Solving Deep Water Wave Dispersion for Storm Distance

  • Thread starter Thread starter ambientdream
  • Start date Start date
  • Tags Tags
    Dispersion Waves
ambientdream
Messages
2
Reaction score
0

Homework Statement


: The dispersion relation for long-wavelength surface waves in
deep water is ω=(gk)^1/2 . Waves of a fixed wavelength (or period) travel at their group velocity.
Surface waves generated by a storm in the mid-Atlantic and having a period of 15 seconds arrive at
the British coast at noon Monday. By noon Tuesday, the period of the waves arriving at the coast has dropped to 13 seconds. How far away did the storm occur?


Homework Equations



ω=(gk)^1/2

The Attempt at a Solution


No idea how to start. Please help :)
 
Physics news on Phys.org
Waves of different periods travel at different group velocity. So they arrive at the cost from the place where the storm occurred at different time. How is the group velocity defined?

ehild
 
The group velocity is defined as follows v=(1/2)(g/k)^(1/2). How would you start calculating how far away the storm is?
 
You know the time period, determine ω. Knowing ω, you find k, knowing k you get the group velocity for both frequencies. The time needed for a wave to arrive from distant x is x/v.
The time delay between the arrival of the two kind of waves is one day. Can you write up am equation for x?


ehild
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top