Set 56 Problem number 13

Problem

What are the frequencies of the first four harmonics of a hole bored into the side of a hill, 13 meters long with its opening clear?  Assume that sound travels in the air inside at 329 m/s.

Solution

The longitudinal pressure wave created in the hole is analogous to, but different from the transverse wave created when a string, free at one end, is randomly plucked.

The main difference is that the pressure wave is longitudinal, making the process of wave formation similar to that of the hanging Slinky.

The similarity is that the pressure wave in the hole has an antinode at one end and a node at the other. Since there is a quarter-wavelength between a node and its adjacent antinode, there can be 1, 3, 5, 7, ... quarter wavelengths between the ends of the hole. From this we can obtain the wavelengths of the first four harmonics.  Then from the wavelengths and speed of wave propagation we can calculate the frequencies of these harmonics.

Since a full cycle of a wave goes from node to (positive) antinode to node to (negative) antinode to node, the full cycle consists of four node-antinode distances. (Note that we are using the terms 'node' and 'antinode' a bit loosely here, as if they could apply to a traveling wave; these terms really only apply to a standing wave, but the meaning here should be clear).

We reason out the wavelengths of the first four harmonics::

• We see that the first harmonic has just a single node-antinode distance in the 13 meters between its ends. This corresponds to 1/4 wavelength, so the wavelength of this harmonic must be 4 * 13 meters = 52 meters.
• The second harmonic goes from node to antinode to node to antinode, a distance of 3/4 wavelength. Thus 3/4 wavelength = 13 meters and the wavelength is 4/3( 13 m ) = 17.33 meters.
• Similarly the third and forth harmonics span 5/4 and 7/4 of a wavelength, respectively, and there respective wavelengths are thus 4/5( 13 m) = 10.4 m and (4/7)( 13 m) = 7.428 m.

Since sound moves 329 m in each second, there will be 329/ 52 = 6.326 peaks in each second of the fundamental harmonic. The frequency is therefore 6.326 Hz.

Similarly the three overtones will have frequencies of

• ( 329 m/s) / ( 17.33 m) = 18.98 Hz,
• ( 329 m/s) / ( 10.4 m) = 31.63 Hz and
• ( 329 m/s) / ( 7.428 m) = 44.29 Hz.

These frequencies could have been found from the fundamental frequency 6.326 Hz and the harmonic ratios 3/1, 5/3 and 7/5 for a standing wave with a node at one end and an antinode at the other.

• In this manner we would obtain the corresponding frequencies  (3/1)( 6.326 Hz) = 18.98 Hz; (5/3)( 18.98 Hz) = 31.63 Hz and (7/5)( 31.63 Hz) = 44.29 Hz.