Example 13-5 Stereo Problems
Your sound system consists of two speakers 2.50 m apart. You sit 2.50 m from one of the speakers so that the two speakers
are at the corners of a right triangle. As a test, you have both speakers emit the same pure tone (that is, a sinusoidal sine
wave). The speakers emit in phase. At your location, what is the lowest frequency for which you will get (a) destructive
interference? (b) constructive interference?
Set Up
There is a path length difference for the two
waves that reach your ear: One travels a distance D1 = 2.50 m; the other travels a longer
distance D2 equal to the hypotenuse of the right
triangle. Since the product of frequency and
wavelength equals the constant speed of sound,
the lowest frequency for each kind of interference corresponds to the longest wavelength.
Constructive interference:
 pl = nl where n = 0, 1, 2, 3,…
speaker 1
Destructive interference:
1
bl
2
where n = 0, 1, 2, 3,…
 pl = an +
L = 2.50 m
(13-15)
D1 = 2.50 m
speaker 2
Propagation speed of a sound wave:
vsound = fl
(13-2)
Speed of sound in dry air at 20°C:
vsound = 343 m>s
(13-13)
Solve
(a) The path length difference  pl is equal to
the difference between the two path lengths
D2 and D1.
Path length from speaker 2 to the listener:
D2 = 2D 21 + L2 = 212.50 m2 2 + 12.50 m2 2
= 3.54 m
Path length difference:
 pl = 3.54 m 2 2.50 m = 1.04 m
The longest wavelength for which there is
destructive interference is the one for which
one half-wavelength fits into the distance  pl.
This corresponds to n = 0 in the destructiveinterference relation in Equations 13-15.
If one half-wavelength fits into the distance  pl,
1
 pl = l
2
l = 2 pl = 2(1.04 m) = 2.08 m
The frequency of a sound wave with this wavelength is
343 m>s
vsound
=
l
2.08 m
21
= 165 s = 165 Hz
f =
If the sound has this frequency, there will be destructive interference
at the listener’s position and the sound level will be diminished.
(b) The longest wavelength for which there is
constructive interference is the one for which
one full wavelength fits into the distance  pl.
This corresponds to n = 1 in the constructiveinterference relation from Equations 13-15.
Note that n = 0 is possible only if the path
length difference is zero, which in this case
it is not.
If one wavelength fits into the distance  pl,
 pl = l = 1.04 m
The frequency of a sound wave with this wavelength is
343 m>s
vsound
=
l
1.04 m
= 330 s21 = 330 Hz
f =
If the sound has this frequency, there will be constructive interference
at the listener’s position and the sound level will be elevated.
Reflect
The frequencies that we found are of musical importance: 165 Hz is E below middle C (E3 in musical nomenclature)
and 330 Hz is E above middle C (or E4). Note that the higher frequency is exactly double that of the lower frequency.
In music, two such tones are said to be an octave apart.
Notice that we didn’t consider the number of wavelengths that fit into the distance between you and either speaker.
All that matters is the difference between the two paths. Can you show that in the destructive case the listener is
1.20 wavelengths away from speaker 1 and 1.70 wavelengths away from speaker 2? Can you also show that in the
constructive case the numbers are 2.40 wavelengths and 3.40 wavelengths?