Phase Difference how do you calculate and why?

[ku1005]

Consider Figure 1. Estimate the phase di®erence between the waves y(x; t) and g(x; t)
at time t0. Select the result closest to your estimate.
A. 2.09 radians B. 0.333 C. 0.167
D. 1.045 radians E. 0.5 m F. None of these
Solution: 2.09 radians

http://img165.imageshack.us/img165/9711/physicsqphasedifferencenf9.png


I know the answer comes from (Distance/Time)2pi = Phase Difference(angle)

And thus Distance = 0.5m(between the same point on both waves in fig 1a) and time has to be 1.5s. The problem is, I don't know where te 1.5 comes from?? From the diagram is anyone able to assist?

 

[ku1005]

mmm just thinking...not sure why it would be (Distance / Time )* 2pi, becasue that makes no sense.

It should be (Distance/Wavelength)*2pi which would therefore give a percentage of the complete cycle and hence an angle!- opr phase diff...is this thinking correct??

 

[HallsofIvy]

For what value of t is y 0? For what value of t is g 0? The difference between the two is the phase difference.

 

[arunbg]

You are correct in saying
Phase Difference = 2pi*(Distance from coinciding of the two waves)/Wavelength
However, you also have to keep in mind that there is more than one way to make the waves coincide - you can move or "shift" a wave forward or backward, and after the first coincidence, you can even shift in integral values of the wavelength, which corresponds to a phase of 2*pi. So the phase difference may be anyone of these values. It is upto you to find which of these values is one of your options.

Can you proceed now? Just to help you on your way I'll give you a little hint,
to match your answer, you should shift wave y(x,t) in fig 1a forward till its first coincidence with g(x,t).

 

[ku1005]

yes i understand now, thanks for your help!