Data : v = 350 m/s, L = 35 cm = 35 × 10-2 m
(i) For a pipe closed at one end, the fundamental frequency is
nc = \(\frac v{4L}\) = \(\frac{350}{4\times35\times10^{-2}}\) = 250 Hz
As only odd harmonics are present in this case, the frequency of pth overtone is np = (p × 2 + 1) nc
∴ The frequency of the 3rd overtone is
n3 = (3 × 2 + 1)nc = 7nc = 7 × 250 = 1750 Hz
(ii) For a pipe open at both ends, the fundamental frequency is
nO = \(\frac v{2L}\) = \(\frac{350}{2\times35\times10^{-2}}\) = 500 Hz
In this case, all harmonics are present.
∴ The frequency of the pth overtone is
np = (p + 1) no
∴ The frequency of the 3rd overtone is n3 = (3 + 1) no = 4no = 4 × 500 = 2000 Hz