/(4L) `
` :. ` Frequency of its nth harmonic ` = (nV)/(4L) `
and it is given that `f_(2) = (nV)/(4L) = n/4 (V/L) = (nf_(1))/4`
It is given that `f_(2) gt f_(1) ` and for a closed pipe only odd harmonics are heard i.e. ` n = 3, 5, 7, .... ` etc.
If ` n = 3 , f_(2) = (3f_(1))/4 `.This is not possible.
But if ` n = 5 , f_(2) = (5f_(1))/4`. This is possible.
This is option (c ) .