physics help
posted by Anonymous .
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!

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 
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 
Whoops, wavelength = 1/2 not 2
so they differ by .25 meter
the method does not change