Simple Pendulum

8 MCQs1 revision card9-step worked example

Source: NCERT Oscillations and WavesPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The question says the bob mass is doubled — what happens to the period? If your instinct is to reach for mass somewhere in the formula, you have walked into the most common simple-pendulum trap in NEET.

The formula and what it excludes. The period of a simple pendulum for small angular amplitudes is

T = 2π√(L/g)

as derived in NCERT Class 11 Physics, Chapter 14 (Oscillations), page 10. Notice: mass does not appear. During the derivation, the restoring force uses the gravitational component mg sin θ, and Newton's second law introduces ma on the other side. Because gravitational mass equals inertial mass, m cancels identically. The period depends only on the pendulum length L and the local gravitational acceleration g.

Conditions for validity. The formula holds when (1) the angular amplitude is small — typically less than 15°, so that sin θ ≈ θ — (2) the string is massless, and (3) the bob is treated as a point mass. When any of these break down (large angle, heavy chain, extended bob), corrections apply, but NEET questions stay within the ideal regime unless explicitly stated otherwise.

What NEET tests. The dominant pattern is: change the length, find the new period (T ∝ √L). A high-frequency distractor variant changes the mass and offers options implying T changes — it does not. A second variant changes g (pendulum on the Moon, in a lift, at altitude) and asks for the new period (T ∝ 1/√g). Both reduce to reading the formula correctly and recognising what is absent from it.

Watch-out. If a problem changes both L and g simultaneously (e.g., pendulum taken to a planet with different g and string length adjusted), handle each variable separately inside the square root before simplifying. Do not assume one change "cancels" the other without computing the ratio.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

A simple pendulum has a time period of 2 s. If the mass of the bob is doubled while keeping the length and location unchanged, the new time period is:

MCQ 2Direct ApplicationPractice

A simple pendulum of length 1.00 m oscillates at a place where g = 9.80 m/s². Its time period is closest to:

MCQ 3Direct ApplicationPractice

The length of a simple pendulum is increased to 4 times its original value. The new time period becomes:

MCQ 4Easy RecallPractice

A seconds pendulum (T = 2.00 s) has a length of approximately:

MCQ 5Direct ApplicationPractice

A simple pendulum is taken from the Earth's surface (g = 9.8 m/s²) to the Moon's surface (g = 1.63 m/s²). If the period on Earth is T, the period on the Moon is approximately:

MCQ 6Easy RecallPractice

Which of the following does NOT affect the time period of a simple pendulum oscillating with small amplitude?

MCQ 7CalculationPractice

A simple pendulum of length L has period T. If the length is increased by 21%, the percentage change in the time period is closest to:

MCQ 8Concept TrapPractice

A simple pendulum oscillates in a lift. When the lift accelerates upward with acceleration a, the effective g becomes (g + a). Compared to the period at rest, the period in the accelerating lift:

Quick recall before you leave

Worked Example

Pattern: Pendulum length-period problem (observed in NEET 2022, 2024)

1
Given
- L₁ = 0.500 m - L₂ = 2.00 m - g = 9.80 m/s² (same location)
2
Required
Ratio T₂/T₁
3
Concept
The simple pendulum period formula T = 2π√(L/g) shows that T ∝ √L when g is constant. To find the ratio, we do not need the absolute value of T — only the length ratio matters.
4
Formula
T₂/T₁ = √(L₂/L₁)
5
Substitution
T₂/T₁ = √(2.00/0.500) = √(4.00)
6
Calculation
T₂/T₁ = √(4.00) = 2.00 Note on exact values: The ratio 4.00 is exact by construction (2.00/0.500 = 4 exactly), and the square root of 4 is the exact integer 2. No significant-figure limitation applies to this ratio.
7
Final answer
T₂/T₁ = 2.00 The new period is exactly twice the original. The 2π and g factors cancel in the ratio, which is why only the length ratio matters.
8
Common trap
A common error is using T₂/T₁ = L₂/L₁ = 4 (linear scaling instead of square-root scaling). The square root is essential: quadrupling the length doubles the period, not quadruples it. A second trap: assuming the mass of the bob matters. If the problem also changes the bob mass, the answer is unchanged — mass does not enter the formula.
9
Similar NEET-style question
A seconds pendulum (T = 2.00 s) is taken to a planet where g is 4 times that on Earth. Find the new time period. *Approach:* T ∝ 1/√g. If g becomes 4g, T' = T/√4 = 2.00/2 = 1.00 s. ---

Before solving, remember these

Formula

Time period of simple pendulum

T = 2π √(L/g), valid for small angular amplitudes (sin θ ≈ θ). Independent of mass and amplitude. Useful for measuring g.

-- NCERT, p. 10

Formulas

10 formulas — click to collapse

Beat frequency

f_{beat} = ∣ f_{1} - f_{2} ∣

When two waves of nearly equal frequencies superpose, amplitude oscillates at the difference frequency.

