A common trap in SHM questions: students assume the period changes when amplitude doubles. It does not. For ideal simple harmonic motion, the period is fixed by the system parameters alone — amplitude plays no role.
NCERT Class 11 Physics Chapter 13 (Part 2, page 261) defines SHM as oscillatory motion where the restoring force is directly proportional to displacement from equilibrium and directed opposite to it: F = −kx. This linear restoring force is the defining condition. Any system satisfying it executes SHM.
The displacement equation that follows from this definition is:
x(t) = A cos(ωt + φ)
where A is the amplitude (maximum displacement), ω = 2πf = 2π/T is the angular frequency, and φ is the initial phase. From this single equation, velocity and acceleration follow by differentiation:
The acceleration expression a = −ω²x is the signature of SHM — acceleration is proportional to displacement and always directed toward equilibrium.
Phase relationships are a high-frequency NEET ask. Velocity leads displacement by π/2. Acceleration leads displacement by π (equivalently, acceleration and displacement are always in antiphase). When displacement is maximum (at the extremes), velocity is zero and acceleration is maximum. When displacement is zero (at equilibrium), velocity is maximum and acceleration is zero.
The amplitude-independence trap: T = 2π/ω depends only on the physical parameters that determine ω (mass and spring constant for a spring; length and g for a pendulum at small angles). Doubling the amplitude stretches the path but also increases the maximum speed proportionally — the period stays the same. NEET distractors exploit this by offering options that scale T with A.
Watch out: the phase constant φ sets where the oscillation starts, not how fast it goes. Changing φ shifts the entire x(t) curve left or right on the time axis without altering T, f, or ω.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
In simple harmonic motion, the acceleration of the particle is
Answer: A. By definition of SHM, a = −ω²x. Acceleration is proportional to displacement from equilibrium and always directed toward it (NCERT Class 11 Physics Chapter 13, Part 2, page 261).
Why B is wrong: B is wrong because acceleration in SHM varies with position (a = −ω²x); it is not constant in either magnitude or direction.
Why C is wrong: C is wrong because SHM acceleration depends on displacement, not velocity. The relation a = −ω²x has no velocity term.
Why D is wrong: D is wrong because at equilibrium x = 0, so a = −ω²(0) = 0. Acceleration is maximum at the extremes where |x| = A. (trap: confusing velocity maximum at equilibrium with acceleration maximum)
The displacement of a particle in SHM is given by x(t) = A cos(ωt + φ). The phase difference between displacement and velocity is
Answer: B. v(t) = −Aω sin(ωt + φ) = Aω cos(ωt + φ + π/2). Velocity leads displacement by π/2 (NCERT Class 11 Physics Chapter 13, Part 2, page 261).
Why A is wrong: A is wrong because velocity is the derivative of displacement; the sine-to-cosine shift introduces a π/2 phase difference, not zero.
Why C is wrong: C is wrong because π/4 has no basis in the differentiation of cos(ωt + φ). The cosine-to-sine conversion gives exactly π/2.
Why D is wrong: D is wrong because π is the phase difference between displacement and acceleration (a = −ω²x), not between displacement and velocity. (trap: confusing the v–x phase with the a–x phase)
The phase difference between the acceleration and displacement of a particle executing SHM is
Answer: C. a = −ω²x. The negative sign means acceleration is in antiphase with displacement — a phase difference of π radians (NCERT Class 11 Physics Chapter 13, Part 2, page 261).
Why A is wrong: A is wrong because the negative sign in a = −ω²x means acceleration opposes displacement. A phase difference of 0 would mean they are in phase, which contradicts the restoring nature of SHM.
Why B is wrong: B is wrong because π/2 is the phase difference between velocity and displacement, not between acceleration and displacement. (trap: confusing v–x phase with a–x phase)
Why D is wrong: D is wrong because a phase difference of 2π is equivalent to 0 (in phase), but acceleration and displacement are in antiphase (π apart).
