If I have an over-damped mechanical system that is excited with a sinusoidal motion. That sinusoidal motion starts with a determined frequency then increases frequency over time. Of course, it is known that there will be a phase shift between the driving force and the motion of the hanging mass.

My question is, how to figure out phase lag of mass motion in relation to driving force?


4 Answers 4


A damped harmonic oscillator with a sinusoidal driving force is represented by the equation

$$\ddot{x} + \gamma\dot{x} + \omega_0^2x = \frac{F_D \sin(\omega_D t)}{m}$$

where $\gamma = b/m$ ($b$ is the damping coefficient, $b=F/v$) and $\omega_0^2 = k/m$ is the resonant frequency of the oscillator. The particular solution to this equation can be determined by taking the imaginary part of the solution to

$$\ddot{x} + \gamma\dot{x} + \omega_0^2x = \frac{F_D}{m}e^{i\omega_D t}$$

If you assume* the solution takes the form

$$x(t) = A e^{i(\omega_D t + \phi)}$$

and plug that in, you get

$$-A \omega_D^2 + \omega_0^2 A = \frac{F_D}{m}\cos(\phi)$$


$$\gamma\omega_D A = \frac{F_D}{m}\sin(\phi)$$

Solving for the phase difference gives

$$\tan\phi = \frac{\gamma\omega_D}{\omega_0^2 - \omega_D^2}$$

This depends on the frequency of the driving force and the resonant frequency of the oscillator, but not on the amplitude of the driving force.

You can express this in terms of the dimensionless variable $x = \omega_D / \omega_0$ as

$$\tan\phi = \frac{\gamma}{\omega_0}\frac{x}{1 - x^2}$$

and if you graph it,

plot of phi versus x

(graph generated by Wolfram Alpha) you'll see how the response of the oscillator jumps from leading to lagging when $\omega_D = \omega_0$ (at $x=1$), that is, when the driving and resonant frequencies are equal.

*The same solution can be obtained from Fourier decomposition without making this assumption.

  • $\begingroup$ @ David's answer: "the response of the oscillator jumps from leading to lagging when ωD=ω0"??? You are simply taking the wrong quadrant when using the arctan of phi! Of course, the phase shift varies CONTINUOUSLY and does not "hop" by an amount of pi when the driving frequency reaches the characteristic frequency. The response is always lagging as you can find in any elementary book on mechanics. Instead the tan^-1 key on your calculator does the hopping for it cannot know in which quadrant your argument is located... you're really a grad student in elementary particle physics??? $\endgroup$
    – user14078
    Oct 15, 2012 at 19:48

You can see this for yourself by swinging a pendulum or waving a flexible curtain rod. If you swing it slower than the resonance frequency, the mass (or the other end of the curtain rod) just follows your hand, so the phase lag is zero. If you swing faster than the resonance frequency, then the mass does the opposite of what your hand does, so the phase lag is 180 degrees. As you approach the resonance frequency from slow to fast, the phase starts going from 0 to 180, and the amplitude increases a lot, hitting a maximum at the resonance frequency and decreasing afterwards. Swinging at exactly the resonance frequency using only your hand is pretty much impossible, but if you could, you would see that the phase lag was 90 degrees - halfway in between 0 and 180.

  • $\begingroup$ I understand that part.. but I meant mathematically, if I know all the characteristics of the driving force, like amplitude and frequency, and number of cycles for each frequency, then how can I (mathematically) calculate the phase difference in radians? $\endgroup$
    – mbadawi23
    Nov 20, 2010 at 20:22

David has the correct answer, with the addition that if $\gamma$ is not known is can be calculated if the system is driven at the natural frequency and the amplitude is measured (you mentioned this is an overdamped system).

$$ A_{x=1}=F_D/(\gamma\,m\,\omega_0) $$


$$ \gamma=F_D/(A_{x=1}\,m\,\omega_0) $$


That the phase $\phi = \arctan \left( \frac{2 \xi \omega \omega_0}{\omega^2 -\omega_0^2} \right)$ becomes equal to $\pi/2$ when $\omega \rightarrow \omega_0$ (here $\omega_0=3$) and $\xi \rightarrow 0$ can be seen from the plot below. For $\omega \rightarrow \omega_0=3$, we get $\phi=\pi/2$ for all nonzero values of $\xi$. When $\xi=0$, we get a step function (not shown in the plot), but we can consider $\phi=\pi/2$ for any value of $\xi$ that is arbitrarily close to zero.

enter image description here

  • 1
    $\begingroup$ An expanded version of this answer is here. $\endgroup$
    – rob
    Oct 23, 2018 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.