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 corrct option

Options

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

Answer

D

Explanation

(d)

`lambda = l`
:. `f_(1) = (v)/(lambda) = (v)/(l)` …(i)
:. `F_(2) = (v)/(lambda) = (nv)/(4l)` …(ii)
Here `n` is a odd number. From (i) and (ii)
`f_(2) = (n)/(4)f_(1)`, For first resonance, `n = 5`, `f_(2) = (5)/(4)f_(1)`