Question

Solve the harmonic oscillator motion for initial conditions x(0) = 0, V(0) = V₀ in the case of (a) underdamped $\gamma < \omega$ (b) overdamped $\gamma > \omega$

Answer 1

$\ddot{x}&plus;2\gamma\dot{x}&plus;\omega^2x=0$

Try the solution of the form

$x=e^{at}$

$a^2&plus;2\gamma a&plus;\omega^2=0$

$a=-\gamma\pm\sqrt{\gamma^2-\omega^2}$

(a) Roots are complex conjugates,

$a=-\gamma\pm i\,\Omega$

where

$\Omega=\sqrt{\omega^2-\gamma^2}$

The general solution is

$x(t)=A'e^{(-\gamma&plus;i\Omega)t}&plus;B'e^{-(\gamma&plus;i\Omega)t}$

I can always adjust A' and B' such that exponential of imaginary quantities are removed using Euler's relation. So the general solution is

$x(t)=e^{-\gamma t}(A\cos\Omega t&plus;B\sin\Omega t)$

For arbitrary constants A and B, to be determined by initial conditions which are

$x(t=0)=0,\ \dot{x}(t=0)=v_0$

which imply

$x(t)=\frac{v_0}{\Omega}e^{-\gamma t}\sin\Omega t$

(b)

Roots are real,

$a=-\gamma\pm \Omega'$

where

$\Omega'=\sqrt{\gamma^2-\omega^2}$

The general solution is

$x(t)=A'e^{(-\gamma&plus;\Omega')t}&plus;B'e^{-(\gamma&plus;\Omega')t}$

As before adjust A' and B' so that the solutions look more stylish! Use hyperbolic functions this time

$x(t)=e^{-\gamma t}(A\cosh\Omega' t&plus;B\sinh\Omega' t)$

For arbitrary constants A and B, to be determined by initial conditions which are

$x(t=0)=0,\ \dot{x}(t=0)=v_0$

which imply

$x(t)=\frac{v_0}{\Omega'}e^{-\gamma t}\sinh\Omega' t$