NDA Physics · Oscillations and Waves

The Simple Pendulum

A simple pendulum is a point mass on a light inextensible string; for small swings it performs simple harmonic motion with a period T = 2π√(L/g) that depends only on the length and the local gravity — never on the mass of the bob.

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Why this matters

The chapter's workhorse — seven PYQs, almost all turning on the one formula T = 2π√(L/g). The bank tests it from every angle: scale the length and the period scales as its square root (×4 length → ×2 period; halve length → period ÷√2), move to weaker gravity and the period lengthens, and — the recurring trap — double the mass and nothing happens at all. A second, subtler thread is amplitude: the period is amplitude-independent only while the swing is small, because only then is the restoring force proportional to displacement. Memorise the formula, spot what each problem changes, and these are reliable marks.

Concept 1 of 3

The pendulum period law T = 2π√(L/g) — mass-independent

Intuition

The time a pendulum takes for one swing depends on just two things: how long the string is and how strong gravity is. A longer string swings more slowly; stronger gravity swings it faster. What it does NOT depend on is the mass of the bob — a heavy bob and a light bob on equal strings keep perfect time together, because gravity pulls harder on the heavy one but also has more inertia to move.

Definition

For small oscillations the period of a simple pendulum is $T = 2\pi\sqrt{\dfrac{L}{g}}$ .

It is **proportional to $\sqrt{L}$ **: quadruple the length and the period doubles; halve the length and the period falls by $\sqrt{2}$ .
It is **inversely proportional to $\sqrt{g}$ **.
It is completely independent of the mass of the bob and of the amplitude (for small swings).

Period of a simple pendulum

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

Tperiod (s)
Llength of the string
gacceleration due to gravity

The restoring force is mg sinθ, directed along the arc back to the mean position. For small θ, sinθ ≈ θ, so the force is proportional to displacement — the simple-harmonic condition behind T = 2π√(L/g).

Worked example

A simple pendulum has period

T

. Its length is increased to nine times its original value while the bob is replaced by one of triple the mass. Find the new period.

Period depends only on length: $T \propto \sqrt{L}$ . The mass change is irrelevant.
$\dfrac{T'}{T} = \sqrt{\dfrac{9L}{L}} = \sqrt{9} = 3$ .
So $T' = 3T$ .

Answer:

T' = 3T

(the mass change has no effect).

Practice this conceptself-check · 4 quick reps

Try it yourself

A pendulum of length

L

has period

T

. The string is shortened to

L/4

and the bob is made twice as heavy. What is the new period?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Length ×4, mass ×3. Ratio of new to old period?
2.
Length halved. New period in terms of old T?
3.
Does doubling the bob's mass change the period?
4.
Approx. period of a 1 m pendulum (g ≈ 10)?

From the bank · past-year question

Example 1Oscillations and WavesMODERATE

The time period of oscillation of a simple pendulum having length

L

and mass of the bob

m

is given as

T

. If the length of the pendulum is increased to

4L

and the mass of the bob is increased to

2m

, then which one of the following is the new time period of oscillation?

[Q91 · Sep · 2018]

Period does NOT depend on the mass of the bob

Every length-and-mass problem in this bank plants a mass change to distract you.

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

has no

m

in it — double, triple or halve the mass and the period is unchanged. Read off only what happens to the LENGTH.

Period scales as √L, not as L

Quadrupling the length doubles the period (

\sqrt{4} = 2

), it does NOT quadruple it. A distractor that gives '4T' for a ×4 length is using the wrong power — the square root is the whole point.

Drill 3 more on the pendulum period law t = 2π√(l/g) — mass-independent

Concept 2 of 3

How gravity changes the period

Intuition

Because gravity is the restoring agent, a weaker pull makes the pendulum sluggish and a stronger pull makes it brisk. Take the same pendulum to a mountaintop, a deep mine, or the Moon, and although its length is unchanged its period shifts — lengthening wherever g is smaller. Since T depends on 1/√g, halving g multiplies the period by √2.

Definition

With length fixed, the period varies as $T \propto \dfrac{1}{\sqrt{g}}$ .

Weaker gravity (high altitude, the Moon) → longer period → a pendulum clock runs slow.
Stronger gravity → shorter period → the clock runs fast.
Quantitatively, if $g$ becomes $g/2$ the period becomes $\sqrt{2}\,T$ ; if $g$ doubles the period becomes $T/\sqrt{2}$ .

Period and gravity (length fixed)

\dfrac{T_2}{T_1} = \sqrt{\dfrac{g_1}{g_2}}

T_1, T_2old and new periods
g_1, g_2old and new gravitational accelerations

Worked example

A pendulum has period

T

where the gravitational acceleration is

g

. It is carried (length unchanged) to a place where the gravitational acceleration is

4g

. Find its new period.

Length fixed, so $T \propto 1/\sqrt{g}$ : $\dfrac{T'}{T} = \sqrt{\dfrac{g}{4g}} = \sqrt{\tfrac{1}{4}} = \tfrac{1}{2}$ .
Stronger gravity → shorter period, as expected.
So $T' = T/2$ .

Answer:

T' = T/2

Practice this conceptself-check · 4 quick reps

Try it yourself

On Earth a pendulum's period is

T

. On the Moon, where

g

is about one-sixth of Earth's, what is its period (same length)?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
g → g/2 (length fixed). New period?
2.
g doubled (length fixed). New period?
3.
Does a pendulum clock run fast or slow on a high mountain?
4.
Where is the period longer: Earth or the Moon?

