If you have a harmonic oscillator with damping $D$ (e.g. small angle pendulum)

$$\ddot{\theta}+D\dot{\theta} + \theta=0$$

then the solution I get in the underdamped case ($D^2-4<0$) is: $$\theta=\theta_0e^{-\gamma t}\cos{(\omega t)}$$ where $\gamma=\frac{D}{2}$ and $\omega=\frac{1}{2}\sqrt{4-D^2}$, and $\theta_0$ is the inital angular displacement. We have assumed the inital angular velocity $\dot{\theta}$ is zero.

The solution I get for the critically damped case ($D^2-4=0$) is: $$\theta=\theta_0(1+t)e^{-\gamma t}$$

Fine. Except that they don't match. When $D^2-4=0$, the underdamped case becomes $\theta=\theta_0e^{-\gamma t}$ which is not the same as the critically damped case.

Furthermore, if the pendulum is close to critically damped, the underdamped equation does not seem to be accurate, as shown by this simulation (using 4th order Runge-Kutta):

If the damping is smaller, the underdamped equation is closer to the simulation, but not quite there:

And if there is no damping, the resulting cosine wave is exactly the same as the simulation.

So does this mean the analytic solution for an underdamped pendulum is actually an approximation only valid for small damping, or have I done something wrong?

  • 2
    $\begingroup$ Your underdamped solution (first graph) seems to have a different starting condition: it starts with finite velocity towards the equilibrium (not a horizontal slope). Can't look at the details of your work right now but that is definitely a red flag. Note the general solution would have a $+\phi$ term... $\endgroup$
    – Floris
    Commented Nov 12, 2015 at 18:14
  • $\begingroup$ Ah, there's the problem: the equation for underdamped motion should have a sine term it seems $\endgroup$
    – binaryfunt
    Commented Nov 12, 2015 at 18:21

1 Answer 1


Your equation for the damped solution is wrong. In order to match the boundary conditions (initial velocity = 0) you have to add either a phase, or a $\sin$ term. I prefer the phase. If initial velocity is zero, the derivative must be zero:

$$A_t = A_0 e^{-t/\tau} \cos(\omega t + \phi)\\ v_t = A_0\left(-\frac{e^{-t/\tau}}{\tau}\cos(\omega t + \phi) - e^{-t/\tau}\omega \sin(\omega t + \phi)\right)$$

Setting $v_t=0$ at $t=0$ we obtain

$$ 0 = A_0\left(-\frac{1}{\tau}\cos(\phi) - \omega \sin(\phi)\right)\\ \tan\phi = -\frac{1}{\omega \tau}$$

Check my math. If I got it right, the solution should then match exactly.

Note that the solution for underdamped will approach the solution for critically damped, but setting $D=0$ in the one will not result in the other because of degeneracy in the solution (which is why there is an additional $+t$ term in the solution for the critically damped case). You should get really close though...


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