- #1

- 3

- 0

Moved from general physics forum

Two loudspeakers are placed facing each 4.0 m apart. The speakers emit sound in phase with a frequency of 266 Hz. An observer at the center (2.0 m from each speaker along the line joining them) experiences constructive interference. The speed of sound is 343 m/s

The lowest frequency at which destructive interference could occur is if 1/2 a wavelength = 1.5 m (the difference between the two)

The speed of sound can be found with the following formula

v = 331 m/s + (0.6 m/s/C)•T

So at 20C you get a speed of 343 m/s.

We know that (1/2)*lambda = 1.5 m, so lambda = 3.0m

The wavelength must be 3 m.

Using the universal wave equation

v = f*lambda

f = v/lambda

f = (343m/s)/(3 m)

= 114 Hz

The lowest freq would be 114 Hz

The next two lowest would be a 3lambda/2 = 1.5 and 5lambda/2 = 1.5

so lambda = 1m and lambda = 0.6m

Those give f = 343 Hz and 572 Hz

i don't know how to find How far toward either speaker must the observer walk to experience destructive interference?

please help me

The lowest frequency at which destructive interference could occur is if 1/2 a wavelength = 1.5 m (the difference between the two)

The speed of sound can be found with the following formula

v = 331 m/s + (0.6 m/s/C)•T

So at 20C you get a speed of 343 m/s.

We know that (1/2)*lambda = 1.5 m, so lambda = 3.0m

The wavelength must be 3 m.

Using the universal wave equation

v = f*lambda

f = v/lambda

f = (343m/s)/(3 m)

= 114 Hz

The lowest freq would be 114 Hz

The next two lowest would be a 3lambda/2 = 1.5 and 5lambda/2 = 1.5

so lambda = 1m and lambda = 0.6m

Those give f = 343 Hz and 572 Hz

i don't know how to find How far toward either speaker must the observer walk to experience destructive interference?

please help me

Last edited: