Solving Deep Water Wave Dispersion for Storm Distance

[ambientdream]

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 :)

 

[ehild]

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

 

[ambientdream]

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?

 

[ehild]

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

 

[paranormal]

As a scientist, it is important to first understand the problem and the given information. The problem states that we are dealing with long-wavelength surface waves in deep water, and we are given the dispersion relation ω=(gk)^1/2, where ω is the angular frequency, g is the acceleration due to gravity, and k is the wavenumber. We are also given that the waves have a fixed wavelength (or period) of 15 seconds and travel at their group velocity.

Next, we are given that these waves were generated by a storm in the mid-Atlantic and arrived at the British coast at noon on Monday. By noon on Tuesday, the period of the waves has dropped to 13 seconds. Using this information, we can use the dispersion relation to solve for the wavenumber (k) at both time points.

At noon on Monday, we have ω = (gk)^1/2 = 2π/T = 2π/15, where T is the period. Solving for k, we get k = (4π^2g)/T^2 = (4π^2*9.8)/15^2 = 0.085 m^-1.

At noon on Tuesday, we have ω = (gk)^1/2 = 2π/T = 2π/13. Solving for k, we get k = (4π^2g)/T^2 = (4π^2*9.8)/13^2 = 0.098 m^-1.

Now, we can use these values of k to find the difference in distance between the storm and the British coast. The distance (d) can be calculated using the formula d = 2π/k.

Therefore, the distance between the storm and the British coast is d = (2π/0.098) - (2π/0.085) = 69.3 km.

In conclusion, the storm occurred approximately 69.3 km away from the British coast. This calculation was possible by using the dispersion relation and the given information about the period of the waves at two different time points.