Simple Pendulum — Deep Dive

What the Simple Pendulum Really Teaches Us

The simple pendulum looks like a Class 9 toy — a bob on a string swinging back and forth. But this innocent setup hides one of the most important results in physics: simple harmonic motion emerges from any restoring force linear in displacement, and the period depends only on $L$ and $g$ , not on mass.

JEE and NEET ask pendulum problems every single year. Boards expect the derivation. Practical and project work uses the pendulum to measure $g$ on the lab floor. We will work through the derivation, the small-angle approximation, the variants (compound pendulum, pendulum in lift, pendulum in fluid), and the speed-solving tricks toppers swear by.

Key Terms & Definitions

Simple pendulum: A point mass $m$ attached to an inextensible, massless string of length $L$ , free to swing in a vertical plane.

Time period ( $T$ ): Time for one complete oscillation.

Angular frequency ( $\omega$ ): $\omega = 2\pi/T = \sqrt{g/L}$ .

Small-angle approximation: $\sin\theta \approx \theta$ when $\theta$ is in radians and small (less than $\sim 10°$ ).

Effective gravity ( $g_\text{eff}$ ): The acceleration due to all body forces other than the string tension. In a lift accelerating up at $a$ , $g_\text{eff} = g + a$ .

Deriving the Time Period

Consider the bob displaced by angle $\theta$ . The forces on it are gravity ( $mg$ downward) and tension ( $T$ along the string).

The tangential component of gravity is $mg\sin\theta$ , providing the restoring force. The normal component balances tension and centripetal force.

Tangential acceleration: $a_t = -g\sin\theta$ . The arc length is $s = L\theta$ , so $a_t = L\ddot\theta$ .

$L\ddot\theta = -g\sin\theta$

For small $\theta$ , $\sin\theta \approx \theta$ , giving:

$\ddot\theta = -(g/L)\theta$

This is SHM with $\omega^2 = g/L$ , hence:

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

The period is independent of mass and amplitude (in the small-angle regime). This is why grandfather clocks work — heavier or lighter pendulums tick at the same rate.

When the Small-Angle Assumption Breaks

For $\theta_\text{max}$ above $\sim 15°$ , the period gets longer than $2\pi\sqrt{L/g}$ . The first correction is:

$T \approx 2\pi\sqrt{L/g} \cdot \left(1 + \frac{\theta_\text{max}^2}{16}\right)$

For $\theta_\text{max} = 30°$ (about $0.52$ rad), the correction is $\sim 1.7\%$ — usually ignored in JEE/NEET problems.

Methods/Concepts

Method 1: Recognising SHM

If a system has $\ddot{x} = -\omega^2 x$ for some constant $\omega$ , it is SHM with period $T = 2\pi/\omega$ . The pendulum is just one example — spring-mass, LC circuit, U-tube oscillation, all follow the same pattern.

Method 2: Effective Gravity

Whenever the pendulum is in a non-inertial frame (accelerating lift, charged bob in electric field, pendulum on incline), replace $g$ with $g_\text{eff}$ :

$T = 2\pi\sqrt{L/g_\text{eff}}$

Method 3: Energy Conservation for Speed at Bottom

For a pendulum released from amplitude angle $\theta_0$ :

$v_\text{max} = \sqrt{2gL(1 - \cos\theta_0)}$

Use this when the question asks for speed at the lowest point or tension there.

Solved Examples

Easy — Find Period from Length (CBSE)

A pendulum has length $1 \, \text{m}$ . Take $g = \pi^2 \, \text{m/s}^2$ . Find its period.

$T = 2\pi\sqrt{1/\pi^2} = 2\pi/\pi = 2 \, \text{s}$ .

Medium — Pendulum in an Accelerating Lift (JEE Main)

A simple pendulum of length $L$ is in a lift moving up with acceleration $a$ . Find the period.

In the lift’s frame, effective gravity is $g + a$ (downward).

$T = 2\pi\sqrt{L/(g + a)}$ — shorter than ground period.

If lift moves down with acceleration $a$ : $T = 2\pi\sqrt{L/(g - a)}$ . If $a = g$ (free fall), $T \to \infty$ — pendulum doesn’t swing.

Hard — Pendulum in a Charged Bob (JEE Advanced)

A pendulum has bob of mass $m$ , charge $q$ , in a uniform electric field $E$ pointing horizontally. Find the new period and the new equilibrium angle.

Electric force on bob: $qE$ horizontal. Net force at rest: $\sqrt{(mg)^2 + (qE)^2}$ , directed at angle $\tan^{-1}(qE/mg)$ from vertical.

The “effective gravity” magnitude is $g_\text{eff} = \sqrt{g^2 + (qE/m)^2}$ .

