Answer to Question #102684 in Mechanics | Relativity for sakshi

Question

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)

Answer 1

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

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

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