# What is the period of a physical pendulum without using small-angle approximation?

What is the expression for the period of a physical pendulum without the $\sin\theta\approx\theta$ approximation? i.e. a pendulum described by this equation: $$mgd\sin(\theta)=-I\ddot\theta$$

Motivation for my question:

I'm asking this because I had an assignment in which I had to measure the period of oscillation of a physical pendulum and we always had to release it precisely from the same angle. So I'm wondering whether that precaution is irrelevant because it feels to me that the measurement error and error created by ignoring air drag is more significant than those that would emerge by not caring about being precise about the initial angle, as long as it is smaller than circa 30°. Am I right or wrong?

-
Thank you, I don't understand the downvote either... I'll check that! –  Schlomo Steinbergerstein Mar 21 at 20:36
Everything's fine :) –  Schlomo Steinbergerstein Mar 21 at 21:05
You might also like this article from John Baez. –  Andrew Gibson Apr 4 at 17:34

Sometimes, a good figure is worth more than a thousand equations :)

I numerically integrated the following equation of motion for a physical pendulum:

$$I\ddot{\theta} - mgL\sin(\theta) + \frac12\mathrm{sgn}({\dot{\theta}})L\rho_{\mathrm{air}}C_DS(L\dot{\theta})^2 - \gamma\theta = 0$$

with $\mathrm{sgn()}$ the signum function. The second term in this equation is the torque exerted by gravity, the third term is due to air drag (which is assumed here to only act on the bob), and the fourth term is a linear drag effect caused by simple bending of the string at the attachment point (therefore, I'm assuming the pendulum was constructed with a string).

I determined the progression of the pendulum's period simply by differencing the zero passes, times 2. I used the following values in the computations (which I think are pretty reasonable):

• $I$: mass moment of inertia of compound system ($mL^2 + 0.2$)
• $m$: mass of the bob (1 kg)
• $g$: gravitational acceleration at sea-level (9.80665 m/s2)
• $L$: length of the pendulum (1 m)
• $\rho_{\mathrm{air}}$: air density at sea level (1.225 kg/m3)
• $C_D$: Combined drag coefficient (spherical bob+string, 0.5)
• $S$: frontal surface area (0.2 m2)
• $\gamma$: drag ratio (0.05)

I normalized the periods thus determined, by dividing them by the period that follows from linearized theory for a compound pendulum (see the wiki, $T=2\pi\sqrt{I/mgL}$), and plotted the results for three cases:

1. no friction, no air drag
2. only friction
3. friction + air drag

for each case, I used three initial starting angles:

1. $\theta_0 = 1^\circ$
2. $\theta_0 = 15\circ$
3. $\theta_0 = 30^\circ$

Here are the results:

So, in conclusion:

• Large angles indeed induce error when compared to the approximating formulae. But actually not very much; for the 30$^\circ$ you stated, the error is in the order of ~2%. You'd have to increase it to $73^\circ$ to reach 10% error.
• The large-angle error is affected by torsional dampening by an overall lowering of the frequency (this should be no surprise if you know your linearized theory well enough).
• It's the air drag that really messes everything up :) Air drag will make doing measurements accurately way more difficult, since the period has a fast rate of change directly after releasing the bob, and when its effects finally fizzle out, the amplitude of the motion (not shown here) is too small to measure accurately.

So I'd say you're right -- although the initial angle matters (which is what I think the exercise was intended to teach you), it doesn't matter as much as neglecting air drag does. It only starts mattering when you repeat the experiment in a vacuum chamber.

-
+1 Your final plot seems to suggest experimentalists should simply assume the small angle theorem and just wait for the real pendulum to come to that conclusion too. –  ejrb May 30 at 13:25
Mother of physics! Now that's what I call a great answer, man, thanks! :) P.S. @ejrb , hahahahaha great comment! –  Schlomo Steinbergerstein May 30 at 15:14
@SchlomoSteinbergerstein: hmm I just noticed there is an $L$ missing in the air drag part of the equation of motion...because of my choice of $L=1$ it will not make any difference, but it will actually increase the influence of air drag on longer pendula... –  Rody Oldenhuis May 30 at 16:01

This is pretty much well explained in the following Mathematica Simulation In general for large $\theta$ the frequency becomes a function of Amplitude and non-linearity sets in. Also the phase space plot becomes distorted.

The next error in small angle approx is keeping upto $\theta^3$ term, which approximately suggests corrections of order $\theta^2/3!$ in frequency. The air drag being proportional to $v=\omega L$, and So if the $\omega$ is small enough then you can neglect the effects of air drag etc.

An adhoc comparison gives which is more important

-
For large amplitudes, the period is corrected by a factor of $K(\sin(\theta/2))$, where $K$ is the complete elliptic integral of the first kind. For most applications, it suffices to take the first few terms of the Taylor series of $K$.
Dissipation may be an issue. Depending on how the pendulum is constructed, it may not be true that dissipation is dominated by drag from the air. It's possible that the main mechanism of dissipation is mechanical friction (if it's hung from a bearing) or transmission of vibrations through a string (if it's hung from a string). If it is dominated by drag from the air, then the drag force is probably proportional to $v^2$, although people usually model this kind of thing with a force proportional to $v$, which produces an exponential decay of the motion.
When moving twice as fast, air will hit you at twice the rate, and twice as hard, therefore, you will experience four times the force, so...not sure where the $v$ comes from. Can you show me an example? –  Rody Oldenhuis May 30 at 15:58