Two loudspeakers, A and B, emit continuous tones of frequency 680 Hz. The sound

Question

Two loudspeakers, A and B, emit continuous tones of frequency 680 Hz. The sound waves

from the two speaker are emitted with no phase difference. The speed of sound in air is 340 m/s.

Initially you are equidistant from the two speakers. Now speaker B is changed so that the tone is

emitted with a phase of ?/2 with respect to speaker A, and the sound amplitude decreases. Keeping

your distance to speaker A constant, how far do you have to walk toward speaker B to restore the sound

amplitude to a maximum level?

Explanation / Answer

wavelength, lamda = v/f

lamda = 340/680 = 0.5 m

phase diference = (2*pi/lamda)*path diference

pi/2 = (2*pi/lamda)*path diference

==> pathe diference = lamda/4 = 0.5/4 = 0.125 m

to make path diference zero, he has to move 0.125m towards B

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