# Change in period of damped pendulum system [closed]

Suppose we have a simple pendulum damped by air resistance, proportional to the velocity of the pendulum. By using the small angle approximation of sin, we are able to solve a second order differential equation and arrive at the conclusion that the angle from the vertical, $\theta$, is equal to a trig function multiplied by a decaying exponential $$\theta(t) = A~\left(e^{-bt/2m}\right)\sin (ft + \omega)$$

It is evident that the amplitude of successive swings become smaller, yet the frequency of the oscillation $f$ remains constant, according to this.

Evidently wrong, how would one be able to quantify such a change in period of such a damped pendulum, as a function of time?

## closed as unclear what you're asking by sammy gerbil, stafusa, M. Enns, Daniel Griscom, Kyle KanosNov 18 '17 at 15:47

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Welcome on Physics SE :) I am having trouble understanding what your question exactly is about - is it how to measure that change or ...? Could you please edit your question to elaborate your last sentence a bit? – Sanya Oct 29 '16 at 9:26
• -1. Not clear what you are asking. First you claim (using the small angle approximation) that the frequency is constant. Then you state without explanation that this is wrong and that the period changes. – sammy gerbil Nov 18 '17 at 0:20
• Possible duplicate of Why is the simple harmonic motion idealization inaccurate? – sammy gerbil Nov 18 '17 at 0:50
• Possible duplicate of Why is the simple harmonic motion idealization inaccurate? – stafusa Nov 18 '17 at 1:14

A damped oscillator (small angle) pendulum is characterised by the following equation of motion:

$$\ddot{x}+2\zeta\omega_0\dot{x}+\omega_0^2x=0\tag{1}$$

Where $\omega_0$ is the natural (undamped) angular velocity:

$$\omega_0=\sqrt{\frac{L}{g}}=2\pi f_0$$

And $\zeta$ is the damping ratio (with $c$ a constant):

$$\zeta=\frac{c}{2\sqrt{Lg}}$$

Underdamped oscillation occurs for $\zeta<1$, in which case the damped angular velocity is:

$$\omega_1=\omega_0\sqrt{1-\zeta^2}=2\pi f_1$$

So the frequency of the underdamped oscillator is smaller than the natural one but doesn't change in time.

The exponential decay function $e^{-\lambda t}$ is defined by:

$$\lambda=\omega_0\zeta$$

For the undamped oscillator, $(1)$ reduces to:

$$\ddot{x}+\omega_0^2x=0$$

Which is the equation of motion of the simple harmonic oscillator.

"the frequency of the oscillation f remains constant" - this statement is dubious: if you consider the Fourier expansion of your formula, you'll see that there is not a single frequency, but some spectrum.

I have rewritten your equation slightly to tie in with @Gert 's analysis:

$\theta(t) = A~\left(e^{-\lambda t}\right)\sin (wt + \phi)$

I have looked at the peaks and troughs of the oscillations.

To proceed one needs to find the turning points of your function.

$\dot \theta = A e^{-\lambda t}\left(-\lambda \sin (wt + \phi) + \omega \cos(\omega t+\phi) \right) = 0$

So either $e^{-\lambda t}=0$ or $-\lambda \sin (wt + \phi) + \omega \cos(\omega t+\phi) = 0$

$e^{-\lambda t}=0$ has no solutions for $t$

$- \lambda \sin (wt + \phi) + \omega \cos(\omega t+\phi) = 0$ has solutions

$t = n\pi +\tan^{-1}\left(\dfrac{\omega}{\lambda} \right )-\phi \quad n \in \mathbb {Z}$

This show that the period (and hence the frequency) of the oscillations is the same but the peaks and troughs occur slightly earlier in time than if the oscillations were undamped.