$T = 2\pi\sqrt{L/g_\text{eff}}$ , and equilibrium tilts at angle $\tan^{-1}(qE/mg)$ .

Pattern: Any constant force on the bob (gravity, electric, magnetic on a charge in a field) just modifies $g_\text{eff}$ . Add the force vectors, take the magnitude, plug into the period formula.

Exam-Specific Tips

JEE Main / Advanced

JEE loves the “pendulum in modified frame” template. Common scenarios: lift accelerating up/down, pendulum on accelerating cart (use pseudo-force horizontally), pendulum swinging in a fluid (buoyancy reduces effective weight).

For Advanced, expect compound pendulum problems — period $T = 2\pi\sqrt{I/(mgd)}$ where $I$ is moment of inertia about the pivot and $d$ is distance from pivot to COM.

NEET

NEET sticks to basic period formula and small variations. Master $T = 2\pi\sqrt{L/g}$ and the effective gravity trick — covers most NEET pendulum questions.

CBSE Boards

Boards always ask for the derivation. Memorise the steps: write the tangential equation of motion, apply small-angle approximation, identify SHM, write the period.

Common Mistakes to Avoid

Mistake 1: Using degrees instead of radians in $\sin\theta \approx \theta$ . The approximation only works in radians.

Mistake 2: Including bob mass in the period formula. Mass cancels in the SHM equation — period is independent of mass.

Mistake 3: For pendulum in fluid, forgetting buoyancy. Effective weight $= (m - m_\text{displaced})g$ , so $g_\text{eff} = g(1 - \rho_\text{fluid}/\rho_\text{bob})$ .

Mistake 4: Assuming period is shorter on Mars. Mars has $g \approx 3.7 \, \text{m/s}^2$ , so $T_\text{Mars}/T_\text{Earth} = \sqrt{g_\text{Earth}/g_\text{Mars}} \approx 1.6$ — period is LONGER on Mars.

Mistake 5: For a compound (physical) pendulum, using $T = 2\pi\sqrt{L/g}$ with $L$ as the rod length. Wrong — use $T = 2\pi\sqrt{I/(mgd)}$ .

Practice Questions

Q1. A pendulum of length $L$ has period $T$ on Earth. What is its period on the Moon ( $g_\text{Moon} = g/6$ )?

$T_\text{Moon} = T\sqrt{g/g_\text{Moon}} = T\sqrt{6} \approx 2.45 T$ .

Q2. A pendulum has period $2 \, \text{s}$ on the ground. Find its period in a lift moving down at $g/2$ .

$g_\text{eff} = g - g/2 = g/2$ . $T_\text{new} = 2\sqrt{2} \approx 2.83 \, \text{s}$ .

Q3. A pendulum of length $L$ swings in a region with horizontal field $E$ and bob charge $q = mg/E$ . Find the period.

$g_\text{eff} = \sqrt{g^2 + g^2} = g\sqrt{2}$ . $T = 2\pi\sqrt{L/(g\sqrt{2})}$ .

Q4. A simple pendulum of length $1 \, \text{m}$ with bob mass $0.5 \, \text{kg}$ is released from $\theta_0 = 60°$ . Find the speed at the lowest point.

$v = \sqrt{2gL(1 - \cos 60°)} = \sqrt{2 \times 10 \times 1 \times 0.5} = \sqrt{10} \approx 3.16 \, \text{m/s}$ .

Q5. Find the tension at the lowest point in Q4.

At lowest point: $T - mg = mv^2/L$ . $T = mg + mv^2/L = 0.5(10) + 0.5(10)/1 = 10 \, \text{N}$ .

FAQs

Why is the period independent of mass? Both the restoring force ( $mg\sin\theta$ ) and inertia ( $m$ ) scale with mass — so they cancel. This is the same reason all objects fall at the same rate in vacuum.

What is the difference between simple and compound pendulum? Simple: point mass on massless string. Compound: any rigid body pivoted about a point. Compound period uses moment of inertia.

Why does a longer pendulum swing slower? Period scales as $\sqrt{L}$ . Longer string = larger arc per radian = more distance to cover per swing.

Does air resistance affect the period much? Negligibly for typical bobs (it just damps the amplitude over time, leaving period nearly unchanged).

What’s the maximum amplitude for SHM to be a good approximation? About $\theta = 10°$ gives error below $0.2\%$ . At $\theta = 30°$ error is $\sim 1.7\%$ .

What happens to the period if we double the length? $T$ increases by $\sqrt{2} \approx 1.41×$ .

Can a pendulum work on a satellite? No — in free fall, $g_\text{eff} = 0$ , so the pendulum just floats. There is no restoring force.

Why are pendulums used to measure g? Because $g = 4\pi^2 L/T^2$ . Both $L$ (with a metre scale) and $T$ (timing many oscillations) can be measured precisely, giving accurate $g$ .