A particle executes SHM with amplitude 5.0 cm and angular frequency ω = 4π rad/s. The maximum speed of the particle is
Answer: B. Maximum speed v_max = Aω = 5.0 cm × 4π rad/s = 20π cm/s. This occurs at x = 0 (equilibrium) where sin(ωt + φ) = ±1 (NCERT Class 11 Physics Chapter 13, Part 2, page 261).
Why A is wrong: A is wrong because this value (10π) would result from using half the given amplitude (2.5 cm) or half the angular frequency. v_max = Aω, and both A and ω must be used as given.
Why C is wrong: C is wrong because 5π = A × π, which corresponds to using ω = π instead of the given ω = 4π. The full angular frequency must be used.
Why D is wrong: D is wrong because 40π would result from doubling either A or ω. No such doubling is justified from the given data.
A block attached to a spring oscillates in SHM with period T. If the amplitude is doubled while the spring constant and mass remain unchanged, the new period is
Answer: D. The period of a spring-mass system depends only on m and k: T = 2π√(m/k). Amplitude does not appear in the formula. Doubling amplitude does not change the period (NCERT Class 11 Physics Chapter 13, Part 2, page 261). (trap: amplitude-independence of SHM period)
Why A is wrong: A is wrong because halving the period would require changing m or k. Amplitude does not affect T. (trap: assuming larger amplitude means the oscillator 'hurries' through a longer path)
Why B is wrong: B is wrong because this assumes T scales as A². There is no such dependence; period is entirely independent of amplitude for ideal SHM.
Why C is wrong: C is wrong because this assumes T is proportional to A. The SHM period formula T = 2π√(m/k) contains no amplitude term. The increased path length is exactly compensated by increased maximum speed. (trap: period–amplitude dependence claim)
A particle in SHM has displacement x = A cos(ωt). At what displacement from equilibrium does the particle have half its maximum speed?
Answer: A. v² = ω²(A² − x²). Setting v = v_max/2 = Aω/2 gives (Aω/2)² = ω²(A² − x²), so A²/4 = A² − x², hence x² = 3A²/4, giving x = A√3/2 (NCERT Class 11 Physics Chapter 13, Part 2).
Why B is wrong: B is wrong because substituting x = A/2 into v² = ω²(A² − x²) gives v = Aω√3/2 ≈ 0.866 v_max, not half the maximum speed.
Why C is wrong: C is wrong because x = A/√2 gives v = Aω/√2 ≈ 0.707 v_max. This is the displacement where KE equals PE, not where speed is half its maximum.
Why D is wrong: D is wrong because x = A√3/4 yields v² = ω²(A² − 3A²/16) = ω²(13A²/16), giving v ≈ 0.90 v_max, which is not half.
Two particles execute SHM with the same amplitude and frequency. Particle 1 starts from the mean position moving in the +x direction; particle 2 starts from the extreme position (+A). The phase difference between them is
Answer: C. Particle 1 at mean position moving in +x: x₁ = A sin(ωt), which equals A cos(ωt − π/2). Particle 2 at +A: x₂ = A cos(ωt). Phase difference = π/2 (NCERT Class 11 Physics Chapter 13, Part 2, page 261).
Why A is wrong: A is wrong because the two particles start at different positions (x = 0 vs x = A). If their phase difference were zero, they would always have identical displacements, which contradicts the given initial conditions.
Why B is wrong: B is wrong because π/4 does not correspond to any standard SHM starting condition pair. The mean-position-to-extreme-position phase gap is exactly π/2 (one quarter-cycle).
Why D is wrong: D is wrong because a phase difference of π would place both particles at extremes (one at +A, other at −A). Particle 1 starts at x = 0, not at −A.
A particle in SHM is at the equilibrium position. At this instant, which of the following is true?
