Solving Two Speaker Intensity Problem: f1, f2, f3

[amatol]

Homework Statement

https://tycho-s.phys.washington.edu/cgi/courses/shell/common/showme.pl?courses/phys123/spring08/homework/01b/twospeaker_INT/02.05.gif
(if image doesn't show up...it is a right angle triangle with S1 at the right angle, 4m between it and O and 3m between S1 and S2)

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 4 W/m2 when either S1 or S2 is sounded alone. They are driven in phase (at the speakers) with various frequencies of sound. Assume that the speed of sound is 325 m/s.

a) Find the three lowest frequencies, f1 < f2 < f3, for which the observer at O will hear an intensity of 16 W/m2 when both speakers are on.

b) Find the three lowest frequencies, f1 < f2 < f3 , for which the observer at O will hear no sound when both speakers are on.

c) Find the lowest frequency for which the observer at O will hear an intensity of 8 W/m2 when both speakers are on.

d) Find the lowest frequency for which the observer at O will hear an intensity of 4 W/m2 when both speakers are on.

Homework Equations

I think these...
f=v/\lambda
Intensity= energy/tA
I also need to figure out the phase difference but not sure what formula

The Attempt at a Solution

so far I know that I need to find the path length difference between the two speakers to the observer...which is 1m. From that I need to find the phase difference which will give me the intensity. How to do this I am clueless on. Then I need to find the wave length from the difference between the two and from that I can find the frequency. (at least this is what I am thinking...but its not working! and I can't figure out the phase difference)

Any help appreciated!

 

[kuruman]

Part (a): You are looking for constructive interference when the path length difference is 1 m. You don't need the relative phase of the interfering waves. All you need to know is that you get constructive interference when you can fit an integer of wavelengths within the path length difference. So in 1 meter the longest wavelength will be 1 m, then comes 1/2 m, then 1/3 m and so on. Can you find the corresponding frequencies? Do part (a) and then we'll worry about the other parts.