Answer to Question #197353 in Quantum Mechanics for asirvad jaif

Question #197353

Waves on deep water with surface tension T and density 𝜌 are governed by the dispersion relation πœ”2=π‘”π‘˜+𝑇 πœŒπ‘˜3. Calculate the phase and group velocities of the waves. Find the wave number kc at which phase velocity reaches a minimum. What is the group velocity for this wave number?


1
Expert's answer
2021-05-23T18:54:48-0400

The phase speed is given by

Ο‰2k2=gk+Tρk⟹cp2=gk+Tρk\frac{Ο‰^2}{k^2}=\frac gk+\frac Tρk⟹c_p^2=\frac gk+\frac Tρk

so Phase speed is

cp=±gk+Tρkc_p=±\sqrt{\frac gk+\frac Tρk}

The group speed is given by

2Ο‰βˆ‚Ο‰βˆ‚k=g+3Tρk22Ο‰\frac{βˆ‚Ο‰}{βˆ‚k}=g+\frac{3T}{ρ}k^2

so we have

vg=g+3Tρk22Ο‰v_g=\frac{g+\frac{3T}{ρ}k^2}{2Ο‰}

To determine the minimum phase speed we are essentially compute

βˆ‚cpβˆ‚k=0\frac{βˆ‚c_p}{βˆ‚k}=0

so we use the first relation

2cpβˆ‚cpβˆ‚k=βˆ’gk2+Tρ2c_p\frac{βˆ‚c_p}{βˆ‚k}=βˆ’\frac{g}{k^2}+\frac Tρ

This is zero for a trial case, cp=0c_p=0 , but in general we require

βˆ’gk2+Tρ=0⟹kc2=gρTβˆ’\frac{g}{k^2}+\frac Tρ=0⟹k_c^2=\frac{gρ}{T}

so the minimum phase speed occurs for

kc=gρTk_c=\sqrt{\frac{gρ}{T}}


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