Question

A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz and 1152 Hz. Calculate the fundamental frequency.

Answer 1

The difference between the given frequencies of the overtones is 256 Hz. This implies that they are consecutive overtones. Let n_c be the fundamental frequency of the closed pipe and n_q, n_q+1, n_q+2 = the frequencies of the qth, (q + 1)th and

(q + 2)th consecutive overtones, where q is an integer.

Given : n_q = 640 Hz, n_q+1 = 896 Hz, n_q+2 = 1152 Hz

Since only odd harmonics are present as overtones,

n_q = (2q + 1)n_c  and n_q+1 = [2(q+ 1) + 1] n_c = (2q +3)n_c

∴ \(\frac{n_{q+1}}{n_q}=\frac{2q+3}{2q+1}=\frac{896}{640}=\frac{7}{5}\)

∴ 14q +7 = 10q + 15

∴ 4q = 8, ∴   q=2

Therefore, the three given frequencies correspond to the second, third and fourth overtones, i.e., the fifth, seventh and ninth harmonics, respectively.

∴ 5n_c = 640  ∴ n_c = 128 Hz