- #1

Two loud speakers are placed on a wall 2.00 m apart. A listener stands 3.00m from the wall direclty infront of one of the speakers. A single ocillator is driving the speakers in phase at a frequency of 300Hz.

A. What is the phase difference between tehe two waves when they reach the observer.

B. What is the frequency closest to 300 hz. to which the oscillator may be adjusted such that the observer hears minimal sound?

This is what I did..

r1= square root of (3.00)^2 = 3

r2= square root of ((3.00)^2 + (2.00)^2) = 3.61

The difference is = r2-r1 = .61

Now this is where i got stuck I wasnt sure if i had to use .61/2pie to find the phase difference.

Also what forumla should I use for part B.

Thank you in advance.