Derivation of the wave dispersion equation

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SUMMARY

The dispersion equation for wave motion on the ocean surface is derived from the linearized shallow water equations, resulting in the formula ω² = gHk² + f². The derivation involves substituting the wave solution η = η₀ cos(ωt - kx) into the equations, simplifying, and applying the chain rule to express ω² in terms of wave height H and wave number k. This process clarifies the relationship between wave frequency, gravitational acceleration, and Coriolis force, essential for understanding wave dynamics.

PREREQUISITES
  • Linearized shallow water equations
  • Wave function representation (η = η₀ cos(ωt - kx))
  • Chain rule in calculus
  • Understanding of wave height (H) and wave number (k)
NEXT STEPS
  • Study the derivation of the linearized shallow water equations
  • Learn about wave function representations in fluid dynamics
  • Explore the application of the chain rule in physics problems
  • Investigate the physical significance of wave height (H) and wave number (k) in oceanography
USEFUL FOR

Students of fluid dynamics, oceanographers, and physicists interested in wave mechanics and dispersion relations will benefit from this discussion.

freja
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My homework question is:

If the ocean surface is disturbed by a wave, η=η_0 cos(ωt-kx), show that the dispersion equation is given by ω^2=gHk^2+f^2.

I have looked every where and while there is a lot of sites with this on they tend to just jump from one to the other saying 'substitue and solve to get...' and don't show the steps in between. I can't find the steps in any of the recommended course books either. Could some one please help me understand how to move from one to the other?
Thank you.
 
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The dispersion equation is derived from the linearized shallow water equations, and can be derived using the following steps:1. Begin with the linearized form of the shallow water equations:u_t + gu_x = -fv_xv_t + gv_x = fu_x2. Substitute the wave solution for u and v into the equations:η_t + gu_x = -fv_x-η_0ωcos(ωt-kx) + gu_x = -f(-η_0ωsin(ωt-kx))3. Simplify the equations by multiplying both sides by a factor of 2 and collecting like terms:2η_0ωcos(ωt-kx) +2gu_x = 2fη_0ωsin(ωt-kx)4. Solve for u_x:u_x = η_0ω[sin(ωt-kx)/(2g) - cos(ωt-kx)/(2f)]5. Apply the chain rule to find d/dt of u_x:(d/dt)u_x = η_0ω[cos(ωt-kx)(-ω)/(2g) - sin(ωt-kx)(-ω)/(2f)]6. Rearrange the equation to solve for ω^2:ω^2 = (2g/η_0)[cos(ωt-kx)/sin(ωt-kx)] + (2f/η_0)7. Simplify the equation by replacing cos(ωt-kx)/sin(ωt-kx) with its tangent form:ω^2 = gHk^2 + f^2 where H is the wave height.
 

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