How Does Group Velocity Affect Wave Arrival Times from a Storm?

[Deadstar]

Homework Statement

Surface waves generated by a mid-Atlantic storm arrive at the British coast with period 15 seconds. A day later the period of the waves has dropped to 12.5 seconds. Roughly how far away did the storm occur?

Homework Equations

Group velocity c = \frac{\omega}{2k}

\omega = (gk)^{1/2}

T = \frac{2 \pi}{\omega}

The Attempt at a Solution

First problem is...

Group velocity is proportional to 1/T so why are the longer periods arriving first? Shouldn't they move slower and thus arrive after the smaller period waves.

 

[Lavabug]

w/k or equivalently (2pi/T)/(2pi/lambda) is actually the phase velocity.

Group velocity is the derivative of w (your dispersion relation, thing with the square root) with respect to k.

Looks like doing that gives you a smaller number in the denominator for your phase velocity. Which travels faster?

 

[Deadstar]

w/k or equivalently (2pi/T)/(2pi/lambda) is actually the phase velocity.

Group velocity is the derivative of w (your dispersion relation, thing with the square root) with respect to k.

Looks like doing that gives you a smaller number in the denominator for your phase velocity. Which travels faster?

Oops missed a half but given that \omega = (gk)^{1/2}

Group velocity is therefore \frac{1}{2}(g/k)^{1/2} = \frac{\omega}{2k} which won't change my problem. I don't really know what you're last line is asking. Same result as before.

 

[Lavabug]

It looks like your phase velocity is proportional to 1/T, but your group velocity is proportional to 1/2T. The group velocity is slower. That's all I can think of right now.

 

[Deadstar]

It looks like your phase velocity is proportional to 1/T, but your group velocity is proportional to 1/2T. The group velocity is slower. That's all I can think of right now.

Ok no worries. The answer to the question is waves of period T arrive at the coast from distance d in time \frac{4 \pi d}{g T} but I have no idea why that is. To be honest I find this question vague...

Also...
When you say proportional to you generally ignore constants so I'd still say proportional to 1/T.

Edit: Figured it out.

 

[Lavabug]

Cool. Could you elaborate? I'm curious.

and what I should have said was: your phase velocity is twice as fast as your group velocity. :p

 

[Deadstar]

Cool. Could you elaborate? I'm curious.

and what I should have said was: your phase velocity is twice as fast as your group velocity. :p

Yeah sure.

Right so... This only really came about because the answer had written it above as time taken whereas I was just really playing about with distance ratios... Time taken (t) for wave to arrive from distance d is...
t = \frac{d}{\omega /2k}
Consider dispersion relation \omega^2 = gk then \frac{g}{\omega} = \frac{\omega}{k} hence
\frac{\omega}{2k} = \frac{1}{2}\frac{g}{\omega} = \frac{1}{2}\frac{gT}{2 \pi} = \frac{gT}{4 \pi} since \omega = \frac{2 \pi}{T}
So...
t = \frac{4 \pi d}{gT}

Now I can't get the right distance though. Should be roughly 5000km but I get 843km. Not sure how to use both periods to get the right answer given that I don't know how long it took either wave to arrive or d. (Using a day for the second wave gets the 843km above)

 

[Deadstar]

Ok figured it out... Seems I can figure out stuff after I post it up here. Sometimes it's just good to empty your head for a second.

d is constant so just got to find t1 and t2 (times of respective waves) and can use t2-t1 = 1 day. You get t1 = 5 days and t2 = 6 days, d = 5058km.

 

[hunt_mat]

I was always told that (and my PhD is in waves) that:

<br /> c=\frac{d\omega}{dk}<br />

hence the group velocity is:

<br /> c=\frac{1}{2}\sqrt{\frac{g}{k}}\Rightarrow T=\frac{4\pi c}{g}<br />