Superposition of Sound Waves

## Question:

An oscillator drives two loudspeakers 35.0 cm apart, which
vibrate in phase at a frequency of 2.00 kHz. What is the smallest
nonzero angle, q, (in degrees) at
which a distant observer hear maximum sound intensity?
## Answer:

For a distant observer, the first maximum in sound level occurs
at an angle such that the angle is the wavelength divided by the
distance between the speakers. That is the condition that makes
the difference in path length from the two speakers to the
observer, one wavelength. We know that the speed of sound is
about 343 m/s so a 2000Hz sound wave has a 343/2000 meter wavelength = 0.1715 meter. Then sine q is 0.1715/0.35 =
0.49, so q is
29.34 degrees.