Question

(30 marks) For water waves with wavelength much longer than Xo, the effect of surface tension can be neglected. These waves are called gravity waves. =%and find its value given σ=0.073 N/m and ρ: 10000 kg/ms fr (a) Show that water at 20°C. (b) Gravity waves with kh » are called deep gravity waves. Deep gravity waves generated by a storm arrived at the coast have a period of about 15 seconds. A day later, the period of the waves arriving dropped to about 12.5 seconds. Roughly how far away from the coast did the storm occur? Hint: Express the group velocity in terms of the period of the waves.]

Answer 1

Ans:

1. For gravity waves,

Equation of motion are,

∇ v = 0

ρ dv/dt + ρ (v ∇) v = - ∇p – ρg e + µ ∇² v,

Where ρ = density, µ = viscosity, g = gravitational acceleration

Let us write

P(r,t) = p₀ - ρg z + p₁ (r, t)……..(1)

Where p0 = atmospheric pressure

P1 = pressure due to wave,

in the absence of the wave, the water pressure a depth h below the surface is p₀ + ρg h

ρdv/dt = - ∇p₁ + µ∇² v

where we have neglected terms that are second order in small quantities,

now neglect viscosity, so our equation becomes,

ρdv/dt = - ∇p₁

taking curl of this equation,

ρdw/dt = 0,

where w = ∇ * v

v = -∇ϕ, ϕ = velocity potential

the velocity field is also divergence free. It follows that the velocity potential satisfies Laplace's equation,

∇ϕ² = 0

p₁= ρdϕ/dt

We now need to derive the physical constraints that must be satisfied at the water's upper and lower boundaries. It is assumed that the water is bounded from below by a solid surface located at z=-d

dϕ/dt = 0

take vertical displacement f

df/dt = v = - dϕ/dt

Accourding to above equation (1)

p₀ = p₀ - ρg z + p₁

Now ρgdf/dt = - ρgdϕ/dz = dp₁/dt

            $$\displaystyle \left.\frac{\partial \phi}{\partial z}\right\vert _{z=0} = -g^{\,-1}\left.\frac{\partial^{\,2}\phi}{\partial t^{\,2}}\right\vert _{z=0}.$$

For a wavelike solution, ϕ(r,t) = F(z)cos(wt – kx)

d²F/dz² – k²F = 0

solution of this eq. exp(+kz) and exp(-kz)

hence

ϕ(x,z,t) = Aexp(kz)cos(wt – kx) + Bexp(-kz)(wt-kx)

A and B are arbitary constant, B = A exp(-2kd)

Φ(x,z,t) = A(exp(kz) + exp(-k(z+2d))cos(wt-kx)

Ak (1-exp(-2kd)cos(wt-kx) = Aw²/g(1 + exp(-2kd)cos(wt-kx)

Which reduces to be the dispersion relation

w² = gktanh(kd)

So k_m = (gρ/T)1/2,

λ = 2π (T/ gρ)^1/2

λ = 0.017 m

2. Deep water limit

kd >> 1

d = depth

now w = (gk)^1/2

Phase velocity = v_p = w/k = (g/k)^1/2

Group velocity = vg = dw/dk

v_g = ½(g/k)^1/2 = 1/2v_p,

v_g = ½(λ/T)

here λ₀/T₁ = λ/T₂, here λ₀ = 0.017m, T₁ = 15 s, T₂ = 12.5 s

λ = 0.014 m