NEET Physics — Simple Harmonic Motion
Factors Affecting Time Period of Simple Pendulum – Complete Notes, Revision, Important Questions & Downloads
This topic covers exactly one subtopic — Effects of Amplitude, Mass, Length, g, and Temperature — examining how each physical variable changes or leaves unchanged the period T = 2π√(l/g) of an ideal simple pendulum. NEET tests this by presenting scenarios such as a pendulum taken to the moon (g decreases → T increases), a heated metallic wire bob (thermal expansion increases l → T increases), or a girl standing up in a swing (centre of mass rises → effective l decreases → T decreases). A landmark NEET trap: replacing a solid bob with a hollow sphere of the same radius but lower mass leaves T unchanged, because T is independent of mass — students who write T ∝ √m select the wrong option every time.
8 SubtopicsTheory + ApplicationNEET High Yield
Expected QuestionsQ
1–2
Tested almost every year; questions are either direct (identify which factor changes T) or scenario-based (compute new T when g or l changes).
Time Required⏱
2–3 hours
One hour for formula derivation and five-factor summary; one hour for scenario drill including pendulum in lifts and on moon/hill.
Difficulty⚡
Easy–Medium
The formulas are simple; the challenge is recognising effective length and effective g in non-standard physical setups.
NRI USA Curriculum GapUS
Low–Medium
AP Physics 1 covers simple pendulum period and the length-g dependence, but does not systematically treat thermal expansion correction or the comparison of solid vs hollow bobs — both of which appear in NEET questions.
8Subtopics
8+Practice Questions
4Free Downloads
2–3 hrsPrep Time
NEET Weightage — Factors Affecting Time Period of Simple Pendulum
Simple Harmonic Motion (Chapter 16)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 1 | 4 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 5–7 | 20–28 |
T is independent of amplitude only for small angles (θ < 15°); for large amplitudes T ≈ T₀[1 + θ₀²/16], which NEET occasionally tests as a 'small difference' numerical.
T ∝ √l is the highest-frequency relationship tested: doubling l increases T by factor √2 ≈ 1.414; quartering l halves T.
Thermal expansion correction ΔT/T = (1/2)αΔθ is a direct NEET question type — a clock based on a steel wire pendulum loses time in summer because T increases as l increases.
T ∝ √l is the highest-frequency relationship tested: doubling l increases T by factor √2 ≈ 1.414; quartering l halves T.
Thermal expansion correction ΔT/T = (1/2)αΔθ is a direct NEET question type — a clock based on a steel wire pendulum loses time in summer because T increases as l increases.
1.0
Avg Questions / Year
24
Total Marks (6 yrs)
Mixed
Pattern
Medium
Difficulty
Exam Strategy for Factors Affecting Time Period
1
Memorise the independence facts first T is independent of (1) amplitude (small angles) and (2) mass. Commit these as binary rules. In NEET this appears as: 'If bob is replaced by another of double mass, T will...' — answer is unchanged. The trap is selecting the option that involves mass.
2
Apply T ∝ √l and T ∝ 1/√g as ratio problems Write T₂/T₁ = √(l₂/l₁) or √(g₁/g₂). If the question says 'pendulum taken to moon where g′ = g/6', T on moon = T√6. Practice substituting before exam day — this numerical format appears in at least 3 of the last 6 NEET papers.
3
Recognise thermal expansion scenarios When temperature rises by Δθ, ΔT/T = (1/2)αΔθ, so a clock based on a metallic pendulum runs slow (T increases). Trap: students confuse 'clock loses time' with 'T decreases'. Clock loses time when T increases because each oscillation now takes longer.
4
Identify effective length in composite scenarios Effective length l is measured from the suspension point to the centre of mass of the bob. When a girl stands up in a swing, her CoM rises, reducing l → T decreases. When water drains from a hollow bob, CoM first moves down (T increases), then returns to centre when empty (T equals original). Never assume l equals string length when the bob has non-negligible size.
Download Study Notes — Factors Affecting Time Period
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Complete coverage of all five factors: amplitude independence derivation, mass independence proof, T ∝ √l scaling with effective-length definition, T ∝ 1/√g with moon/hill examples, and temperature correction formula with clock-error calculation.
Download PDFFormula Sheet
T = 2π√(l/g), ratio form T₂ = T₁√(l₂/l₁), thermal correction ΔT/T = (1/2)αΔθ, large-amplitude correction T ≈ T₀[1 + θ₀²/16], and the independence statements for amplitude and mass.
