# Homework Help: What is the wavelength of the sound wave?

1. Mar 16, 2005

### Physically Impaired

Here's the problem:

Two speaks side by side are separated by 1 meter. If the sound intensity is found to reach its first maximum when someone stands 1.0 meter directly in front of one of the speakers ( ie. perpendicular to the line between the speakers) what is the wavelength of the sound wave?

I know that wavelength is velocity over frequency but I don't think that applies here. If I draw a right triangle I can get the hypotenuse which is the distance between the person and the other speaker which is the square root of 2. Then take the difference 1.41 minus 1 = .41 meters. Is this the wavelength?

I'm not sure of my methodology!!!! Help!!!!

2. Mar 16, 2005

### Sojourner01

Use the formula wavelength=dsintheta where theta is the angle that the line between the two sources makes with the sound 'ray' heading toward the source.

Note that you're fudging it a little here - the formula assumes that theta is small and as such, both rays are parallel. Here they're clearly not, but you can get away with it because you're directly in front and the path length is the same for -theta as it is for theta.

3. Mar 16, 2005

### xanthym

Your methodology is correct for this case where distances are not large with respect to speaker separation distance. Shown in the drawing below, interference will cause First Max to occur where the path length difference between "A" and "B" is 1.0 λ. The right triangle geometry yields:
λ = √2 - 1.0 = (0.414 meters)

Code (Text):

First Max
/|
/ |
/  |
/   |
[COLOR=Red][B]A[/B][/COLOR]  /    | [COLOR=Red][B]B[/B][/COLOR] = 1.0 Meter
/     |
/      |
/       |
/________|
∪        ∪
1.0 Meter

Interference will cause First Max to occur where
path length difference between [COLOR=Red]"A"[/COLOR] and [COLOR=Red]"B"[/COLOR] is 1.0 λ.

~~

Last edited: Mar 16, 2005