(i) Velocity is zero
(ii) Acceleration is zero
(iii) Kinetic energy is maximum
(iv) Potential energy is maximum
Answer: D. At equilibrium, x = 0. So a = −ω²(0) = 0 — statement (ii) is true. Velocity is maximum at equilibrium (v_max = Aω), so KE = ½mv² is maximum — statement (iii) is true. Velocity is NOT zero (that happens at extremes), and PE is minimum at equilibrium, not maximum. (NCERT Class 11 Physics Chapter 13, Part 2, pages 261 and 7).
Why A is wrong: A is wrong because at equilibrium, velocity is maximum, not zero. Velocity is zero at the extremes (x = ±A). Similarly, PE is minimum at equilibrium, not maximum. (trap: confusing equilibrium properties with extreme-position properties)
Why B is wrong: B is wrong because while (ii) is correct (acceleration is zero at equilibrium), statement (iv) is false — PE is minimum at equilibrium since PE depends on x² and x = 0.
Why C is wrong: C is wrong because while (iii) is correct (KE is maximum at equilibrium), statement (i) is false — velocity is maximum, not zero, at x = 0.
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Pattern: Phase difference between displacement, velocity, and acceleration in SHM (based on NEET pattern: shm phase difference — frequency count 3, years observed 2020, 2021, 2023, 2024).
Given
• x(t) = 0.02 cos(4πt + π/6) m• A = 0.02 m, ω = 4π rad/s, φ = π/6 rad• t = 1/8 s (for part b)
Required
• (a) Phase difference between v and x• (b) x and a at t = 1/8 s
Concept
SHM displacement x = A cos(ωt + φ). Velocity is the time derivative: v = −Aω sin(ωt + φ) = Aω cos(ωt + φ + π/2). Acceleration: a = −ω²x. (NCERT Class 11 Physics Chapter 13, Part 2, page 261.)
Formula
• v(t) = −Aω sin(ωt + φ)• a(t) = −ω²x(t)• Phase of x: (ωt + φ); Phase of v: (ωt + φ + π/2). Difference = π/2.
Substitution (part b)
• Argument at t = 1/8 s: ωt + φ = 4π(1/8) + π/6 = π/2 + π/6 = 2π/3• x = 0.02 cos(2π/3) = 0.02 × (−1/2) = −0.01 m• a = −ω²x = −(4π)²(−0.01) = +16π² × 0.01
Calculation
• x = −0.01 m = −1.0 cm• a = 0.16π² m/s²
Note: π is a mathematical constant (exact); 1/8 and 1/2 are exact fractions from the problem definition. These do not limit significant figures. The amplitude 0.02 m has 1 significant figure (or 2 if the trailing zero is significant), so we express the answer to 2 significant figures.
• a ≈ 0.16 × 9.8696 ≈ 1.6 m/s²
Final answer
• (a) Velocity leads displacement by π/2 radians.• (b) At t = 1/8 s: displacement x = −1.0 × 10⁻² m; acceleration a ≈ 1.6 m/s² (directed toward equilibrium, since a is positive while x is negative).
Note: The factor π² is an exact mathematical constant and does not constrain the significant-figure count.
Common trap
Confusing the velocity–displacement phase difference (π/2) with the acceleration–displacement phase difference (π). NEET distractors frequently offer π as the answer when the question asks about velocity. Read the question carefully: velocity → π/2; acceleration → π.
Similar NEET-style question
A particle's SHM is given by x = 0.05 sin(2πt) m. What is the phase difference between acceleration and velocity? At what time does the particle first reach maximum speed?
---
Oscillation in which the restoring force is directly proportional to displacement from equilibrium and directed back toward equilibrium: F = -k x. Equation: a = -ω² x. Solution: x(t) = A cos(ω t + φ).
-- NCERT Class 11 Physics, Ch. 13, p. 260Displacement 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 |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
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.
Question gives changes in amplitude and asks for new period.
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.
Root cause: concept gap
Ideal SHM: T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — no amplitude dependence. Doubling amplitude does not change period.
More in Oscillations and Waves: 4 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
uses pi 2 instead of pi
Confuses v-x phase with a-x phase
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →