Question

A progressive wave of height η1 = 0.05 m and wave length λ1 = 0.1 m is generated at r1 = 0.5 m away the center of a pond with a constant water depth of h = 2 m. As the waves propagate away from the origin (where they are generated), they become gravity–capillary waves. The dispersion relation of gravity-capillary waves in deep water is given as follows:

o2 = gk + d P

where σ ′ = 0.073 N/m is the surface tension for air–water interface , and ρ = 1000 kg/m^3 is the density of water.

It is given that at a distance r2 = 4 m from the origin, the wave length is λ2 = 1.7 cm and the wave is gravity-capillary. Compute the wave height at a distance r2 = 4 m away from the origin. Neglect energy dissipation in the wave propagation.

Find the following:

A) What is the wave number at point 1? Write the answer to the nearest tenth in units of: rad/m

B) What is the group velocity at point 1? Write the answer to the nearest hundredth in units of: m/s

C) What is the wave number at point 2? Write the answer to the nearest one (e.g. round 22.8 to 23) in units of: rad/m

D) What is the group velocity at point 2? Write the answer to the nearest hundredth in units of: m/s

E) What is the wave amplitude at point 2? Write the answer to the nearest tenth in units of mm.

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Answer #1

A) The wave number at point 1 is k_{1}=(2\pi /\lambda _{1}) and \lambda_{1} =0.1 \; m . So \mathbf{k_{1}=62.83\; rad/m} .

B) Group velocity is  \frac{d\sigma }{dk}. So taking derivative of the dispersion relation we get, \frac{d\sigma }{dk}=\frac{g+3\sigma {}'k^{2}/\rho }{2\sqrt{gk+\sigma {}'k^{3}/\rho }} . Putting the value of the constants we get \mathbf{\frac{d\sigma }{dk}\approx 0.218\; m/s} at point 1.

C)The wave number at point 2 is k_{2}=(2\pi /\lambda _{2}) where \lambda_{2} =0.017 \; m .\mathbf{k_{2}=369.6\; rad/m}.

D) Group velocity is  \frac{d\sigma }{dk}. So taking derivative of the dispersion relation we get, \frac{d\sigma }{dk}=\frac{g+3\sigma {}'k^{2}/\rho }{2\sqrt{gk+\sigma {}'k^{3}/\rho }} . Putting the value of the constants we get \mathbf{\frac{d\sigma }{dk}\approx 0.232\; m/s}

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