Simple pendulum — derive time period T = 2π√(l/g) and factors affecting it

Question

Derive the expression for the time period of a simple pendulum: $T = 2\pi\sqrt{l/g}$ . What factors affect the time period? Does the mass of the bob matter?

(NCERT Class 11, Chapter 14)

Solution — Step by Step

A simple pendulum consists of a mass $m$ (bob) suspended by a massless, inextensible string of length $l$ . When displaced by a small angle $\theta$ from the vertical, the restoring force along the arc is:

F = -mg\sin\theta

For small angles ( $\theta < 15°$ ), $\sin\theta \approx \theta$ (in radians):

F \approx -mg\theta = -mg\frac{x}{l}

where $x = l\theta$ is the arc displacement.

Since $F = -\frac{mg}{l} \cdot x$ , this has the form $F = -kx$ with effective spring constant $k = mg/l$ .

This is SHM. The angular frequency:

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{mg/l}{m}} = \sqrt{\frac{g}{l}}

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

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

Notice: $T$ depends on $l$ and $g$ only. It does not depend on the mass $m$ or the amplitude (for small oscillations).

Length $l$ : $T \propto \sqrt{l}$ — longer pendulum oscillates slower.
Gravity $g$ : $T \propto 1/\sqrt{g}$ — on the Moon ( $g$ is smaller), the pendulum swings slower.
Mass: No effect — the $m$ cancels out in the derivation.
Amplitude: No effect (for small angles). For large angles, the formula needs correction.

Why This Works

The key physics is the small-angle approximation $\sin\theta \approx \theta$ . This linearises the equation of motion, turning it into SHM. Without this approximation, the pendulum equation is nonlinear and the period depends on amplitude — but for angles below about 15°, the SHM approximation is excellent (error < 1%).

The mass-independence is physically intuitive: gravity provides both the restoring force (proportional to $m$ ) and the inertia (also proportional to $m$ ). These effects cancel, just like in free fall.

Alternative Method — Using Torque Approach

Taking torques about the pivot:

\tau = -mgl\sin\theta \approx -mgl\theta

Since $\tau = I\alpha$ and $I = ml^2$ for a point mass at distance $l$ :

ml^2 \ddot{\theta} = -mgl\theta \implies \ddot{\theta} = -\frac{g}{l}\theta

This is SHM with $\omega^2 = g/l$ , giving the same result.

A favourite JEE trick: “A pendulum clock is taken to the top of a mountain. Does it run fast or slow?” At altitude, $g$ decreases, so $T$ increases — the clock runs slow. Similarly, taking it to a deep mine (where $g$ also decreases below Earth’s surface) also makes it slow.

Common Mistake

Students frequently write $T = 2\pi\sqrt{g/l}$ (inverting the fraction). Quick sanity check: a longer string should give a longer period. With $T = 2\pi\sqrt{l/g}$ , increasing $l$ increases $T$ — correct. With the inverted formula, increasing $l$ would decrease $T$ , which makes no physical sense.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Torque Approach

Common Mistake

Try These Next