Question #102684

Write down the differential equation for a damped harmonic oscillator. What is the basis of representing the damping force in terms of velocity? Show that the average energy of a weakly damped oscillator is given by:

< E > = E0 exp (- 2bt)

< E > = E0 exp (- 2bt)

Expert's answer

1)

"\\sum F=ma"

"F_{damp}=-bv=-b\\frac{dx(t)}{dt}"

"F_{el}=-kx(t)"

"-b\\frac{dx(t)}{dt}-kx(t)=m\\frac{d^2x(t)}{dt^2}"

"\\frac{d^2x(t)}{dt^2}+\\frac{b}{m}\\frac{dx(t)}{dt}+\\frac{k}{m}x(t)=0"

So, the differential equation of a damped haronic oscillator

"\\frac{d^2x(t)}{dt^2}+2\\beta\\frac{dx(t)}{dt}+\\omega^2_0 x(t)=0" ,

where "\\beta=\\frac{b}{2m}" and "\\omega_0=\\sqrt{\\frac{k}{m}}"

2)

The forces of resistance and friction are always directed against the direction of the velocity vector and reduce the kinetic energy of the body

3)

"E(t)=KE(t)+PE(t)=\\frac{1}{2}m(\\frac{dx}{dt})^2+\\frac{1}{2}kx^2(t)"

"x(t)=A_0e^{-\\beta t}\\cos(\\omega t+\\phi)"

"\\frac{dx}{dt}=-A_0e^{-\\beta t}[\\beta \\cos(\\omega t+\\phi)+\\omega \\sin(\\omega t+\\phi)]"

"E(t)=\\frac{1}{2}mA^2_0e^{-2\\beta t}[(\\beta^2+\\omega^2_0) \\cos^2(\\omega t+\\phi)+"

"+\\omega^2 \\sin^2(\\omega t+\\phi)+\\beta \\omega \\sin2(\\omega t+\\phi)]"

We have

"\\langle E(t) \\rangle=\\frac{1}{2}mA^2_0e^{-2\\beta t}[\\frac{\\beta^2+\\omega^2_0}{2}+\\frac{\\omega^2}{2}]="

"=\\frac{1}{2}mA^2_0e^{-2\\beta t}\\omega^2_0=E_0e^{-2\\beta t}"

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