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A storm generates large waves in the Pacific Ocean. On november 25, 208, 6pm, swell with a wave height of 5m reaches an exposed location at the coast of Southern California. At 6 am of the next day, the wave height is 4m. The water depth within 100km of the coast is 1000m and 3000m everywhere else. All waves have a steepness of ak=0.25.

a) How far offshore were the waves generated by that storm?

b) At what date and time do waves with a height of 2m reach the same location at the coast?

c) How long does it take a tsunami with a wave length of 120km to travel the same distance (travelling perpendicular towards shore)?

To solve these problems you have to be familiar with the speed of the various waves involved. I assume you have course materials that you can refer to so I won't go into great detail.

Swell waves are deepwater waves for this example. The wave dispersion relation is ω = sqrt(gk), where ω = frequency (rad/sec), g = gravitational constant (= 10m/s^2)and k = wavenumber (rad/m). The energy in the waves propagates at the group velocity, Cg, which is the one we are interested in (as opposed to the phase velocity, Cp). From the dispersion relation, we have

Cg = ∂ω/∂k = (1/2)sqrt(g/k)

The wave steepness is ak = (wave amplitude)(wavenumber) which allows us to solve for k for both waves.

Plugging in numbers, for the first wave, W1, to hit the coast, Cg1 = 7.1m/s and for the 2nd, W2, Cg2 = 6.3m/s. The 1st question is how far away is the source of both waves, call it D. Then

Time for W1 to traverse D is T1 = D/Cg1 and for W2 is T2 = D/Cg2. To find D, we know that the difference in time between when the waves hit the coast is

∆T = (time when W2 hits) - (time when W1 hits) = D( (1/Cg2) - (1/Cg1) ) = D(Cg1-Cg2)/(Cg2Cg1) = 12 hours = 43200 secs. Thus

D = 2415420 m ≈ 2415 km.

b) For 2m waves we can again use the steepness to calculate k3 = 0.25/2 = 0.125 and Cg3 = 4.47m/s. The time to traverse D for these waves is

T3 = 2415420m/4.47m/s = 540367 sec.

The time for W1 to hit was 2415420m/7.1m/s = 340200. The difference in these times needs to be added to the date of W1 to get

Date W3 = 11/25/2008 + 540367sec - 340200 = 11/25/2008 + 2 days 7 hrs 36 min.

c) Tsunamis are really long waves and so are called shallow-water waves (they "feel" the botom) and have a different dispersion relation and group velocity.

Shallow water (tsunami)) group velocity CgT = sqrt(gH) where H is the depth of the water. The distance to the coast has 2 depths covering the distances D1 and D2. For each depth H we have the transit time over the distance D

T = D/CgT.

If you calculate D1 and D2 from the information given and apply this formula for each distance and add them up you should get ≈ 4 hours.

Let me know if this makes sense.

Randy

Physical oceanography, surface and internal wave characteristics, ocean currents, fluid mechanics, geophysical fluid dynamics, ocean optics, coastal dynamics, modeling and simulation, data analysis, El Nino and related large scale dynamics Not an expert in marine biology (some in bioluminescence) or chemical oceanography

26 years as professional scientist for research company working mostly on Navy and other government contracts. Projects included modeling, simulations and data analysis related to Non-acoustic Anti-submarine Warfare (NAASW). Other projects included remote sensing of ocean features, statistical analysis of ship tracks, ocean optics instrumentation development, synthetic aperture radar (SAR) and sonar (SAS).**Publications**

Journal of Physical Oceanography, 1984, "A Numerical Model for Low-Frequency Equatorial Dynamics" (with M. Cane)**Education/Credentials**

MS Physical Oceanography, MIT, 1981
BS Applied Math, UC Berkeley, 1976**Past/Present Clients**

Currently an Expert for All Experts in Advanced Math