Symbol	Quantity	SI Unit
f_beat	beat frequency	Hz
f1, f2	superposed frequencies	Hz

Valid when

Linear superposition
f1, f2 close in value

Period of simple pendulum (small angle)

T = 2 π \frac{L}{g}

Period of simple pendulum of length L. Holds for small amplitudes (sin theta ~ theta).

Symbol	Quantity	SI Unit
T	period	s
L	pendulum length	m
g	gravity	m/s^2

Valid when

Small angular amplitude (typically <15°)
Massless string
Point bob

SHM displacement

x (t) = A cos (ω t + ϕ); ω = 2 π f

Displacement in simple harmonic motion. Velocity = -A*omega*sin(omega*t+phi); a = -omega^2 * x.

Symbol	Quantity	SI Unit
A	amplitude	m
omega	angular frequency	rad/s
phi	phase	rad
T	period	s
f	frequency	Hz

Valid when

Restoring force linear (F = -kx)
No damping

Total energy in SHM

E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}

Total mechanical energy is constant. Oscillates between KE (max at x=0) and PE (max at x=±A).

Symbol	Quantity	SI Unit
E	total energy	J
k	spring constant	N/m
A	amplitude	m
m	mass	kg
omega	angular frequency	rad/s

Valid when

Conservative SHM (no damping)
Elastic regime

Period of mass-spring oscillator

T = 2 π \frac{m}{k}

Period of horizontal spring with mass m, spring constant k. Independent of amplitude.

Symbol	Quantity	SI Unit
T	period	s
m	mass	kg
k	spring constant	N/m

Valid when

Hooke's law spring
No damping
Small enough amplitude to stay in elastic regime

Standing wave in closed-end pipe

f_{n} = \frac{( 2 n - 1 ) v}{4 L}

Pipe closed at one end has only odd harmonics: f, 3f, 5f, ...

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	sound speed	m/s
L	pipe length	m

Valid when

Closed at one end (open at other)
End correction neglected

Standing wave in open-open pipe

f_{n} = \frac{n v}{2 L}

Pipe open at both ends has all harmonics. Same formula as string.

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	sound speed	m/s
L	pipe length	m

Valid when

Open at both ends
End correction neglected

Standing wave frequencies on fixed-fixed string

f_{n} = \frac{n v}{2 L}

Allowed frequencies on string fixed at both ends. n=1 fundamental; harmonics 2f, 3f, ...

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	wave speed on string	m/s
L	string length	m
n	harmonic number	-

Valid when

String fixed at both ends
Wave speed v as defined above

Speed of sound in gas (Newton-Laplace)

v = \frac{γ P}{ρ}

Speed of sound in gas. Adiabatic index gamma, pressure P, density rho. Increases with sqrt(T).

Symbol	Quantity	SI Unit
v	speed of sound	m/s
gamma	adiabatic index	-
P	pressure	Pa
rho	density	kg/m^3

Valid when

Ideal gas
Adiabatic compression/expansion of sound waves

Wave speed on string

v = \frac{T}{μ}

Speed of transverse wave on string under tension T, linear mass density mu.

Symbol	Quantity	SI Unit
v	wave speed	m/s
T	tension	N
mu	linear mass density	kg/m

Valid when

Stretched uniform string
Small amplitude

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

6 items — click to collapse

Trap

Pendulum: mass-dependence claim

Category: Overthinking

Student writes T as depending on bob mass. Simple pendulum T = 2π√(L/g); independent of m.

When it triggers

Question changes pendulum bob mass and asks for new period.

How to avoid

Mass cancels in derivation (gravitational mass = inertial mass). Mass changes the bob's KE and PE proportionally; period unaffected.

Trap

SHM period: amplitude dependence claim

Category: Overthinking

Student claims SHM period depends on amplitude. For ideal SHM (Hooke's law spring or simple pendulum at small angle), period is INDEPENDENT of amplitude.

When it triggers

Question gives changes in amplitude and asks for new period.

How to avoid

T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — neither depends on A. Only at large pendulum angles does T pick up a small amplitude correction.

Trap

Pipe harmonics: open vs closed end

Category: Similar Terms

Student includes even harmonics in a closed-end pipe. Closed pipe has only ODD harmonics (f, 3f, 5f, ...).

When it triggers

Question describes pipe closed at one end (e.g. resonance tube).

How to avoid

Open both ends: all harmonics, f_n = nv/(2L). Closed one end: odd only, f_n = (2n-1)v/(4L). Fundamental of closed pipe is HALF that of open pipe of same L.

Mistake

Includes bob mass in simple pendulum period formula.

Root cause: concept gap

Correction

Simple pendulum T = 2π√(L/g) — independent of mass. Equivalence of inertial and gravitational mass cancels m.

Mistake

Claims SHM period depends on amplitude.

Root cause: concept gap

Correction

Ideal SHM: T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — no amplitude dependence. Doubling amplitude does not change period.

Mistake

Includes even harmonics in a closed-end pipe.

Root cause: concept gap

Correction

Closed-end pipe has only ODD harmonics (f, 3f, 5f, ...). Open-both-ends pipe has all (f, 2f, 3f, ...). Reason: closed end has displacement node and pressure antinode.