Question

Two adjacent natural frequencies of an organ pipe are found to be 550 Hz and 650 Hz. a.) Calculate the fundamental frequency of the pipe. b.) Is the pipe open at both ends or open at only one end? c.) What is the length of the pipe.

Please show your work and provide an answer thank you.

Answer 1

The simple formula for the frequencies that may be obtained from an open organ pipe, that is, one with both ends open, is given by

f = n.v/2.L

where f is the frequency in Hz, v the velocity of sound, 340 m/s, n the harmonic produced, equal to 1, 2, 3 ... upwards, and L is the length of the pipe in metres. Since two adjacent harmonics occur at 550 and 650 Hz

550 = n x 340/2.L and 650 = (n + 1) x 340/2.L

Subtracting gives L = 340/(2 x 100) = 1.7 m.

The fundamental frequency for which n = 1, is at 340/(2 x 1.7) = 100 Hz.

For a closed organ pipe, which has one end (only) closed,

f = n.v/4L

with only odd harmonics supported ie n = 1, 3, 5 ... For this

550 = n x 340/4.L and 650 = (n + 2) x 340/4.L

Subtracting again we obtain L = 2 x 340/(4 x 100) = 1.7 m

The fundamental is at 340/(4 x 1.7) = 50 Hz.

Which applies in this case? Well, the frequencies found, 550 and 650 Hz, are not harmonics of the fundamental frequency, 100 Hz, for the open pipe, so it cannot be of this form. On the other hand, they do form the 11th and 13th harmonics of the fundamental (50 Hz) for the closed pipe case, so this must apply here.

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f(n) = 550 Hz = nf, where f is the fundamental frequency
f(n + 1) = 650 Hz = (n + 1)f
=> (n + 1)f - nf = 650 50 = 100 Hz
=>fundamental frequency, f = 100 Hz.

Length of the pipe, L = v/2f = 340 / 200 = 1.7 m.