# Homework Help: Phase Difference - Homework Help Needed!

1. Apr 13, 2007

### ml_lulu

1. The problem statement, all variables and given/known data

Two sound fonts make plain armonic waves with the same amplitude and frequency. If f=200 Hz and V=340 m/s, the phase difference in a point situated 8m away from one font and 25 m away from the other one will be...

2. Relevant equations

V=Yf
Y: Wave longitude
Phase Difference=kr2-kr1 (Found it online and is so far the only one I know about)

3. The attempt at a solution

Y=V/f=340/200=17/100
TT=Pi
K=2TT/Y
K=2TT/(17/100)
Phase Difference=k(r2-r1)
P.D=(2TT/(17/100))*17
P.d=2TT*100

Nut in the answer sheet it says the answer is: The waves are in phase. Please explain why?

2. Apr 13, 2007

### e(ho0n3

Here is how I would reason about the problem: Let A be the source of sound 8 meters away and B the one 25 meters away. The time it takes for you to hear A is 8/340 seconds. The time it takes for you to hear B is 25/340 seconds which is roughly 3*(8/340) seconds. Hence, the phase difference would be very small.

3. Apr 13, 2007

### Dick

340/200 is NOT 17/100. Please correct that, ok? You have a phase difference of 100*pi (it's not really that, but you'll correct it). 2*pi is one whole wave length. A wave shifted by 2*pi in phase is the same wave again since the wave pattern repeats every 2*pi. So a shift of 2*pi is the same as no shift at all. So if your phase shift is a multiple of 2*pi then the two sounds are 'in phase'. If you get a phase shift bigger than 2*pi then just take the remainder after dividing by 2*pi to get the phase shift.