From the bank · past-year question

Example 2Oscillations and WavesMODERATE

A pendulum clock is lifted to a height where the gravitational acceleration has a certain value

g

. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is

\frac{g}{2}

. The time period of the second pendulum would be: (in terms of period T of the first pendulum)

[Q83 · Sep · 2019]

Smaller g gives a LONGER period (and a slow clock)

Because

T \propto 1/\sqrt{g}

, weaker gravity LENGTHENS the period — the clock loses time. Don't let the mass in the problem mislead you: the bob's mass never enters, only the change in g (and length) does.

Use the √(g₁/g₂) ratio, not g₁/g₂

Halving g multiplies the period by √2, not by 2. Gravity sits under a square root, so always take the square root of the gravity ratio when comparing periods.

Concept 3 of 3

Amplitude-independence holds only for small swings

Intuition

The neat formula T = 2π√(L/g) is built on one approximation: for a small angle, sinθ ≈ θ, so the restoring force mg sinθ is very nearly proportional to the displacement — exactly the SHM condition. Push the swing out to a large angle and that approximation breaks: sinθ falls below θ, the restoring force is a little weaker than SHM predicts, and the real period comes out slightly LONGER than T₀.

Definition

The restoring force on the bob is $mg\sin\theta$ .

For small angles, $\sin\theta \approx \theta$ , so the force is proportional to displacement → true SHM → period $T_0 = 2\pi\sqrt{L/g}$ , independent of amplitude.
For large angles, $\sin\theta < \theta$ , so the restoring force is smaller than the SHM value → the motion is slower → the actual period $T > T_0$ .

So amplitude-independence is a small-angle property, not a universal one.

Restoring force and the small-angle condition

F = -mg\sin\theta \;\approx\; -mg\,\theta \quad (\theta \text{ small})

Frestoring force along the arc
\thetaangular displacement from the vertical
mmass of the bob

Worked example

A pendulum is set swinging first through a tiny angle and then through a large angle of 50°. Compared with the small-angle period

T_0 = 2\pi\sqrt{L/g}

, is the large-angle period longer, shorter, or the same? Why?

The exact restoring force is $mg\sin\theta$ ; the SHM formula assumes $mg\theta$ .
At 50°, $\sin 50° \approx 0.766$ is less than $\theta = 0.873$ rad, so the real restoring force is weaker than the SHM value.
A weaker restoring force means slower motion, so the actual period exceeds $T_0$ .

Answer:Longer than

T_0

— at large angles

\sin\theta < \theta

weakens the restoring force.

Practice this conceptself-check · 4 quick reps

Try it yourself

A simple pendulum is said to have an amplitude-independent period. Under what condition is this statement true, and why?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Is the pendulum period amplitude-independent at large angles?
2.
At a large angle, is the real period larger or smaller than T₀?
3.
Which approximation gives SHM for a pendulum?
4.
What is the exact restoring force on the bob?

From the bank · past-year question

Example 3Oscillations and WavesMODERATE

Which one of the following statements regarding simple pendulum is correct ? Simple pendulum has a time period independent of amplitude :

[Q87 · Apr · 2024]

Amplitude-independence is a SMALL-angle property

The period is independent of amplitude only while the swing is small, because only then is sinθ ≈ θ and the restoring force proportional to displacement. At a large amplitude the period is no longer constant — it grows. Picking 'for any amplitude' is the trap.

Large amplitude → period gets LONGER, not shorter

Because sinθ < θ for large angles, the restoring force is weaker than the SHM value, so the bob moves more slowly and the period exceeds T₀. An option saying the large-angle period is slightly SMALLER than T₀ has the inequality backwards.

Drill 1 more on amplitude-independence holds only for small swings

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)

The pendulum period law T = 2π√(L/g) — mass-independent
Period of a simple pendulum
$T = 2\pi\sqrt{\dfrac{L}{g}}$
How gravity changes the period
Period and gravity (length fixed)
$\dfrac{T_2}{T_1} = \sqrt{\dfrac{g_1}{g_2}}$
Amplitude-independence holds only for small swings
Restoring force and the small-angle condition
$F = -mg\sin\theta \;\approx\; -mg\,\theta \quad (\theta \text{ small})$

Watch out for (6)

Period does NOT depend on the mass of the bob
→ The pendulum period law T = 2π√(L/g) — mass-independent
Period scales as √L, not as L
→ The pendulum period law T = 2π√(L/g) — mass-independent
Smaller g gives a LONGER period (and a slow clock)
→ How gravity changes the period
Use the √(g₁/g₂) ratio, not g₁/g₂
→ How gravity changes the period
Amplitude-independence is a SMALL-angle property
→ Amplitude-independence holds only for small swings
Large amplitude → period gets LONGER, not shorter
→ Amplitude-independence holds only for small swings

Mastery check — 4 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Oscillations and WavesMODERATE

The length of a simple pendulum is increased four times to its previous value while the mass is doubled. What is the ratio of the new and previous time period of the pendulum?

[Q134 · Apr · 2025]

Example 2Oscillations and WavesMODERATE

A pendulum of length

L

oscillates with an angular amplitude of

\theta = 60°

and time period

T

. Let

T_0 = 2\pi\sqrt{\frac{L}{g}}

be the time period for small angle oscillations. If air resistance is negligibly small and the string remains straight, then which one of the following is correct?

[Q51 · Apr · 2026]

Example 3Oscillations and WavesMODERATE

A simple pendulum having bob of mass

m

and length of string

l

has time period of

T

. If the mass of the bob is doubled and the length of the string is halved, then the time period of this pendulum will be

[Q135 · Sep · 2022]

Example 4Oscillations and WavesEASY

The time period of a 1 m long pendulum approximates to

[Q70 · Apr · 2022]

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