Posted by **Anonymous** on Monday, April 19, 2010 at 6:32pm.

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. If you are 3.00m 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? What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers? I know the answer to the first part is yes but I'm not sure how to find the answer for the second part. Please help!

- physics help -
**Damon**, Monday, April 19, 2010 at 6:43pm
wavelength = T*c = c/f = 344/688 = 2 meters

We want the distance to differ by half a wavelength, 1 meter

sqrt(3.5^2 +x^2) - sqrt(3^2+x^2) = 1

- physics help -
**Damon**, Monday, April 19, 2010 at 6:49pm
3.5^2 + x^2 = 1 + 2sqrt(9+x^2) + 9 + x^2

12.25 = 10 + 2 sqrt(9+x^2)

sqrt (9+x^2) = 2.25/2

square both sides and get x

- physics help -
**Damon**, Monday, April 19, 2010 at 6:54pm
Whoops, wavelength = 1/2 not 2

so they differ by .25 meter

the method does not change

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