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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 m s^-1`.

A

25cm

B

100cm

C

50cm

D

12.5cm

Text Solution

Verified by Experts

The correct Answer is:
A
`n_2-n_1 = 2592 - 1944 = 648`
`n=(v)/(2l)`
`therefore l=(v)/(2n) = (324)/(2xx648)`
0.25cm
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