For a certain organ pipe, three successive resonance frequencies are observed at 425, 595 and 765 Hz. The…
For a certain organ pipe, three successive resonance frequencies are observed at 425, 595 and 765 Hz. The speed of sound in air is 340 $\text{ms}^{-1}$. The pipe is a
closed pipe of length 1 m
closed pipe of length 2 m
open pipe of length 1 m
open pipe of length 2 m
Solution
Ratio of three successive resonance frequencies,
$f_1 : f_2 : f_3 = 425 : 595 : 765 = 5 : 7 : 9$
Since, these are odd harmonics, so pipe will be closed.
Further, $5 \left(\frac{v}{4l}\right) = 425$
$\therefore$ The length of closed pipe, $l = \frac{425 \times 4}{5v} = \frac{425 \times 4}{5 \times 340} = 1\text{ m}$