Relation Between Length and Period

In this segment, we will derive the period of a simple pendulum of length $\ell$. First, we say the bob has mass $m$ and we draw a free-body diagram. A simple one would look like this.

After some more inspection on the object, we can label some components and we get something that looks like this.

The weight component, $mg \sin \theta$, is accelerating the mass towards equilibrium along the arc of the circle. This component is the restoring force of the pendulum. Now, this about how you might use Newton's second law to get some equations we can solve.

Set the restoring force equal to mass times the tangential acceleration to get $$mg \sin \theta = m \cdot a_{\text{tang}}.$$What do you think this behaves like?

A simple harmonic oscillator! We will leave it as an exercise for the reader to use the concepts of simple harmonic motion to prove that the equivalent spring constant of the restoring force for the pendulum is $\frac {mg}{\ell}$. Once you get this, how would you finish?

Recall that, for simple harmonic motion, we have $$T = 2\pi \cdot \sqrt{\frac{m}{k}},$$ so, here, we get $T = 2\pi \cdot \sqrt {\frac{\ell}{g}}$, which is the period of a pendulum with length $\ell$.

Before you leave this page, ask yourself something $-$ don't you see something remarkable? All $m$s cancelled! The period does not depend on the mass. Did we do something wrong? Actually not. Read the next section on this to discover more.

As in the previous explanation, the period fo a simple pendulum, $T=2\pi \sqrt{\frac{l}{g}}$ can be derived by making the small angle approximation ($\sin\theta \approx \theta$) which the reader can verify himself / herself.

But what about large angle pendulums? It turns out that the differential equation has no solution that can be expressed in the form of elementary functions, and what we get instead is an elliptic integral of the first kind. This can be solved numerically.

So, be careful when you use the formula for the period of a simple pendulum!! However, the deviation from the actual period is actually very small for angles below $\frac{\pi}{6}$.