# Phase difference equation.

Jenn_Lane2001
Hi i posted here before and got great help, and I ran into another problem maybe you can help me this is the question.

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.

Hi Jenn_Lane2001,
I think you got the difference in distance correctly. Next you should find out what the wavelength is for a 300Hz sonic wave.

Jenn_Lane2001
R u sure..?

I didnt think you need to know the wavelength. more the velocity but that is what i wasnt sure calculating. I am not sure of the equation to use. Thanks

[lamb] = v/f,
where
[lamb] is wavelength
v is velocity
f is frequency
I think, for sonic waves, v = 340 m/s or something.

Jenn_Lane2001
Oh ok so sonic is the speed of sound

the speed of sound is 343v and then you will divide that by frequency which is 300 hz which equals wavelength= 1.143. now that i have the wavelenght where would i put taht into use as to determine the answers. Thanks alot for the help.

Right. Since the progress in phase is 2[pi] for each wavelength travelled, your phase difference is
[psi] = .61m / 1.143m * 2[pi].