Download PDFMCQ Practice
30 scenario-based MCQs covering pendulum on moon/hill, thermal expansion of metallic wire, girl swinging and standing, solid vs hollow bob replacement, and water-draining bob.
Download PDFPYQ
Previous year NEET questions on factors affecting pendulum time period, with year tags, full solution steps, and wrong-option explanations.
Download PDFSubtopics
2-Column TableRevision Cards
Concept → Trap → Example1) Effects of Amplitude, Mass, Length, g, and Temperature
Core TheoryT = 2π√(l/g); independent of amplitude (small θ) and independent of mass. T ∝ √l; T ∝ 1/√g. ΔT/T = (1/2)αΔθ for thermal correction.
- T is independent of amplitude for small angles and independent of mass of the bob — replacing a solid bob with a hollow sphere of the same radius leaves T unchanged.
- T ∝ √l: doubling the effective length (distance from pivot to CoM of bob) multiplies T by √2 ≈ 1.41. The effective length is NOT always the string length when the bob has size.
- Thermal expansion trap: when temperature rises, metallic wire bob's effective l increases → T increases → a pendulum clock runs slow (loses time), not fast.
Example (NEET-style)A pendulum of l = 1 m on Earth (g = 10 m/s²) has T = 2π√(0.1) ≈ 1.99 s. On the moon (g = 10/6 m/s²), T_moon = 2π√(0.6) ≈ 4.87 s. Ratio T_moon/T_earth = √6 ≈ 2.45 — so the pendulum clock on the moon loses more than half its rate.
US Curriculum Gap
For students coming from the US curriculum, here are the key conceptual differences you will encounter in NEET Physics.Thermal Expansion of Pendulum Wire
AP Physics 1 (College Board) does not cover the thermal expansion correction ΔT/T = (1/2)αΔθ for metallic pendulum wires. NEET routinely tests clock-gain/loss calculations using this formula.
- AP Physics 1 treats the simple pendulum period as T = 2π√(l/g) with l as a given constant, not a thermally variable quantity.
- NEET questions require knowing that a steel wire pendulum (α ≈ 12×10⁻⁶ /°C) loses about 6 s/day per 1°C temperature rise.
- US students must memorise the expansion formula l = l₀(1 + αΔθ) and its effect on T before tackling this category of NEET problems.
Effective Length vs String Length
AP Physics 1 equates pendulum length with string length and treats the bob as a point mass. NEET systematically tests the concept of effective length (suspension point to CoM) with hollow bobs, water-draining bobs, and swinging girls who stand up.
- When a hollow sphere partially filled with water oscillates, effective l increases as water drains, reaches maximum when water level equals the bob centre, then decreases — producing a non-monotonic T versus time graph.
- US Physics courses do not cover this compound CoM analysis of pendulum length, making it an easy loss-of-marks scenario for NRI students.
- NEET MCQs on this sub-concept appear in roughly 1 out of every 4 SHM question sets.
NEET-style Practice Questions
8 questions1A simple pendulum has a solid metallic bob of mass 50 g suspended from a steel wire of length 1 m at 0°C. The coefficient of linear expansion of steel is 1.2 × 10⁻⁵ /°C. If the temperature rises to 50°C, the fractional change in the time period ΔT/T is approximately:Thermal Correction
3.0 × 10⁻⁴
6.0 × 10⁻⁴
3.0 × 10⁻³
6.0 × 10⁻³
Using ΔT/T = (1/2)αΔθ: ΔT/T = (1/2)(1.2 × 10⁻⁵)(50) = (1/2)(6.0 × 10⁻⁴) = 3.0 × 10⁻⁴. Wait — let me restate: (1/2)(1.2×10⁻⁵)(50) = 0.5 × 6×10⁻⁴ = 3×10⁻⁴. Option A would be correct with these values — however the question is designed to test whether students remember the (1/2) factor: without it, they get 6×10⁻⁴ (option B). The correct answer following the formula is 3×10⁻⁴. NEET 2019-style: the distractor at 6×10⁻⁴ exploits forgetting the factor of 1/2. Mass (50 g) is irrelevant — T is independent of mass. Wrong options: 6.0×10⁻³ and 3.0×10⁻³ arise from misreading Δθ as 500 instead of 50.
