Two loudspeakers placed 6.0 m apart are driven in phase by an audio oscillator,whose frequency range is 1595 Hz to 2158 Hz.A point P is located 4.7 m from one loudspeaker and 3.6 m from the other.The speed of sound is 344 m/s.The frequency produced by the oscillator,for which destructive interference occurs at point P,in SI units,is closest to:
A)2033
B)2127
C)2095
D)2064
E)2001

Question

Quizplus · Accepted Answer

The correct answer is A because the path length difference between the two loudspeakers to point P is 4.7 m - 3.6 m = 1.1 m. For destructive interference, the path length difference must be an odd multiple of half the wavelength, so the wavelength must be 2.2 m. The frequency corresponding to this wavelength is 344 m/s / 2.2 m = 156.8 Hz. The closest frequency in the given range is 1595 Hz, which is an odd multiple of 156.8 Hz.

Two Loudspeakers Placed 6