Calculating Minima Spots Between Two In-Phase Speakers

[NotMrX]

Hello,

Please help me figure this problem out.

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?

 

[Andrew Mason]

Hello,

Please help me figure this problem out.

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?

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

 

[NotMrX]

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

wavelength = L = v/F

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

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 argument in the cosine.

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

Does this seem right so far?

 

[Andrew Mason]

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 argument in the cosine.

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

Does this seem right so far?

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

 

[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 = (2m+1)\lamba/4 + d/2[\tex]

 

[turbo]

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

http://mitworld.mit.edu/video/291/

 

[NotMrX]

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