I have a forced oscillating system, with driving force as $f_0\cos\omega_0 t \cos \delta t$ giving the equation of motion:

$$\ddot{x}(t) +\Gamma \dot{x}(t) +\omega_0^2x(t) = f_0\cos\omega_0 t \cos \delta t$$

Notation is standard, i.e $\Gamma$ is the damping coefficient, $\omega_0$ is the natural frequency of the undamped system, i.e $ \omega_0= \sqrt{k/m}$

It is given that $\delta \ll \omega_0$ and $\Gamma =0$

The question asks for the displacement $x$ for non-zero $\delta$ to leading order in $\delta/\omega_0$. The next statement is to express it in the for $\alpha(t)\cos\omega_0 t + \beta(t) \sin\omega_0 t $

The following hint is provided: $$\cos\omega_0 t \cos \delta t = \frac{1}{2} Re\Big(e^{-i(\omega+\delta)t}+e^{-i(\omega-\delta)t}\Big)$$

I don't want the solution. I cannot understand how to approach the problem.

I have the equation of motion

$$\ddot{z}(t) - \omega^2 z(t) = \frac{f_0}{2} \Big(e^{-i(\omega+\delta)t}+e^{-i(\omega-\delta)t}\Big)$$

I make use of the approximation $\delta \ll \omega_0$ I conduct the expansion to get

$$\ddot{z}(t) - \omega^2 z(t) = f_0 \Big(1-i t\omega_0 - t^2\omega_0^2 \Big(1+\Big(\frac{\delta}{\omega_0}\Big)^2\Big)\Big)$$

But this hasn't simplified it enough to allow me to solve it.

The methods I now know of solving the original equation (Green's function and variation of parameters) have not been introduced, so the author must require us to work under certain approximations.

Source is Howard Georgi's physics of waves. This is not homework.

  • $\begingroup$ The expansion of the exponentials in Taylor series is making things worse, not better. I don't think this is the way to make use of the approximation. Why not just assume a solution for z, with unknown amplitude and phase, and plug it in? $\endgroup$
    – user4552
    Commented Aug 20, 2011 at 3:20
  • $\begingroup$ Your approximation can't work. Just keep the $\cos \omega_0 t$ term (this doesn't hurt since we know how to deal with harmonic forcing) and expand only the $\cos \delta t$ part. In total you will get polynomial times harmonic as forcing and there are standard methods to deal with this. $\endgroup$
    – Marek
    Commented Aug 20, 2011 at 11:18

1 Answer 1


Do you know what the solutions to the equation of motion look like in the case where there's only one sinusoidal term on the right-hand side, i.e., $$ \ddot z-\omega^2z={f_0\over 2}e^{-i(\omega+\delta)t}, $$ and the corresponding equation with $\omega-\delta$? If so, then you're most of the way there. When you have an inhomogeneous linear differential equation with multiple terms on the right side (which is what you've got), you can solve it for each term separately and add the solutions together.

(If this isn't a fact you already know from either a physics or a mathematics class, you can probably convince yourself of it.)

To be precise, your solution can be written in the form $$ z(t)=z_+(t)+z_-(t)+z_g(t), $$ where $z_+$ is a particular solution to the equation containing just the first term on the right, $z_-$ is a particular solution with just the second term on the right, and $z_g$ is the general solution to the homogeneous equation (zero on the right).

(In case you don't know how to solve the two inhomogeneous equations, I'll mention that you can do it by guessing that the solution is of the form $Ae^{-i\alpha t}$ where $\alpha=\omega\pm\delta$ is the driving frequency, and working out what value of $A$ works. But I suspect you probably already knew that.)


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