Digital Sound & Music: Concepts, Applications, & Science, Chapter 4, last updated 6/25/2013
46
Let’s assume again that the speed of sound is 1000 ft/s. Imagine a sound wave emanating from
the center of the room. The sound waves reflecting off the walls either constructively or
destructively interfere with each other at any given location in the room, depending on the
relative phase of the sound waves at that point in time and space. If the sound wave has a
wavelength that is exactly twice the width of the room, then the sound waves reflecting off
opposite walls cancel each other in the center of the room but reinforce each other at the walls.
Thus, the center of the room is a node for this sound wavelength and the walls are antinodes.
We can again apply the wavelength equation, , to find a frequency f that
corresponds to a wavelength that is exactly twice the width of the room, 2*10 = 20 feet.
At the antinodes, the signals are reinforced by their reflections, so that the 50 Hz sound is
unnaturally loud at the walls. At the node in the center, the signals reflecting off the walls
cancel out the signal from the loudspeaker. Similar cancellations and reinforcements occur with
harmonic frequencies at 100 Hz, 150 Hz, 200 Hz, and so forth, whose wavelengths fit evenly
between the two parallel walls. If listeners are scattered around the room, standing closer to
either the nodes or antinodes, some hear the harmonic frequencies very well and others do not.
Figure 4.33 illustrates the node and antinode positions for room modes when the frequency of the
sound wave is 50 Hz, 100 Hz, 150 Hz, and 200 Hz. Table 4.6 shows the relationships among
frequency, wavelength, number of nodes and antinodes, and number of harmonics.
Cancelling and reinforcement of frequencies in the room mode phenomenon is also an
example of comb filtering.
Room mode when the
frequency of sound wave is 50 Hz,
1st
harmonic
Room mode when the
frequency of sound wave is 100 Hz,
2nd
harmonic
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