17.6 Simple Pendulum

What is a Simple Pendulum?

An ideal simple pendulum consists of a heavy point mass (called the bob) suspended from a rigid, frictionless support by a weightless, inextensible string of length $l$. When displaced from its equilibrium position and released, it oscillates back and forth.

Restoring Force and SHM Condition

Consider a bob of mass $m$ displaced through a small angle $\theta$ from the vertical equilibrium position.

The forces acting on the bob are:

• Tension $T$ along the string
• Weight $mg$ vertically downward

The tangential component of weight acts as the restoring force, directed back toward the equilibrium:

$F=-mg\sin \theta$

The negative sign indicates the force opposes the displacement.

Small Angle Approximation

For small angles ($\theta <10°$), $\sin \theta \approx \theta$ (in radians), so:

$F\approx -mg\theta$

Since the arc length $s=l\theta$, we have $\theta =s/l$, giving:

$F=-\frac{mg}{l}\cdot s$

This shows that the restoring force is directly proportional to the displacement $s$ and directed toward the mean position — the defining condition for Simple Harmonic Motion (SHM).

Acceleration in SHM

Applying Newton's second law ($F=ma$):

$ma=-\frac{mg}{l}\cdot s$

$a=-\frac{g}{l}\cdot s$

Comparing with the standard SHM equation $a=-{\omega }^{2}x$:

${\omega }^2=\frac{g}{l}$

$\omega =\sqrt{\frac{g}{l}}$

This confirms the pendulum executes SHM (for small angles) with $a\propto -x$.

Time Period of a Simple Pendulum

Using $\omega =2\pi /T$:

$\frac{2\pi }{T}=\sqrt{\frac{g}{l}}$

$\boxed{T=2\pi \sqrt{\frac{l}{g}}}$

where:

• $T$ = time period (s)
• $l$ = length of the pendulum (m)
• $g$ = acceleration due to gravity (m s${}^{-2}$)

Key Observations

Worked Example

Problem: A simple pendulum has a length of 1.0 m. Calculate its time period on Earth where $g=9.8$ m s${}^{-2}$.

Solution:

$T=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{1.0}{9.8}}=2\pi \times 0.319\approx 2.0\text{ s}$

Seconds Pendulum

A seconds pendulum is a pendulum with a time period of exactly 2 seconds (it takes 1 second to swing from one side to the other — one beat).

Its length on Earth ($g=9.8$ m s${}^{-2}$):

$l=\frac{g{T}^2}{4{\pi }^2}=\frac{9.8\times 4}{4{\pi }^2}\approx 0.99\text{ m}\approx 1\text{ m}$

Using $a=-{\omega }^{2}x$ for the Pendulum

For a simple pendulum, $\omega =\sqrt{g/l}$, so:

$a=-{\omega }^{2}x=-\frac{g}{l}\cdot x$

Example: A pendulum of length 0.25 m is displaced 0.02 m from equilibrium. Find the acceleration.

${\omega }^2=\frac{g}{l}=\frac{9.8}{0.25}=39.2{\text{ rad}}^2{\text{s}}^{-2}$

$a=-39.2\times 0.02=-0.784{\text{ m s}}^{-2}$

The negative sign confirms the acceleration is directed toward the equilibrium position.