Group and Phase velocity of wave packet

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Homework Help Overview

The discussion revolves around the analysis of a wave packet formed by the superposition of two cosine waves, focusing on the concepts of phase velocity and group velocity. Participants are examining the relationships between the wave parameters and how to derive the velocities from the given equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating phase velocity using the formula Vp = w/k and question the correctness of their approach. They also explore the calculation of group velocity through the superposition principle and the derivative of frequency with respect to wave number.

Discussion Status

Some participants have provided guidance on the calculations, suggesting simpler methods for finding group velocity by approximating the slope of the w-k curve. There is an ongoing exploration of the implications of the results, particularly regarding the relationship between phase and group velocities.

Contextual Notes

Participants note confusion regarding the partial derivatives and the interpretation of group velocity as a function of both position and time. There is also a discussion about the conditions under which group velocity can be negative while phase velocity remains positive.

Quelsita
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OK, I think I understand the problem, I'm just a little confused on some pieces in the middle.

Problem:
Consider a wave packet formed by the superposition of two waves
psi1=cos(1.00x-2.00t) <--1.00=k, 2.00=w
psi2=cos(1.01x-2.03t) <--1.01=k, 2.03=w
where x and t are measured in meters and seconds respectively. What is the Phase velocity?
What is the group velocity?

We know:
Vp= w/k
Vg=dw/dk

-To find the Vp, can we simply say that Vp=2.00/1.00=2.00m/s?

Now, to find the Group Velocity, we can use the suerposition principle and find the net wave by the sum of the two individual waves (psi1 + psi2) which gives us

psi=2cos(1.00*1.01)x*cos(-2.00*-2.03)t = 2cos(1.01)x*cos(4.06)t

thus Vg=dw/dk= -(2/1.01)sin(1.01)x*cos(4.06)t - (2/4.06)doc(1.01)x*sin(.06)t?

Is this correct? Is the partial derivative correct?

Thanks.
 
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Quelsita said:
-To find the Vp, can we simply say that Vp=2.00/1.00=2.00m/s?

Good enough. For the 2nd wave it's slightly different, but 2.00 m/s is fine.

Now, to find the Group Velocity, we can use the suerposition principle and find the net wave by the sum of the two individual waves (psi1 + psi2) which gives us

psi=2cos(1.00*1.01)x*cos(-2.00*-2.03)t = 2cos(1.01)x*cos(4.06)t

thus Vg=dw/dk= -(2/1.01)sin(1.01)x*cos(4.06)t - (2/4.06)doc(1.01)x*sin(.06)t?

It doesn't make sense for Vg to be a function of x and t.

There's a simpler approach to find Vg. You have w and k at two points. What's a good approximation for the slope of the w-k curve near those two points?
 
Hmm...I'm not sure why I put that but I think I meant to say cos(1.01)x.

I see what you mean by a slope being much easier, but I'm not sure how to apporximate the curve. Our text only shows that the Vg=delta(w)/delta(k) which can be approximated by differentials.

-OK, w and k are at two points, one in the first wave and nother in the second. So can you say that the slope is just (-2.03--2.00)/(1.01-1.00)= -3?

-Can group velocity be negative if the phase velocity is positive?
 
That's the right idea, but use the values of w you had in post #1: +2.03 and +2.00.
 
Ah, I see. Thank you for your help!
 

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