# Is angular frequency dependent on time in damped harmonic motion?

I have a doubt regarding the angular frequency of a harmonic oscillator when there is damping involved. The frequency of the oscillation changes with time in the case of damping, but I haven't seen mention of this anywhere. I would like to find how the angular frequency depends on time (I'm guessing there must be some function $\omega=\omega(\omega_0,t,\beta)$ or something like that, where $\beta$ refers to the damping coefficient and the $\omega$'s refer to frequencies).

I checked with Landau and Taylor; neither of them, as far as I can see, discuss this phenomenon (although of course they talk about the decrease in amplitude and all that).

I'm pretty sure this phenomenon of frequency decreasing with time does occur (I checked quickly with a mish-mash harmonic oscillator), so why doesn't anyone mention it?

Could someone explain to me the time dependence of frequency when there are damped oscillations? Or maybe point out resources I could check out that do talk about this?

• I think Laudau had discussed this, see section 25 in Mechanics and formula 25.4 therein. It says: the amplitude will decrease with time but the frequency is decreased but independent of time actually. – an offer can't refuse Sep 21 '14 at 11:01

An (undriven) damped harmonic oscillator (satisfying $m\ddot{x} + \gamma \dot{x} + \omega_0^2 x=0$) can be solved by the solution(s) $x_0e^{i \omega t}$. For an underdamped oscillator these solutions represent pure oscillations mixed with exponential decay(/growth). Because both solutions for $\omega$ oscillate with the same period, all combinations of them also oscillate with the same period.

I suppose that your confusion arises from an intuitive idea of why a restoring force leads to periodic motion. In the undamped case the phase space trajectories of the particle are closed (i.e. the particle always returns to the same position(s)). In the damped case the trajectories spiral towards rest. But both of these motions are periodic in the sense that they reach their relative extrema (and zero crossing) after a specific interval of time. How can you see this? Perhaps you can take some inspiration from the undamped case: notice that the period is independent of the size of the orbit. Because the equation is linear any change in amplitude is exactly accounted for by a change in the force. In fact all linear (homogenous) ODE's can be satisfied by the anstaz above (i.e. solutions are periodic, damped, or both).

Damping an oscillation changes the frequency in two ways.

The oscillation can no longer be said to take place at a single frequency, but covers a continuous distribution of frequencies characterised by a Lorentzian profile, with a width that increases with the damping.

Second, the peak of this distribution occurs at a lower frequency than the natural frequency of the system, by an amount which increases with the damping.

i.e. the frequency does not depend on time.

The above applies to linear oscillators where the restoring force and damping terms are linearly dependent on displacement and velocity respectively. In non-linear oscillators things can be different. One can produce a relationship between frequency and amplitude (and hence time). e.g a pendulum actually has a frequency that decreases with amplitude and therefore increases with time as the amplitude is damped. Here is an animation of large and small amplitude pendulums that enables a comparison to be made.

This simple pendulum is an example of a "soft spring" situation - where the restoring force becomes less than an extrapolation of a linear relationship with displacement (or angle in the case of a pendulum) at large amplitudes. To get a frequency which decreases as the amplitude is damped requires a "hard spring" - for example a spring where the restoring force varies as $\alpha x + \beta x^3$ with $\alpha, \beta>0$. These non-linear springs are often referred to as Duffing oscillators.

• When I take out the tweezers from my Swiss Army knife, they oscillate (the tongs or whatever it's called) and if they're next to my ears I can hear a hum, whose frequency decreases with time. I guess this isn't a linear oscillator. Do you know what it could be? – Physics Llama Sep 21 '14 at 19:45
• Here is an applet showing a pendulum. $d^2\theta/dt^2 + \sin(\theta) = 0$. peter-junglas.de/fh/physbeans/applets/fricmathpendulum.html You can see that the period of oscillation decreases with time, because of the non-linear behaviour (i.e. the frequency increases with time). I don't know enough about this to say whether if the restoring force varied in the complementary way that this would produce a decreasing frequency. – Rob Jeffries Sep 21 '14 at 22:37
• @Physics Llama I think you are looking for something called a "Duffing oscillator" with a positive, non-linear term in the restoring force. – Rob Jeffries Sep 27 '14 at 20:42
• I looked into it, and here's the paper that will be most helpful to me: Dynamics of Transversely Vibrating Beams Using Four Engineering Theories csxe.rutgers.edu/research/vibration/51.pdf – Physics Llama Sep 27 '14 at 22:45