Q.When will the motion of a simple pendulum be simple harmonic?

Concept: Simple Pendulum Period — Simple harmonic motion (SHM) requires a restoring force proportional to displacement and directed toward equilibrium.

For a simple pendulum, the restoring force is $F = -mg \sin\theta$, where $\theta$ is the angular displacement. This is proportional to $\sin\theta$, not $\theta$ itself.

Only when $\theta$ is small (typically $< 10^\circ$ or $< 0.17$ rad) can we use the approximation $\sin\theta \approx \theta$ (in radians). Then $F \approx -mg\theta$, which is proportional to displacement $\theta$, giving SHM. …

Why the pendulum is not always simple harmonic

A simple pendulum consists of a point mass $m$ attached to a massless, inextensible string of length $L$, swinging under gravity. The restoring force that pulls the bob back toward the equilibrium position comes from the tangential component of gravity.

When the bob is displaced by an angle $\theta$ from the vertical, the gravitational force $mg$ splits into two components:

• Radial: $mg\cos\theta$ (tension balances this)
• Tangential: $mg\sin\theta$ (this is the restoring force)

The tangential force is $F = -mg\sin\theta$, where the minus sign indicates it always points opposite to the displacement. For the motion to be simple harmonic, the restoring force must be directly proportional to the displacement — that is, $F \propto -\theta$ (or $F \propto -x$ for linear displacement).

Here lies the catch: $\sin\theta$ is not proportional to $\theta$ for large angles.

The small-angle approximation

For small angles measured in radians, the Taylor expansion of $\sin\theta$ gives:

When $\theta \ll 1$ radian, the higher-order terms become negligible, and we can write:

This approximation is excellent for $\theta < 0.17$ rad (about $10^\circ$), where the error is less than 0.5%.

Since the arc length $s = L\theta$, the linear displacement $x \approx s = L\theta$ (for small angles, the arc is nearly straight), so $\theta = x/L$. Substituting:

This is exactly Hooke's law: $F = -kx$ with effective spring constant $k = mg/L$. The motion is therefore simple harmonic.

Deriving the period

Using Newton's second law for rotational motion, the torque about the pivot is:

For small $\theta$, $\tau \approx -mgL\theta$. Since $\tau = I\alpha$ and the moment of inertia of a point mass at distance $L$ is $I = mL^2$:

This is the SHM equation $\frac{d^2\theta}{dt^2} = -\omega^2\theta$, where $\omega = \sqrt{g/L}$.

Step 1: For a pendulum bob displaced through angle $\theta$, the tangential (restoring) component of gravity is $F=-mg\sin\theta$.

Step 2: For SHM we need $F\propto -\theta$ (a linear restoring force), but $\sin\theta \approx \theta$ (radians) only when $\theta$ is small ($\theta\lesssim10^\circ$, from the Taylor series $\sin\theta=\theta-\theta^3/6+\cdots$).

Step 3: Under that approximation, $F\approx -mg\theta = -mg(x/L) = -(mg/L)x$ — Hooke's-law form — so the pendulum's motion is SHM only for small $\theta$. …