For a driven damped oscillation, if the driven force $F = F_0 \cos(\omega t)$, then the solution to the motion is $$x = A \cos(\omega t+\varphi ) \, .$$

  • Why must the the oscillation and the driven force have the same frequency but can be out of phase?

The solution to $\varphi$ is $\tan(\varphi )=\omega \gamma / (\omega_0^2 - \omega^2)$. ($\gamma$ is the damping coefficient divided by mass for a spring system)

  • Why would $\varphi$ goes to zero (meaning $F$ and $x$ are in phase) if $\omega$ goes to zero?
  • $\begingroup$ The symbol $\gamma$ is not defined anywhere. Please define all symbols used in Physics.SE posts. $\endgroup$
    – DanielSank
    Jan 12, 2016 at 8:00

1 Answer 1


I'll start with the second part of your question:

If the frequency of the force, $\omega$, goes to zero then it is just a constant force - there won't be any oscillation! Since there is no oscillation the force and the motion will be in phase (by default).

To answer why the force and motion can HAVE separate phases in the first place we look at the differential equation that describes the motion. Let $x$ be the displacement of the object about the central (rest) point. For the system you describe we will have:

  • A restoring force, $F_R$, with equation $$F_R=-kx$$ where $k$ is a positive constant (often called spring constant).

  • A damping force, $F_D$ proportional to the objects velocity, which takes the form $$F_D=-c\frac{dx}{dt}$$ where $c$ is also a positive constant.

  • and a driving force, which in your question takes the form $$F(t)=F_0\cos(\omega t) \, .$$

Applying Newton's second law ($F=ma$): $$\Sigma F=m\frac{d^2x}{dt^2}\\ \implies F_0\cos(\omega t)-c\frac{dx}{dt}-kx=m\frac{d^2x}{dt^2} \, .$$ Solving this differential equation isn't too hard, and I encourage you to look up the details on solving it. The result is: $$x(t)=x_0\cos(\omega t+\phi)$$ where $x_0$ and $\phi$ depend on the other parameters given.

  • $\begingroup$ Yes you're correct @RobJeffries, I'll edit. $\endgroup$
    – Robbie
    Jan 12, 2016 at 6:41
  • $\begingroup$ Just stating the solution to the differential equation is sort of begging the question, don't you think? $\endgroup$
    – DanielSank
    Jan 12, 2016 at 8:01

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