Question

Two closed organ pipes of length 100 cm and 101 cm produce 16 beats in 20 sec.

Answer 1

The frequency of fundamental mode of vibrations of 1st closed organ pipe is following – v1 = v/ 4l1 = v/(4 * 100) – (i).

On the other hand, the frequency of the fundamental mode of vibrations of 2nd closed organ pipe is following – v2 = v/ 4l2 = v/(4 * 101) – (ii).

We can put by (i) – (ii) as per the question, v1 – v2 = 16/ 20. Putting the value of v1 and v2 we get, v = 323. 2 m/s.