I have the problem of a dampened harmonic oscillation (more concrete a "Pohl wheel" (here is an illustration of it)) whose motion is given by the following differential equation
$$J\frac{d^2 \alpha}{d t^2} + R \frac{d \alpha}{dt} + D \alpha = M_0e^{i\omega t},$$
where $J$ is the moment of inerta, $R$ a friction coefficient (The friction is assumed to be proportional to the angular speed of the oscillator) and $D$ is the feather constant (of the respective spiral fether that tries to force the oscillator back into it's equilibrium position).
As can be seen in the equation there is an external, periodic torque applied to the oscillator of frequency $\omega$ and with the amplitude $M_0$.

The solution to this differential equation is the complex function
$$\alpha(t) = c_1 \cdot exp\left(\frac{-R + \sqrt{R^2 - 4JD}}{2J}\right) + c_2 \cdot exp\left(\frac{-R - \sqrt{R^2 - 4JD}}{2J}\right)$$ $$ + \frac{M_0}{-J\omega + D + iR\omega} \cdot exp(i\omega t)$$

So far so good. The problem is solved mathematically but how do I have to interpret this complex function in order to describe the angle $\alpha(t)$ so that $\alpha(t)$ would represent the angle of such an oscillator in the "real world"?
Or in other words: What does it physically mean for this function to be complex?

And something else: The physics book of mine doesn't describe the external torque as complex via $M_0 \cdot e^{i\omega t}$ but as it's real part $M_0 \cos(\omega t)$. If the problem is solved with that approach are there any solutions missing or would that lead to the same physical result (by that I mean to the same result as when interpreting above complex function physically)?

I forgot to mention what $\alpha(t)$ should represent: It is the angle the oscillator is currently shifted away from it's equilibrium position.

  • $\begingroup$ Once you solve something like this, you can see if any solution in the complex form can be simplified to only real terms, then in the real world, that is what describes the system. $\endgroup$ Aug 14, 2017 at 15:09
  • $\begingroup$ Is that by applying initial conditions or is that something else? Like making an experiment and then try to force the complex function down to a real one? If that is so is there now way of directly deriving the "real function" from the solution of the differential equation (e.g. by doing $Re(\alpha)$)? $\endgroup$
    – Raven
    Aug 14, 2017 at 15:13
  • $\begingroup$ The problems seems to stem from the definition of the $\alpha(t)$ function or the driving function. What are they ? Either the driving function should be real (not complex) or the $\alpha(t)$ function should be complex and it's meaning follows from it's definition. $\endgroup$ Aug 14, 2017 at 15:18
  • $\begingroup$ Oh did I forget to mention that? My bad... $\alpha(t)$ is the angle the oscillator is being shifted from the equilibrium position. And the driving function (I assume you are referring to the $M_0e^{\omega t}$) is the external torque that is being forced on the oscillator $\endgroup$
    – Raven
    Aug 14, 2017 at 15:23
  • $\begingroup$ This is explained in e.g. physics.stackexchange.com/q/228279/2451 $\endgroup$
    – Qmechanic
    Aug 15, 2017 at 8:21

1 Answer 1


The actual differential equation corresponding to the model you describe is

$$ J\frac{d^2 x}{dt^2} + R \frac{dx}{dt} + Dx = M_0 \cos(\omega t)$$

Each term has a direct physical interpretation, everything is real-valued, and nothing should be confusing. You could solve it like this, and presumably all of your conceptual difficulties would go away.

From a mathematical point of view, however, the cosine on the right hand side is annoying; this becomes apparent when one includes the particular solution to the differential equation. It's not an insurmountable difficulty, but it's not convenient.

Consider a related differential equation: $$J \frac{d^2 \alpha}{dt^2} + R \frac{d\alpha}{dt} + D\alpha = M_0e^{i\omega t}$$

This is actually two equations in one. If we let $\alpha(t) = x(t) + i\cdot y(t)$, then it's not difficult to see that the real part of $\alpha$ obeys precisely the equation we wrote above, while the imaginary part obeys a similar equation with a sine instead of a cosine.

Therefore, we are free to solve for the complex function $\alpha(t)$ and then take the real part of that to get the solution which obeys the physical equation we wrote down in the first place. This turns out to be easier, so we typically do it, but you're perfectly entitled to keep the trig functions in from the start.

  • $\begingroup$ Thanks for the answer! Bit I have one more question... If I use the first differential equation (the "real one") I run into problems when $R^2 < 4JD$ because then the equation wouldn't have a real solution... Does this mean that at this point there is no oscillation happening? $\endgroup$
    – Raven
    Aug 15, 2017 at 9:26
  • $\begingroup$ @Raven Check again. If your initial conditions are real, then your solution will be real. $\endgroup$
    – J. Murray
    Aug 15, 2017 at 9:36
  • $\begingroup$ @Raven Alternatively, if you don't want to see any complex numbers whatsoever, then try a solution of the form $\alpha = e^{rt}\cos(\omega_0 t)$ while demanding that $r$ and $\omega_0$ are real. It'll take some extra work, but you'll find that under those constraints, $R^2<4JD$ implies that $r=-R/2J$ and $\omega_0 = \frac{\sqrt{4JD-R^2}}{2J}$. $\endgroup$
    – J. Murray
    Aug 15, 2017 at 9:51
  • $\begingroup$ Thanks a lot! I'll have a look at this and see how far I'll get :) $\endgroup$
    – Raven
    Aug 15, 2017 at 10:03

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