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
25 cm
Generalizing n(λ2) = L λ = vf
f = nv2L
Next resonant frequency = (n+1)v2L
Difference =(v2L) = 2592 − 1944
L = 25 cm