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I was told that the motion of a simple pendulum is harmonic only when its amplitude is much less than the length of the pendulum (let's take this as first statement). Why it is so?

Also my teacher told me that a simple pendulum is SHM for small angular displacement and that's the reason for the above statement. But I felt that both the statements are implying the same idea. What could be the reason for the first statement?


marked as duplicate by user36790, AccidentalFourierTransform, Bill N, Jon Custer, Floris Dec 21 '16 at 20:28

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  • $\begingroup$ Oh?! So the statement itself is wrong? $\endgroup$ – Harini Dec 21 '16 at 18:27
  • $\begingroup$ The statement that I stated there was given by my teacher and what's that you have typed? Theta...I couldn't decipher it :/ @accidentalfouriertransform $\endgroup$ – Harini Dec 21 '16 at 18:30

There are many ways to find the equation of the motion of a simple pendulum, but the easiest way seems to be using the conservation of angular momentum.

Let's consider a pendulum made of a mass $M$ and a rope of length $L$, and be $\theta$ the angle between the rope and a vertical line. Then, the only force applied on $M$ whose torque isn't null is the weight of $M$, so we get

$$ ML^2\ddot{\theta} = - Mg \sin(\theta) $$

or $$\ddot{\theta} = -\omega^2 \sin(\theta) $$

where $$\omega = \sqrt{\frac{g}{L}}$$

Now, we have to solve this non-linear differential equation. However, we prefer to deal with linear equation... So we can study a particular case: when $\theta \ll 1 $. Then, we know that $\sin(\theta) \approx \theta$, so the dynamic equation becomes

$$\ddot{\theta} \approx -\omega^2\theta$$

so $$\theta \approx \theta_0 \cos(\omega t + \varphi)$$

We find that, for small angles, the pendulum has an almost harmonic motion.

For bigger angles, the motion can also be found, but this requires harder calculations.

Also, you can try to find the dynamic equation using other methods (energy, Newton's second law, ...), this is a good exercise.


Both conditions are equivalent, the amplitude in the arc the pendulum describes is $L\theta_{max}$ where $L$ is the length of the pendulum and $\theta_{max}$ is the maximum angular displacement. Clearly, $\theta_{max}\ll 1$ holds if and only if $L\theta_{max}\ll L$.

The reason for the condition $\theta_{max}\ll 1$ for the motion to be harmonic is that, if it holds, the equation of motion of the pendulum $\ddot{\theta}=-\omega^2\sin\theta$ simplifies to $\ddot{\theta}=-\omega^2\theta$, the equation for harmonic motion.


In the derivation to show that a simple pendulum undergoes simple harmonic motion one comes to a point where an approximation has to be made to solve the equation of motion resulting from the use of Newton's second law with relative ease.
The approximation concerns the angle of swing of the pendulum bob from the vertical, $\theta$ and is that $\sin \theta \approx \theta$ where $\theta$ is measured in radians.
This comes from taking the first term of an expansion for $\sin \theta = \theta -\dfrac{\theta^3}{3!} + \dfrac{\theta^5}{5!} - ......$ and is good with an error less than about $1\%$ if $\theta< 0.2$ radian or $\theta< 10^\circ$, $(\sin0.2^c = 0.1987)$.

In the derivation the angle $\theta$ is the displacement of the bob from the equilibrium position $x$ along an arc of a circle of radius the length of the pendulum $l$.
So a long as the amplitude of the pendulum $x$ is less than about a fifth of the length of the pendulum the motion will be simple harmonic.


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