Group and Phase velocity of wave packet

Quelsita

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.

Redbelly98

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

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

There's a simpler approach to find V_g. 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?

Quelsita

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?

Redbelly98

That's the right idea, but use the values of w you had in post #1: +2.03 and +2.00.

Quelsita

Ah, I see. Thank you for your help!