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 3) 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
The displacement of a particle in SHM is given by x(t) = A cos(ωt + φ). The phase difference between displacement and velocity is
The phase difference between the acceleration and displacement of a particle executing SHM is
A particle executes SHM with amplitude 5.0 cm and angular frequency ω = 4π rad/s. The maximum speed of the particle is
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
A particle in SHM has displacement x = A cos(ωt). At what displacement from equilibrium does the particle have half its maximum speed?
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
A particle in SHM is at the equilibrium position. At this instant, which of the following is true?
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 3.)
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, p. 3When 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 | - |
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 |
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 |
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 writes T as depending on bob mass. Simple pendulum T = 2π√(L/g); independent of m.
Question changes pendulum bob mass and asks for new period.
Mass cancels in derivation (gravitational mass = inertial mass). Mass changes the bob's KE and PE proportionally; period unaffected.
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.
Category: Similar Terms
Student includes even harmonics in a closed-end pipe. Closed pipe has only ODD harmonics (f, 3f, 5f, ...).
Question describes pipe closed at one end (e.g. resonance tube).
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.
Root cause: concept gap
Simple pendulum T = 2π√(L/g) — independent of mass. Equivalence of inertial and gravitational mass cancels m.
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.
Root cause: concept gap
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.
treats beat as sum
Adds frequencies instead of subtracting
expects mass dependence
Assumes T depends on bob mass
treats closed pipe like open
Includes even harmonics in closed pipe
uses pi 2 instead of pi
Confuses v-x phase with a-x phase
forgets 2pi factor
Drops 2*pi
uses amplitude in period
Believes T depends on amplitude
uses linear tension scaling
Treats v ∝ T not sqrt(T)
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