Question

Two successive resonance frequencies in an open organ pipe are 1620 Hz and 2268 Hz. Find the length of the tube. The speed of sound in air is $324m{s}^{-1}$.

Answer 1

Let the length of the resonating column will be =I

Here, v =320 m/s

Then the two successive resonance frequencies are $\frac{(n+1)}{4I}$and $\frac{nv}{4I}$

Here given, $\frac{(n+1)}{4I}=2592$

$\lambda =\frac{mv}{4I}=1944$

$\Rightarrow \frac{(n+1)v}{4I}-\frac{nv}{4I}=2592-1944=548$

$\Rightarrow \frac{v}{4I}=548$

$\Rightarrow I=\frac{320\times 100}{4\times 548}cm=25cm$