We are given that the oscillator obeys the following DE: $\ddot{\theta} + \omega_0^2\theta = \text{cos}(\omega t)$. The solution is $\theta = Ae^{i\omega_0t} + Be^{-i\omega_0t} + \frac{\text{cos}(\omega t)}{\omega_0^2 - \omega^2}$. Using the fact (given to us in the problem) that the oscillator is at rest at t = 0, we have that $\theta = 2A\text{cos}(\omega_0 t) + \frac{\text{cos}(\omega t)}{\omega_0^2 - \omega^2}$.

Here is where I am stuck. The question then asks us to find the RMS angle of the oscillator. I know that this requires us to evaluate $\theta^2$ and integrate it between $t = 0$ and $t = \frac{2\pi}{\omega_0}$. We are also allowed to assume that the average is taken over a time much larger than $\frac{1}{\mid \omega_0 - \omega\mid}$. (I don't really know what this means). I ended up with this expression:

$\displaystyle\int_{0}^{T}\theta^2 dt = \frac{1}{2(\omega_0^2 - \omega^2)^2}(\frac{\text{sin}(2\omega T)}{2\omega} + T) - \frac{4\omega A\text{sin}(\omega T)} {(\omega_0^2 - \omega^2)^2} + 2A^2T$ (where $T = \frac{2\pi}{\omega_0}$).

Now, I'm not even sure if I have evaluated the integral properly but I would like to know how I am supposed to use the previously mentioned assumption that they have provided in the question. Also, the question says that the oscillator (which is a pendulum) is in equilibrium at $t = 0$ - does this mean that $\theta = 0$ at $t = 0$? This would give $A = \frac{1}{2(\omega_0^2 - \omega^2)}$. We are also supposed to discuss the motion as $\mid \omega_0 - \omega\mid \rightarrow 0$. Can anyone help?

  • $\begingroup$ as you have defined it above $\theta$ is not the phase angle of the oscillator thus its mean square will not get you the rms angle $\endgroup$
    – hyportnex
    Commented Jan 4, 2016 at 17:55

1 Answer 1


The expression you use for the RMS angle, $$ \theta_\mathrm{rms}^2=\frac 1T\int_0^T\theta(t)^2\mathrm dt, $$ is only valid for a periodic function, with period $T$. Your solution is not periodic, since there is in general no expected relationship between the two frequencies $\omega$ and $\omega_0$, so that expression is not really valid.

In these cases the correct expression is the long-time limit of the average, $$ \theta_\mathrm{rms}^2=\lim_{T\to\infty}\frac 1T\int_0^T\theta(t)^2\mathrm dt, $$ which is insensitive to edge effects.

Regarding the $\omega\to\omega_0$ limit, it should be evident that your solution breaks down at the limit itself. However, there are standard tools for solving the equation of motion in that case (e.g. here). How does the behaviour change in the resonant case? How with this square with the limit of the $\theta_\mathrm{rms}$ you found previously when you take $\omega\to\omega_0$?


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