# Destructive Interference in Sound Waves

1. Aug 25, 2008

### cmerickson21

1. The problem statement, all variables and given/known data
Two speakers are driven by the same oscillator whose frequency is 140 Hz. They are located on a vertical pole a distance of 4.45 m apart from each other. A man walks straight toward the lower speaker in a direction perpendicular to the pole.
(a) How many times will he hear a minimum in sound intensity?
(b) How far is he from the pole at these moments? Take the speed of sound to be 330 m/s and ignore any sound reflections coming off the ground. (Give the first distances at which this happens.)

2. Aug 25, 2008

### LowlyPion

Welcome to PF. You might want to read this first:
https://www.physicsforums.com/showpost.php?p=785408&postcount=1

What formulas do you think apply?

3. Aug 27, 2008

### cmerickson21

I figured out that he hears a minimum twice. I have tried finding the wavelength and then using the Peak to Valley overlap equation which should apply to destructive interference where PVO=(m+.5)lamda m=0,1,2,3,... but that isn't working. What am I doing wrong should I not be using that equation? Do you have any ideas what I should be doing?

4. Aug 27, 2008

### LowlyPion

First of all what do you calculate for the wave length of the 140 hz sound wave?

Second of all at what point will the sound from two sources cancel each other out? They must be out of phase by how much? Which is what part of a wavelength?

5. Aug 27, 2008

### cmerickson21

I found the wavelength to be 2.36 by taking the velocity=330/the frequency=140. The sound will cancel out when the peak of one wave overlaps a valley of the other. Which will be when they are out of phase by half a wavelength.

6. Aug 27, 2008

### LowlyPion

That's what I get. OK. So if you move away from speaker A say, chosen for convenience as the problem is symmetrical, at what point is it that the sound from speaker A is exactly 1/2 wavelength out of phase with the two speakers?