Question

If two successive resonance frequencies in an open organ pipe are 1944 and 2592 hz find tge length of the tube

Answer 1

Answer:

The successive resonance frequencies in an open organ pipe are 1944 Hz and 2600 Hz. The length of the pipe if the .

Answer 2

Dear Student,

◆ Answer -

l = 26.23 cm

● Explanation -

Two successive resonance frequencies are expressed as nv/4l and (n+2)v/4l.

nv/4l = 1944

(n+2)v/4l = 2592

Then,

(n+2)v/4l - nv/4l = 2592 - 1944

v/2l = 648

l = 340 / (2×648)

l = 0.2623

l = 26.23 cm

Hence, length of the tube is 26.23 cm.

Thanks dear. Hope this helps you...