Question

Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air is 324 ms−1.

Answer 1

Given:

Velocity of sound in air v = 324 ms−1

Let l be the length of the resonating column.

Then, the frequencies of the two successive resonances will be  (n+2)v4I and nv4I .

As per the question,

(n+2)v4l = 2592

  nv4l = 1944

So,

(n+2)v4l−nv4I=2592−1944=648⇒2v4l=648⇒l=2×324×1004×648cm=25 cm

Hence, the length of the tube is 25 cm.