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

"\\frac{\u03c9^2}{k^2}=\\frac gk+\\frac T\u03c1k\u27f9c_p^2=\\frac gk+\\frac T\u03c1k"

so Phase speed is

"c_p=\u00b1\\sqrt{\\frac gk+\\frac T\u03c1k}"

The group speed is given by

"2\u03c9\\frac{\u2202\u03c9}{\u2202k}=g+\\frac{3T}{\u03c1}k^2"

so we have

"v_g=\\frac{g+\\frac{3T}{\u03c1}k^2}{2\u03c9}"

To determine the minimum phase speed we are essentially compute

"\\frac{\u2202c_p}{\u2202k}=0"

so we use the first relation

"2c_p\\frac{\u2202c_p}{\u2202k}=\u2212\\frac{g}{k^2}+\\frac T\u03c1"

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

"\u2212\\frac{g}{k^2}+\\frac T\u03c1=0\u27f9k_c^2=\\frac{g\u03c1}{T}"

so the minimum phase speed occurs for

"k_c=\\sqrt{\\frac{g\u03c1}{T}}"


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