# Inbetween Two Speakers

1. Aug 17, 2006

### NotMrX

Hello,

Suppose that 2 speakers face each other with a distance D inbetweem them and that both emit the same frequency f. Also they hooked up to the same amp so they are in phase. How do we determine the spots of minima?

2. Aug 18, 2006

### Andrew Mason

First of all, determine the wavelength from the frequency and the speed of sound in air (which depends somewhat on temperature and pressure).

Then determine where you will have destructive interference. What is the condition for destructive interference? (think of sound as a transverse wave - it is easier to visualize).

AM

3. Aug 20, 2006

### NotMrX

wavelength = L = v/F

For destructive interfence the waves are out of phase by 180 degrees.

Maybe the waves could be described mathmatically by
Y1 = A* cos(K*x-w*t)
Y2 =A*cos[K*(x-d) +w*t]

then odd integer multiples of 180 or pi is the difference of arguement in the cosine.

[k*x-w*t] - [K*(x-d) + w*t] = n*pi

Does this seem right so far?

4. Aug 21, 2006

### Andrew Mason

This will give points of constructive and destructive interference. It is much simpler to just look at the path difference being odd multiples of half wavelengths:

$$\Delta x = (2m+1)\lamba/2$$ where m = 0, 1, 2 ....

The distance between the speakers will limit the number of points where this can occur.

AM

5. Aug 21, 2006

### NotMrX

Here is another approach then:

$$\Delta$$ X = $$X_1$$ - $$X_2$$
$$X_1$$ = x
$$X_2$$ = D - x
$$\Delta$$ X = 2x - d
$$\Delta x = (2m+1)$$$$\lamba$$/2 = 2x-d
x = [tex](2m+1)\lamba/4 + d/2[\tex]

Last edited: Aug 21, 2006
6. Aug 21, 2006

### turbo

NotMrX, you may wish to view this lecture by Walter Lewin of MIT.

http://mitworld.mit.edu/video/291/ [Broken]

Last edited by a moderator: May 2, 2017
7. Aug 22, 2006

### NotMrX

thank you all for help, I got it to work out.