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Suppose we have a simple pendulum executing SHM about its mean position. We can easily notice that the bob of the pendulum swings back and forth making an angle, say $\theta$ about its central perpendicular (mean position). Let $\theta_{max}$ be the maximum angle subtended on either side when the bob reaches the maximum height.

The total time period, as the bob swings across to reach the same angle on the other side is $T$. However consider two angles $\alpha$ and $\beta$, that the bob of the pendulum subtends at some time during the SHM. How can we calculate the time spent by the bob between these angles as a fraction of the total time period?

My initial idea was, SHM can be represented as a circle, and the total time period of each cycle in a sinusoidal motion is $2\pi$. In a single cycle, the bob traverses the path between the two angles twice (back and forth). Thus we can use the following formula to find the fraction of the total time period that the bob spends in this region.

$$t = \frac{2(\alpha - \beta)}{2\pi} = \frac{\alpha-\beta}{\pi}$$

The time spent between two angles is the difference between the two angles divided by the total angle traversed, which is $2\pi$.

However, I was looking for a more rigorous approach to the problem, maybe I could solve it by integration. Any help would be appreciated.

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  • $\begingroup$ I suggest adopting the method explained here $\endgroup$
    – Roger V.
    Commented Jul 22, 2021 at 9:54
  • $\begingroup$ U can calculate the time required for θ(max) and then subtract the time required for θ-dθ and then limit as dθ approaches 0. This is for θ max. For other angle range just increase dθ upto which u want answer. $\endgroup$
    – Samyak Marathe
    Commented Jul 22, 2021 at 10:34

3 Answers 3

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Your idea of representing simple harmonic motion as a circle is the right concept. However, you should note that the $\alpha$ and $\beta$ in your equation is the phase angle $\omega t$ and not the actual angle of the pendulum. The denominator should also be $2\pi$ instead of $\pi$, since one oscillation is always $2\pi$.

Suppose we have a simple pendulum executing SHM about its mean position.

However, the issue here is that the pendulum is not simple harmonic.

I was looking for a more rigorous approach to the problem

If you are looking for a more rigorous approach, you have to drop the small-angle approximation. The differential equation for the pendulum is $$\ddot{\theta} = -\frac{g}{l} \sin\theta$$ which can be manipulated into an integral expression for the exact time taken. It is a fairly easy exercise; the final result is $$t = \sqrt{\frac{l}{2g}} \int_{\theta_1}^{\theta_2} \frac{\text{d}\theta}{\sqrt{\cos\theta - \cos\theta_0}}$$ where $\theta_1$ and $\theta_2$ are the initial and final angles of the pendulum respectively. This is an elliptic integral which has no closed-form expression. You could try numerical integration or a series expansion, depending on what you had in mind.

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  • $\begingroup$ Elliptic integral can be used very easily with a software as Maple, Mathematica, or Python. $\endgroup$
    – Vincent Fraticelli
    Commented Jul 22, 2021 at 10:22
  • $\begingroup$ Thanks, this is exactly what I was looking for. I had made the mistake of thinking the phase angle to be the same as the 'real' angle. Thank you so much. $\endgroup$
    – Nakshatra Gangopadhay
    Commented Jul 22, 2021 at 12:32
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Even in small angle approximation the velocity of the pendulum is variable and not a constant. The time depends on the actual values of the angles and not just on the difference. It takes longer to swing by 1 degree when closer to an end point than when closer to equilibrium. And this is so even for the SHO approximation. You need to use the formula with inverse sin given in gandalf61's answer.

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If we remain in the context of SHM (and so avoid the complications of elliptic integrals) then the angle $\theta(t)$ that the pendulum makes with the vertical at time $t$ is given by

$$\displaystyle \theta(t) = \theta_{max} \sin \left(\frac{2\pi t}T\right)$$

where $T$ is the period of the pendulum (note that this is the time taken for the bob to swing from one extreme to the other and return) and I am assuming that $\theta=0$ (i.e. the pendulum is vertical) at $t=0$. So if $\theta=\alpha$ at time $t_\alpha$ then we have

$$\displaystyle \alpha = \theta_{max} \sin\left(\frac{2\pi t_\alpha}T\right) \\ \displaystyle \Rightarrow t_\alpha = \frac T {2\pi} \sin^{-1} \left( \frac \alpha {\theta_{max}}\right)$$

If $\theta=\beta$ at time $t_\beta$ then we have

$$\displaystyle t_\beta - t_\alpha = \frac T {2\pi} \left( \sin^{-1} \left( \frac \beta {\theta_{max}}\right) - \sin^{-1} \left( \frac \alpha {\theta_{max}}\right) \right)$$

If (and only if) $|\alpha|$ and $|\beta|$ are small compared to $\theta_{max}$ (i.e. $|\alpha|, |\beta| \ll \theta_{max} \ll \pi$) then we can use the small angle approximation $\sin(x) \approx x$ to get

$$\displaystyle t_\beta - t_\alpha \approx \frac {T(\beta - \alpha)} {2\pi \theta_{max}}$$

Note that if $\alpha$ and $\beta$ are on opposite sides of the vertical line $\theta=0$ then one is positive and the other is negative.

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    $\begingroup$ Both $\alpha$ and $\theta_{max}$ are small but their ratio is not. It can be anything between 0 and 1. The last approximation is not justified. It assumes uniform angular velocity. $\endgroup$
    – nasu
    Commented Jul 22, 2021 at 12:24
  • $\begingroup$ @nasu Yes, $\theta_{max}$ is small if we assume SHM, but note that I said the small angle approximation is only valid if, in addition, $\alpha$ and $\beta$ are small compared to $\theta_{max}$. This is still possible - we could have $\alpha, \beta \ll \theta_{max} \ll \pi$ $\endgroup$
    – gandalf61
    Commented Jul 22, 2021 at 12:35
  • $\begingroup$ But to approximate the inverse sin you need the ratio to be small and this is not true even when the angles are small. The approximation will work only if both angles are very close to equilibrium but not if one of them is close to $\theta_{max}$. $\endgroup$
    – nasu
    Commented Jul 22, 2021 at 17:27
  • $\begingroup$ @nasu Yes, that’s exactly what I said.We can use the small angle approximation if and only if $\alpha$ and $\beta$ are both much smaller than $\theta_{max}$ (which is possible). If they are not ... then we cannot use that approximation and so we have to use the full $\sin^{-1}$ version. You seem to think that it is impossible for $\alpha$ and $\beta$ to both be much smaller than $\theta_{max}$, but I don’t understand why you think that. $\endgroup$
    – gandalf61
    Commented Jul 22, 2021 at 18:04
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    $\begingroup$ But this second condition is not assumed in the OP. You already make an approximation ($\theta_{max}<<1$ in order to use the SHO equation for the motion. Then you make some sort of second order approximation assuming that the angles are also much smaller than $\theta_{max}$. Why would you do this? If you take a portion of trajectory small enough to consider constant speed, then this works for any motion and has nothing to do with pendulum motion anymore. You just get the definition of instantaneous speed, $\endgroup$
    – nasu
    Commented Jul 23, 2021 at 16:25

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