MHT-CET Physics · Oscillations

The Simple Pendulum: Period, Effective g, Speed and Tension

A simple pendulum of length L swings with T = 2π√(L/g), independent of its mass and (for small swings) its amplitude; in a lift, on an accelerating support or in orbit, g is replaced by the effective gravity, and energy conservation gives the bob's speed and the string's tension along the swing.

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Why this matters

24 PYQs, 5 HARD. Ten are the period and length — ratios of lengths and frequencies, a length change of 20%, a pendulum of length L₁ − L₂, a bob that leaks water, a mass hung from two strings; eight are effective g — lifts accelerating up or down, a support moving as y = kt², a pendulum in orbit, and a sonometer wire at the poles and the equator; six are speed and tension — the speed at 60°, the maximum tension, the angle where the maximum tension is four times the minimum. Three cards.

Concept 1 of 3: Period and Length

T = 2π√(L/g): the period goes as the square root of the length and has nothing to do with the bob's mass or material. Four times the length doubles the period; frequencies 4 : 3 mean lengths 9 : 16; a 20% longer period needs 1.44 times the length. Lengths subtract as T²: a pendulum of length L₁ − L₂ has period √(T₁² − T₂²). The length is measured to the bob's centre of mass, so a leaking bob first lengthens its pendulum and then, once nearly empty, shortens it back.

Definition

$T = 2\pi\sqrt{\dfrac{L}{g}}$ , $T \propto \sqrt{L}$ ; independent of the bob's mass and density.
$\dfrac{L_1}{L_2} = \left(\dfrac{f_2}{f_1}\right)^2$ ; T + 20% ⇒ L × 1.44; T × 2 ⇒ L × 4.
$T_{L_1 - L_2} = \sqrt{T_1^2 - T_2^2}$ .
Leaking bob: T first increases, then decreases back to its first value.
Hung from two strings of length L from points 2d apart, swinging out of their plane: effective length $\sqrt{L^2 - d^2}$ .

Simple pendulum

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

Worked example

Pendulums of periods 5 s and 3 s. Period of a pendulum whose length is the difference of theirs?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q18Moderate

Example 1 · Oscillations · Simple Pendulum — Period, Lift, Weightlessness

A simple pendulum of length

l_1

has time period

T_1

. Another simple pendulum of length

l_2\,(l_1>l_2)

has time period

T_2

. Then the time period of the pendulum of length

l_1 - l_2

will be

Changing the bob to change the period

A heavier or denser bob of the same size leaves T unchanged. Only the length (to the centre of mass) and g matter.

Drill 9 more on period and length

Concept 2 of 3: Effective g: Lifts, Moving Supports and Orbit

In an accelerating frame the pendulum feels g_eff = g + a when its support accelerates upward and g − a when downward, so T′ = T√(g/g_eff). A support moving as y = kt² has acceleration 2k. In free fall or in orbit g_eff = 0 and the pendulum does not swing at all — its period is infinite. The same g changes a sonometer: a wire stretched by a hanging mass has less tension where g is smaller, at the equator, so it must be shortened to keep its frequency.

Definition

Up at a: $T' = T\sqrt{\dfrac{g}{g + a}}$ ; down: $T' = T\sqrt{\dfrac{g}{g - a}}$ . Up at g/3 ⇒ $\tfrac{\sqrt{3}}{2}T$ ; down at g/4 ⇒ $\tfrac{2}{\sqrt{3}}T$ ; to halve T accelerate up at 3g.
Support $y = kt^2$ : a = 2k; with k = 1, g = 10: $\dfrac{T_1^2}{T_2^2} = \dfrac{12}{10} = \dfrac{6}{5}$ .
In orbit or free fall: $g_{\text{eff}} = 0$ , period infinite.
Sonometer (tension Mg): at the equator g is smaller, so the resonating length must be decreased.

Accelerating support

T' = 2\pi\sqrt{\frac{L}{g \pm a}}

Worked example

A pendulum has period 2 s at rest. The lift accelerates down at g/2. New period?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift I · Q38Moderate

Example 2 · Oscillations · Simple Pendulum — Period, Lift, Weightlessness

A simple pendulum has time period '

T_{1}

'. The point of suspension is now moved upward according to equation

y =kt^{2}

where

k = 1\text{ }m/s^{2}

. If new time period is '

T_{2}

' then

T_{1}^{2}/T_{2}^{2}

will be (

g = 10\text{ }m/s^{2}

)

Adding the acceleration in the wrong direction

Accelerating UP presses the bob down harder — larger g_eff, shorter period. Accelerating down (or falling) lightens it — longer period, infinite in free fall.

Drill 7 more on effective g: lifts, moving supports and orbit

Concept 3 of 3: Speed and Tension Along the Swing

Energy conservation gives the speed: dropping from angle θ, the bob falls L(1 − cos θ), so it passes the bottom at √(2gL(1 − cos θ)). The string's tension is least at the extremes, mg cos θ, where the bob is momentarily still, and greatest at the bottom, mg + mv²/L = mg(3 − 2 cos θ). Setting the maximum to four times the minimum gives cos θ = ½. For a small amplitude A the bottom speed is A√(g/L), so the maximum tension is mg(1 + A²/L²).

Definition

Speed at the bottom from θ: $v = \sqrt{2gL(1 - \cos\theta)}$ ; from the bottom at speed u up to angle φ: $v^2 = u^2 - 2gL(1 - \cos\varphi)$ (4 m/s, 1 m, 60° ⇒ √6 m/s).
$T_{\min} = mg\cos\theta$ , $T_{\max} = mg(3 - 2\cos\theta)$ ; $T_{\max} = 4T_{\min}$ ⇒ $\theta = \cos^{-1}(0.5)$ .
Small amplitude A: $T_{\max} = mg\left(1 + \dfrac{A^2}{L^2}\right)$ .
Equal total energies, equal masses, $L_1 = 2L_2$ : $\dfrac{A^2}{L}$ equal, so the SHORTER pendulum has the smaller amplitude.

Tension along the swing

T_{\max} = mg(3 - 2\cos\theta), \qquad T_{\min} = mg\cos\theta

Worked example

A 2 m pendulum is released from 60°. Speed at the bottom and the maximum tension for a 0.5 kg bob (g = 10)?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q23Hard

Example 3 · Oscillations · Simple Pendulum — Period, Lift, Weightlessness

A simple pendulum oscillates with an angular amplitude

\theta

. If the maximum tension in the string is 4 times the minimum tension then the value of

\theta

Taking the minimum tension as mg

At the extreme the bob is still but the string is slanted: only mg cos θ is balanced by the tension. mg is the tension of a pendulum at rest.

Drill 5 more on speed and tension along the swing

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Period and Length
Simple pendulum
$T = 2\pi\sqrt{\frac{L}{g}}$
Effective g: Lifts, Moving Supports and Orbit
Accelerating support
$T' = 2\pi\sqrt{\frac{L}{g \pm a}}$
Speed and Tension Along the Swing
Tension along the swing
$T_{\max} = mg(3 - 2\cos\theta), \qquad T_{\min} = mg\cos\theta$

Watch out for (3)

Changing the bob to change the period
→ Period and Length
Adding the acceleration in the wrong direction
→ Effective g: Lifts, Moving Supports and Orbit
Taking the minimum tension as mg
→ Speed and Tension Along the Swing

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