2A girl is sitting in a swing of length l and time period T. She stands up, raising her centre of mass by 0.3l. The new time period T′ is approximately:Effective Length
T√(1.3)
T/√(1.3)
T√(0.7)
T (unchanged)
When the girl stands, the effective length decreases from l to l − 0.3l = 0.7l. Using T ∝ √l: T′/T = √(0.7l/l) = √0.7. So T′ = T√0.7. Option D is the classic mass-independence trap — students assume 'nothing about mass changed, so T is unchanged', but what changed is the effective length (position of CoM). Option A applies the increase factor instead of the decrease. Option B is the reciprocal of option A. The correct reasoning: effective length fell from l to 0.7l, so T decreased by factor √0.7 ≈ 0.837.
3A simple pendulum beats seconds (T = 2 s) on Earth. When taken to the surface of the moon where g_moon = g_earth/6, the time period becomes:Effect of g
2/√6 s
2√6 s
12 s
2 × 6 s
T = 2π√(l/g). For the same pendulum (same l), T_moon/T_earth = √(g_earth/g_moon) = √6. So T_moon = 2√6 ≈ 4.9 s. Option A (2/√6) arises from inverting the ratio — a sign-error trap. Option C comes from multiplying by 6 instead of √6. Option D from multiplying by 6 directly, ignoring the square root. The key check: T increases as g decreases, which is physically correct (slower restoring force → longer period).
4A hollow sphere filled with water acts as the bob of a simple pendulum. As water slowly leaks out from a small hole at the bottom, the time period of the pendulum:Effective Length — CoM
Continuously increases
Continuously decreases
First increases, then decreases, and finally equals the initial value
Remains constant throughout
As water leaks, the centre of mass of the sphere + water system descends from the centre, increasing effective length l → T increases. When half the water has leaked, CoM is below the original centre. As water continues to drain, CoM rises back toward the centre of the now lighter sphere. When completely empty, CoM is again at the sphere's geometric centre (same as when full), so T returns to its initial value. This non-monotonic behaviour is a classic NEET scenario: options A and B are wrong because they assume a monotonic change. Option D ignores the CoM shift entirely.
Practice Questions
Click "Reveal Answer" after attempting1A pendulum clock keeps correct time at 25°C. If the temperature falls to 0°C and the coefficient of linear expansion of the wire is 1.8 × 10⁻⁵ /°C, how many seconds does the clock gain per day?
Approx. 8.6 s/day
Approx. 17.3 s/day
Approx. 19.4 s/day
Approx. 4.3 s/day
👁 Reveal Answer
Option C is incorrect; correct answer is Option C — wait, let me recalculate. ΔT/T = (1/2)αΔθ = (1/2)(1.8×10⁻⁵)(25) = 2.25×10⁻⁴. Time gained per day = ΔT/T × 86400 s = 2.25×10⁻⁴ × 86400 ≈ 19.4 s. Since temperature fell, l decreases → T decreases → clock gains time. Answer: Option C. Steps: compute (1/2)(1.8×10⁻⁵)(25) = 2.25×10⁻⁴; multiply by seconds in a day (86400); result ≈ 19.4 s/day gained.
2The length of a seconds pendulum (T = 2 s) is 0.993 m at Earth's surface. At an altitude h = 1000 km above Earth's surface (R = 6400 km), the time period becomes approximately (assume g decreases as 1/r²):
2.32 s
2.16 s
2.00 s
1.85 s
👁 Reveal Answer
At altitude h, g′ = g(R/(R+h))² = g(6400/7400)² = g(0.864)² = 0.747g. T′ = 2π√(l/g′) = T/√0.747 = 2/√0.747 ≈ 2/0.864 ≈ 2.31 s ≈ 2.32 s. Answer: Option A. Key formula: T ∝ 1/√g; as altitude increases g decreases → T increases.
3Two simple pendulums have lengths in the ratio 4:9. Their time periods are in the ratio:
2:3
4:9
9:4
3:2
👁 Reveal Answer
T ∝ √l, so T₁/T₂ = √(l₁/l₂) = √(4/9) = 2/3. Answer: Option A. Trap: ratio of lengths is 4:9 but ratio of periods is 2:3, not 4:9 — students who forget the square root select option B.
4A simple pendulum has a bob of mass 100 g. Keeping everything else constant, the bob is replaced by another of mass 400 g. The time period:
Doubles
Becomes half
Increases by factor √4 = 2
Remains unchanged
👁 Reveal Answer
T = 2π√(l/g) has no mass term. Time period is independent of bob mass. Answer: Option D. Options A, B, and C all involve mass dependence, which is the central misconception this question tests.
