# Destructive sound waves

1. Sep 21, 2008

### Foxhound101

1. The problem statement, all variables and given/known data
Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound waves in phase at 688 Hz. Take the speed of sound in air to be 344 m/s.

What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers?

2. Relevant equations

f1= V/2L
lambda1 = 2L

3. The attempt at a solution

So...this is what I did...

688 = 344/2L

L = .25

lambda1 = 2L
lambda = 2(.25)
lambda = .5

Destructive interference occurs at a distance that is half lambda. I said that the answer was .25 meters, but according to the program that is not the correct answer. So...where is my error? (It is always possible the program is wrong, but normally it is me) Please help me identify my mistake.

2. Sep 22, 2008

### Foxhound101

Unless I'm supposed to walk beyond one of the speakers...but that doesn't make sense either...

Well, using another formula...

lambda=V/f
lambda=344/688
lambda=.5

So I still believe that lambda is .5.

3. Sep 22, 2008

### Redbelly98

Staff Emeritus
That is not really true. Review the part of your textbook or lecture notes that discusses interference more carefully. What is the actual condition?

4. Sep 22, 2008

### Foxhound101

"Destructive interference occurs when the path-length difference is...

r=(m+.5)*lambda
where m = 0, 1, 2, 3, ...

Constructive interference occurs when the path-length is...
r = m*lambda
where m = 0, 1, 2, 3, ..."

So...I still don't really understand. The book does not do a good job explaining this.

5. Sep 22, 2008

### Redbelly98

Staff Emeritus
Yes, that's it. The difference in the path lengths from you to each speaker should be a "half-integer" multiple of the wavelength ... i.e., (1/2)λ, (3/2)λ, (5/2)λ, etc. etc. In other words, (m+0.5)λ as the formula says.

You'll have to use some geometry to figure out where you can be, so that the distances from you to the two speakers differ by the minimum amount of λ/2 (= 0.25 m).

By the way, some things are not clear from your original post:
1. How far apart are the speakers?
2. What is your location, when you start to walk forward?

6. Sep 22, 2008

### Foxhound101

Woops...silly me. That information was in part A, which I had already answered and didn't include in the original post.

*restates problem*

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound waves in phase at 688 Hz. Take the speed of sound in air to be 344 m/s.

Part A
If you are 3.00 m from speaker A directly to your right and 3.50m from speaker B directly to your left, will the sound that you hear be louder than the sound you would hear if only one speaker were in use?
YES

Part B
What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers?

7. Sep 22, 2008

### Redbelly98

Staff Emeritus
Okay, good.

At what locations can you be (in between the 2 speakers) so that your distance to 1 speaker is (1/2)λ, (3/2)λ, (5/2)λ, etc greater than or less than your distance to the 2nd speaker?