M3Q10

Consider a storm in the Southern Ocean at about 300 km from the coast of Australia. The sea state is characterised by a wave height of 10.3 m and a wave period is 13s. Waves propagate towards Australia at an angle of 28 degrees. Considering a uniform depth of 72m, how long do waves take to reach Australia? (Answer in hours).

Question

M3Q10

Consider a storm in the Southern Ocean at about 300 km from the coast of Australia. The sea state is characterised by a wave height of 10.3 m and a wave period is 13s. Waves propagate towards Australia at an angle of 28 degrees. Considering a uniform depth of 72m, how long do waves take to reach Australia? (Answer in hours).

BlackTom AI · Accepted Answer

We need to estimate the travel time of ocean waves from a storm in the Southern Ocean to Australia, given the distance to the coast, the wave period, the depth, and the propagation angle.

First, convert the wave period to angular frequency: T = 13 s, so ω = 2π / T ≈ 0.483 rad/s.

Next, use the linear dispersion relation for finite depth to solve for the wavenumber k, given depth h = 72 m: ω^2 = g k tanh(kh).
Let kh = x, so k = x/h and ω^2 = g (x/h) tanh(x). With g ≈ 9.81 m/s^2 and h = 72 m, we have ω^2 ≈ 0.2336. The equation becomes 0.2336 ≈ 0.13625 · x · tanh(x).
Solve this numerically to find x ≈ 1.81, which gives kh ≈ 1.81 and k = x/h ≈ 1.81 / 72 ≈ 0.0251 1/m.

Compute the phase speed c_p using c_p = ω / k: c_p ≈ 0.483 / 0.0251 ≈ 19.2 m/s.
To get the group velocity c_g in finite depth, use c_g = (1/2) c_p [1 + (2kh)/sinh(2kh)]. Here, 2kh = 3.62 and sinh(2kh) ≈ sinh(3.62) ≈ 18.6. Thus (2kh)/sinh(2kh) ≈ 3.62 / 18.6 ≈ 0.194. Then c_g ≈ 0.5 · 19.2 · (1 + 0.194) ≈ 9.6 · 1.194 ≈ 11.5 m/s.

Now determine the distance the wave crests must travel to reach the coast. The problem states the storm is about 300 km from the coast and that waves propagate toward Australia at a 28° angle. Assuming the given 300 km is the straight-line distance along the beam toward the coast, the path length to shore is approximately 300 km. (If one projects onto the coastline, the along-beam distance would be 300 km / cos(28°), but the standard interpretation here uses the direct distance along the wave crests toward shore.)

Convert distance to meters: 300 km = 300,000 m. The travel time along the ray is t = distance / c_g ≈ 300,000 m / 11.5 m/s ≈ 26,100 s ≈ 7.25 hours.

If one instead uses the longer projected distance due to the 28° angle (distance / cos 28°), the time would increase modestly to about 7.3–8 hours, depending on the exact values used for c_g and the projection. The value around 7 hours is consistent with the provided problem parameters.

In summary: the key steps are solving the dispersion relation for kh in finite depth to get k, deriving the group velocity from k and h, and dividing the along-beam distance by c_g to obtain the travel time. The numerical result emerges from those relationships and yields a travel time on the order of about 7 hours.

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