hey everyone

 

hmm...can't seem to work out how to attack this question.... any suggestions guys?

 

 

Cheers

Sarah

use the euler relation

 

 

i'm completely guessing, but [math]e^{i\pi}=-1[/math], and it said to use that, so maybe [math]k_1{x}-w_1{t}=\pi=k_2{x}-w_2{t}[/math]

why set [math]

k_1{x}-w_1{t}=\pi=k_2{x}-w_2{t}

[/math] ?

 

why is it to pi? i'm not quite following...

i just set each = to [math]e^{i\pi}[/math]......i kinda just BSed the answer. i'm only in AP Physics B. [math]{\Delta}kx-{\Delta}wt=\pi[/math] looks like a good answer to me, except it is missing the "time-dependant modulating factor." if that is A, then it might be [math]A({\Delta}kx-{\Delta}wt)=\pi[/math]. keep in mind i have no idea what i'm doing.

hmmm ok so i get:

 

[math] 2A e^{i(kx-wt)} cos( \frac{k_1 x-k_2 x-w_1 t+w_2 t}{2}) [/math]

 

i think thats ok.... but as for the phase and group velocities...???

What are the equations for phase and group velocities?

i think they are:

 

[math] v_{phase} = \frac{w}{k} [/math]

 

and

 

[math] v_{group} = \frac{dw}{dk} [/math]

Does the frequency vary with k? (You haven't described any such relationship, so if this is an EM wave in free space the answer is no, and [math]\frac {w}{k} = c[/math])

you've lost me i think...

 

to get my above answer i set

 

[math] w = \frac{w_1+w_2}{2} [/math]

 

and

 

[math] k = \frac{k_1+k_2}{2} [/math]

 

and its a plane wave?

You get beats, which are the wave packets. look here for a visual, and see here for the explanation.

 

If the phase and group velocities are equal, the individual oscillations don't move with respect to the packet. If you go away from the plane wave description, so you have only one wave packet, then when the velocities aren't equal you get dispersion, and the wave packet spreads out. (with the plane waves, all dispersion does is move an oscillation from one beat to the next)

oh ok i get it now!

 

thanks swantsont

 

-Sarah