Determining the possible time interval between two sine waves

1. Feb 1, 2013

randoreds

ok I just got stuck half way into a problem, I would like it if someone explained it!

Ok the question says, two identical sinusodial waves with a wavelength of 3.0 m and traveling in the same direction with a v of 2m/s. Starting from the same point, just the second waves starts later. and the amplitude of the resultant wave is the same as each of the initial waves.

Find the min. time interval between the two waves

So, I would suppose I would need the phase difference, so I found it and it is 2/3 pi. now I asked my teacher and he said to use this equation to find the time interval.

T/3 = 1/3f

so I get I would just sub for f with v and wavelength

what I don't get is the 3. I see it comes from my phase difference, but could someone explain that a little further. Like what if my phase shift was 4/5 pi? would the equation become T/5 =1/5f ?

because I don't really get where the two goes and how I would know to put a 3 under T

Last edited: Feb 1, 2013
2. Feb 1, 2013

rollingstein

I see emptiness....

3. Feb 1, 2013

Fixed it!

4. Feb 1, 2013

Staff: Mentor

You know that 1 cycle corresponds to 2 radians.
So then x cycles correspond to (2/3). Find x.

5. Feb 1, 2013

haruspex

Yes, period = 1/frequency, and you can calculate the frequency from wavelength and speed. What remains is to say what one whole period corresponds to in terms of phase, and what fraction of that is the phase shift you found.

6. Feb 2, 2013

randoreds

T

Thanks, I think I get it. just to be sure. Theoretically if I found the phase shift to be 3/5 pi and all the other variable were the same as in the problem. My T would be 2/5 T because one sine wave is 2/5 pi ahead of the other, right?

7. Feb 2, 2013

haruspex

No. Try answering my question first: what, in radians, would a phase shift of a whole period be? Then, what fraction of that would 3/5 pi be?