JEE Mains Physics · Oscillations

Simple Pendulum and Effective g

A simple pendulum swings with T = 2π√(L/g), whatever the mass of its bob; a height, a planet, a lift, an incline or a liquid changes only the g in that formula.

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Why this matters

Twenty-six PYQs, twenty-three of them multiple choice, and two from 2026. Twelve use T = 2π√(L/g) directly: a changed length, a T²–L graph, the seconds pendulum, a clock's daily error, a large swing. Fourteen change g: a height above the earth, another planet, a lift, a vehicle on an incline, a bob in a liquid.

Concept 1 of 2: Period of a simple pendulum and how it depends on length

For small swings the pull back towards the lowest point is

mg\theta

, and the bob's mass cancels. So the period depends only on the length and on g, and grows as the square root of the length: four times the length, twice the period.

Definition

$T = 2\pi\sqrt{L/g}$ , independent of the bob's mass and of the (small) amplitude.
$T^{2} = \dfrac{4\pi^{2}}{g}L$ : the $T^{2}$ –L graph is a straight line through the origin, and $g = 4\pi^{2}L/T^{2}$ .
n oscillations in time t: $T = t/n$ . Twice as many in the same time needs a quarter of the length.
Seconds pendulum: $T = 2$ s, so 1 s from one extreme to the other. Its length is $g/\pi^{2}$ , about 1 m.
Clock error: $\dfrac{\Delta T}{T} = \dfrac{1}{2}\dfrac{\Delta L}{L}$ . Time lost per day $= \dfrac{1}{2}\dfrac{\Delta L}{L} \times 86400$ s; a longer pendulum runs slow.
Angular acceleration $= -\dfrac{g}{L}\theta$ .
Large swings (amplitude $\theta_0$ ): at the extreme the acceleration is $g\sin\theta_0$ , all tangential; at the lowest point it is $v^{2}/L = 2g(1 - \cos\theta_0)$ , all towards the pivot. The tension is largest at the lowest point: $mg(3 - 2\cos\theta_0)$ .

Simple pendulum

T = 2\pi\sqrt{\frac{L}{g}} \qquad \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta L}{L}

Worked example

A pendulum 100 cm long makes 30 oscillations in one minute. (a) Find its period. (b) What length makes 60 oscillations in one minute? (c) Find g at that place.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q6Moderate

Example 1 · Oscillations · Simple Pendulum and Effective g

A simple pendulum of string length 30 cm performs 20 oscillations in 10s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is

\_\_\_\_

cm. [Assume that the mass of the pendulum remains same.]

A seconds pendulum has a period of 2 s

It takes 1 s to go from one extreme to the other, and 2 s for a full oscillation. Taking T = 1 s gives a length a quarter of the right one.

The bob's mass does not enter

Changing the mass of the bob, or keeping it the same, does not change T = 2π√(L/g). Only L and g matter.

A longer pendulum makes a clock slow

A longer pendulum has a longer period, so the clock ticks less often and loses time. Heat lengthens the pendulum, so a clock loses time in summer.

Large swings: the two accelerations point different ways

At the extreme the speed is zero, so the acceleration is only tangential, g sin θ₀. At the lowest point the tangential part is zero and only v²/L remains.

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Concept 2 of 2: Effective g for a pendulum at a height, on a planet, in a lift or in a liquid

Whatever changes, the pendulum still obeys

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

with g replaced by the acceleration the bob would have if it were let go, measured from the support. Find that effective g first, then use the formula.

Definition

Height h above the earth: $g_h = g\dfrac{R^{2}}{(R + h)^{2}}$ , so $T \propto R + h$ . At $h = R$ , g falls to $g/4$ and T doubles.
Mountain: g is smaller, T is longer, so a pendulum clock runs slow.
Planet: $g \propto M/R^{2}$ . Four times the mass and twice the radius give the same g.
Lift accelerating upward at a (or moving down and slowing): $g + a$ . Accelerating downward at a: $g - a$ . In free fall: 0, and the pendulum does not swing.
Vehicle sliding freely down a smooth incline of angle α: $g\cos\alpha$ . Vehicle accelerating horizontally at a: $\sqrt{g^{2} + a^{2}}$ .
Bob in a liquid (ignoring drag): buoyancy reduces the pull, $g_{eff} = g(1 - \rho_{liquid}/\rho_{bob})$ .

Pendulum with an effective g

T = 2\pi\sqrt{\frac{L}{g_{eff}}} \qquad g_h = g\left(\frac{R}{R + h}\right)^{2}

Worked example

A pendulum has period 2 s in a lift at rest. Find its period (a) when the lift accelerates upward at

g/3

, (b) when it accelerates downward at

g/3

, and (c) on the ground of a planet with twice the earth's mass and twice its radius.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 July 2022 · Q96Moderate

Example 2 · Oscillations · Simple Pendulum and Effective g

Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height

h

above the earth surface. Clock-1 and Clock-2 operate at time periods

4\text{ }s

and

6\text{ }s

respectively. Then the value of

h

is (consider radius of earth

R_{E}= 6400\text{ }km

and

g

on earth

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

)

A height R puts the pendulum 2R from the centre

"At a height equal to the earth's radius" means a distance R + R = 2R from the centre, so g is g/4 and T doubles. Using distance R from the centre gives no change at all.

The lift's acceleration decides, not its velocity

A lift moving down but slowing has an upward acceleration, so g_eff = g + a and the period is shorter. Look at the direction of a, not of v.

A clock on a mountain runs slow

g is smaller at a height, so T is longer and the clock ticks less often. It runs slow, not fast.

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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 (2)

Period of a simple pendulum and how it depends on length
Simple pendulum
$T = 2\pi\sqrt{\frac{L}{g}} \qquad \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta L}{L}$
Effective g for a pendulum at a height, on a planet, in a lift or in a liquid
Pendulum with an effective g
$T = 2\pi\sqrt{\frac{L}{g_{eff}}} \qquad g_h = g\left(\frac{R}{R + h}\right)^{2}$

Watch out for (7)

A seconds pendulum has a period of 2 s
→ Period of a simple pendulum and how it depends on length
The bob's mass does not enter
→ Period of a simple pendulum and how it depends on length
A longer pendulum makes a clock slow
→ Period of a simple pendulum and how it depends on length
Large swings: the two accelerations point different ways
→ Period of a simple pendulum and how it depends on length
A height R puts the pendulum 2R from the centre
→ Effective g for a pendulum at a height, on a planet, in a lift or in a liquid
The lift's acceleration decides, not its velocity
→ Effective g for a pendulum at a height, on a planet, in a lift or in a liquid
A clock on a mountain runs slow
→ Effective g for a pendulum at a height, on a planet, in a lift or in a liquid

Test yourself on Oscillations

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