5A simple pendulum executes SHM with amplitude θ₀ = 0.2 radians. Compared to a pendulum with amplitude θ₀ → 0, the fractional increase in time period is approximately:
θ₀²/8
θ₀²/16
2θ₀²
θ₀/2
👁 Reveal Answer
Using T ≈ T₀[1 + θ₀²/16], ΔT/T₀ ≈ θ₀²/16 = (0.2)²/16 = 0.04/16 = 2.5×10⁻³. Answer: Option B. The formula for the first-order correction is T ≈ T₀(1 + θ₀²/16), so ΔT/T₀ = θ₀²/16. Options A and C are incorrect multiples; option D is the first power of θ₀, which appears nowhere in the Taylor expansion.
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Frequently Asked Questions
Notes · Downloads · Revision · Important QuestionsWhy is the time period of a simple pendulum independent of the mass of the bob?
When you write the equation of motion for a pendulum bob, the restoring torque is τ = −mgl sin θ ≈ −mglθ, and the rotational equivalent of Newton's second law gives ml²α = −mglθ, which simplifies to α = −(g/l)θ. The mass m cancels from both sides before you extract ω² = g/l and compute T = 2π/ω = 2π√(l/g). Because m never appears in the final expression, doubling or halving the mass has zero effect on the period.
For what amplitude range is the period truly independent of amplitude?
The period is strictly independent of amplitude only in the limit θ → 0. For practical purposes, independence holds for θ₀ < 5° (where the error is below 0.1%). For θ₀ up to 15°, the first-order correction T ≈ T₀[1 + θ₀²/16] gives a fractional error of about 0.35%, acceptable for most experiments. Beyond 20°, the motion is no longer simple harmonic and the correction becomes significant — this is the regime where NEET occasionally asks for the approximate increase using θ₀²/16.
A pendulum clock shows the correct time at 20°C. Will it gain or lose time at 40°C?
At 40°C the metallic suspension wire expands thermally: l = l₀(1 + αΔθ) where Δθ = 20°C. Since l increases, T = 2π√(l/g) increases. Each oscillation now takes slightly longer, so fewer oscillations are completed per hour — the clock runs slow and loses time. The fractional loss per day = (1/2)α × 20 × 86400 s. For steel (α = 12 × 10⁻⁶ /°C), this is about 10.4 s lost per day.
How does effective length differ from string length?
Effective length l is the distance from the fixed pivot point to the centre of mass of the bob, not to the centre of the string. For a point-mass bob on a massless string, effective length = string length. But for a hollow bob, a physical bob, or a situation where the CoM shifts (e.g., water draining from inside the bob), effective length changes even though string length stays fixed. NEET exploits this in multi-step questions about girls standing in swings or hollow bobs with water.
Why does a pendulum at the centre of the earth stop oscillating?
At the centre of the earth, the net gravitational force on the bob is zero (contributions from all sides cancel), so g_eff = 0. Since T = 2π√(l/g_eff), T → ∞ as g_eff → 0, meaning the oscillation period becomes infinite — the bob has no restoring force and remains stationary in any displaced position. This is the same physics as in a freely falling satellite, where the pendulum also fails to oscillate.
How does the time period change when a pendulum is taken to a hill vs taken to a mine?
Both at height above Earth (hill) and depth below Earth (mine), the value of g decreases relative to the surface value. Therefore T = 2π√(l/g) increases in both cases. The functional forms differ: at altitude h, g′ = g(R/(R+h))² (decreases with height); at depth d inside Earth, g′ = g(1 − d/R) (decreases linearly with depth). In both cases T increases and a pendulum clock loses time.
Does a pendulum clock gain or lose time in summer compared to winter?
In summer, ambient temperature is higher, so the metallic pendulum wire expands thermally, increasing effective length l. Since T ∝ √l, T increases → the clock beats more slowly → it loses time in summer. A brass pendulum (α ≈ 19 × 10⁻⁶ /°C) loses more time than a steel pendulum (α ≈ 12 × 10⁻⁶ /°C) for the same temperature change. Invar (an iron-nickel alloy) is used in precision clocks because α ≈ 10⁻⁶ /°C, minimising the seasonal drift.
Why is the large-amplitude correction T ≈ T₀[1 + θ₀²/16] relevant for NEET?
NEET occasionally presents questions where two pendulums of the same length but different amplitudes are compared, and asks whether T₁ = T₂ or which is larger. The correction factor θ₀²/16 shows that the larger-amplitude pendulum has a slightly longer period. For θ₀ = 0.3 rad, the correction is (0.09)/16 ≈ 5.6 × 10⁻³, so T increases by about 0.56% — a testable quantity. This formula also appears in matching-type questions where students must identify the correct expression for T at finite amplitude.
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