Question

Two identical loudspeakers are some distance apart. A person stands 4.40 m from one speaker and 3.50 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 343m/s.

Answer 1

wavelength =

Destructive interference occurs when the path length difference = 1/2

The lowest frequency at which destructive interference could occur is if 1/2 = 0.9 m (the difference between the two)

The speed of sound in air = 343m/s

We know that (1/2)*= 0.9m ,so = 1.8 m

The wavelength must be 1.8 m

Using the universal wave equation

v = f*  
f = v /
f = (343m/s)/(1.8m)
= 190.55Hz

The lowest freq would be 190.55Hz

The next two lowest would be a 3* / 2 = 0.9 and 5* / 2 = 0.9

so = 0.6 m and = 0.36 m

so that f₂ = 571.66Hz and f₃ = 952.77 Hz

so third lowest frequency f₃ = 952.77 Hz