Question

Two out of phase loudspeakers are some distance apart. A person stands 5.30 m from one speaker and 3.10 m from the other. What is the third lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 339 m/s. (answer in Hz)

Answer 1

Please note that the destructive interference of two identical sound waves occurs, if the sound waves are exactly shifted by one half of their wavelength.

Means, / 2 = Distance between the two loudspeakers

=> / 2 = (5.30 - 3.10) m = 2.20 m

=> = 2 x 2.20 = 4.40 m

First lowest frequency will occur at /2

Second lowest frequency will occur at (3)/2

Third lowest frequency will occur at (5)/2

Speed of the sound, v = 339 m/s

Therefore, the third lowest frequency at which destructive interference will occur = v / ((2)/5)

= 339 / ((2x4.40)/5) = 192.6 Hz (Answer)