Physics- Sound; constructive and destructive interference?

Question

"Two identical loudspeakers 2.0 m apart are emitting 1800 Hz sound waves into a room where the speed of sound is 340 m/s."

"Is the point 4.0 m directly in front of one of the speakers (perpendicular to the plane of the speakers), a point of maximum constructive interference, perfect destructive interference, or something in between?"

Can you please tell me how to go about figuring this out? Thanks in advance for your help! : D

Answer 1

Find the wavelength of the sound: lambda = c/f

Find the path-length difference to the point from the two speakers, and divide it by the wavelength.

L1 = 4 m

L2 = sqrt(4^2+2^2) m

deltaL = L2-L1

x = deltaL/lambda

If the result is nearly an integer, the waves reinforce at the point. If it is nearly an integer + 0.5, the waves interfere destructively at the point. If it is neither, the point is "something in between".

Solving, x = (sqrt(4^2+2^2)-4)/(340/1800) = 2.49954, very close to 2.5, an integer + 0.5. So it's a point of destructive interference.

Answer 2

The other speaker is a distance of

d^2 = 2^2 + 4^2

d = sqrt(20) = 2sqrt(5) m

Path difference from the two spwakers to the point is

2sqrt(5) - 4 = 4.47 - 4 = 0.47 m

If this path difference is a whole number of wavelengths we get constructive interference and if it is a whole number of half wavelengths we get descructive interference.

wavelength = v/f = 340/1800 = 0.188 m

0.47/0.188 = 2.5

0.47/0.094 = 5

so it is destructive interference.

Answer 3

The correct answer is B.

Answer 4

The answer is C. Increased frequency means the wavelengths are smaller. And since the smallest value for destructive interference to occur is at 1/2*lambda, as the wavelengths decrease, the separation will decrease as well.