Solving Sound Interference: Find Diff. Freq. Destructive Interference

[kahless2005]

Given for the problem:
A speaker sends out two sound waves with equal Amplitudes but the frequencies are f(1) and f(2) respectively. The motion of sound as w = A * cos(k*x - t*(Omega)). The wave number's and the angular frequency's definition are the same for light.

Find for the problem:
Show that at a distance x directly in front of the speaker, there is destructive interference between the waves with a frequency f(1) - f(2).

My solution so far:
w(1) = A * cos((2PI/(Lambda(1))) * x - (2PI * f(1) * t)
w(2) = A * cos((2PI/(Lambda(2))) * x - (2PI * f(2) * t)

I assume that the final equation will be in the form of:
dt = (x / v) - t
where v is the speed of sound

A little advice please!

 

[kahless2005]

I hope i put this in the right section... It is a Sophmre level physics class...

 

[topsquark]

Given for the problem:
A speaker sends out two sound waves with equal Amplitudes but the frequencies are f(1) and f(2) respectively. The motion of sound as w = A * cos(k*x - t*(Omega)). The wave number's and the angular frequency's definition are the same for light.

Find for the problem:
Show that at a distance x directly in front of the speaker, there is destructive interference between the waves with a frequency f(1) - f(2).

My solution so far:
w(1) = A * cos((2PI/(Lambda(1))) * x - (2PI * f(1) * t)
w(2) = A * cos((2PI/(Lambda(2))) * x - (2PI * f(2) * t)

I assume that the final equation will be in the form of:
dt = (x / v) - t
where v is the speed of sound

A little advice please!

Hint: You are looking for a point, x, where the two waves are out of phase by pi radians.

-Dan

 

[lightgrav]

"Destructive Interference" in this case is time-dependent cancellation of the total amplitude (that means add the wave functions), at any location.
This is in contrast to location-dependent cancellation of the total amplitude
(an interference pattern) at all time.

Choose an x-value, and add the wave forms ; see when (time) they cancel.