Periodic motion is any motion that repeats at regular time intervals. The time for one complete cycle is the period (T); the number of cycles per second is the frequency (f = 1/T). NCERT Class 11 Physics Chapter 13, page 260 defines periodic motion and distinguishes it from oscillatory motion — every oscillation is periodic, but not every periodic motion is oscillatory (uniform circular motion is periodic but not oscillatory about a mean position).
Simple harmonic motion (SHM) is the simplest oscillatory motion. The displacement follows x(t) = A cos(ωt + φ), where ω = 2π/T. Two properties that trap aspirants repeatedly:
Trap 1 — Period is independent of amplitude. For an ideal spring (T = 2π√(m/k)) or a simple pendulum at small angles (T = 2π√(L/g)), doubling the amplitude does NOT change the period. The restoring force scales linearly with displacement, so larger swings produce proportionally larger restoring forces, keeping the timing unchanged. NEET exploits this by giving amplitude changes and asking for the new period — the answer is "period stays the same."
Trap 2 — Pendulum period is independent of bob mass. Because gravitational mass equals inertial mass, m cancels in the pendulum derivation. Questions that change the bob from iron to lead are testing whether you fall for a mass-dependence that does not exist.
Trap 3 — Closed-pipe harmonics are odd only. A pipe closed at one end supports only odd harmonics (f, 3f, 5f, …) because the closed end forces a displacement node. An open-open pipe supports all harmonics. Including even harmonics in a closed pipe is a common negative-marking error.
Watch out: these three traps appear in straightforward "what changes?" questions. The calculation is trivial — the conceptual clarity is what earns the mark.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following motions is periodic but NOT simple harmonic?
Answer: D. D is correct. Uniform circular motion repeats every revolution, so it is periodic, but it is not a to-and-fro motion about an equilibrium position and there is no restoring force proportional to displacement. Its PROJECTION on a diameter is simple harmonic, which is how SHM is often introduced.
Why A is wrong: A is wrong because small oscillations of a simple pendulum are simple harmonic to a good approximation, because for small angles the restoring force is proportional to the displacement.
Why B is wrong: B is wrong because a mass on an ideal spring is the model case of SHM: Hooke's law gives a restoring force F = −kx.
Why C is wrong: C is wrong because x = A sin ωt is the defining equation of simple harmonic motion.
A mass-spring system oscillates with amplitude A and period T. If the amplitude is increased to 2A (same spring, same mass), the new period is:
Answer: A. For ideal SHM, T = 2π√(m/k) — the period is independent of amplitude (NCERT Class 11 Physics Chapter 13, page 263). Changing amplitude does not alter the period.
Why B is wrong: B is wrong because halving the period would require increasing k by a factor of 4. Amplitude does not appear in the period formula T = 2π√(m/k).
Why C is wrong: C is wrong because this assumes T is proportional to A. The period formula for SHM contains no amplitude term.
Why D is wrong: D is wrong because this assumes T is proportional to A². No such dependence exists in ideal SHM.
Which of the following is NOT an example of periodic motion?
Answer: C. A single throw of a ball goes up and comes down once — it does not repeat at regular intervals, so it is not periodic. The other three repeat their motion pattern at fixed time intervals (NCERT Class 11 Physics Chapter 13, page 260).
Why A is wrong: A is wrong because Earth's revolution around the Sun repeats every year — this is periodic motion by definition.
Why B is wrong: B is wrong because a tuning fork executes oscillatory (and hence periodic) motion at a fixed frequency.
Why D is wrong: D is wrong because a pendulum at small amplitude executes SHM, which is periodic with period T = 2π√(L/g).
A pipe closed at one end has a fundamental frequency of 200 Hz. The frequency of its second overtone is:
Answer: B. A closed pipe supports only odd harmonics: f, 3f, 5f, … The fundamental is f = 200 Hz. The first overtone is 3f = 600 Hz. The second overtone is 5f = 5 × 200 = 1000 Hz (NCERT Class 11 Physics Chapter 13, page 271).
Why A is wrong: A is wrong because 400 Hz = 2f, which is the second harmonic. A closed pipe does not support even harmonics — only odd (trap: treating closed pipe like open pipe).
Why C is wrong: C is wrong because 800 Hz = 4f is the fourth harmonic, which does not exist in a closed pipe. Only odd multiples of the fundamental are allowed.
Why D is wrong: D is wrong because 600 Hz = 3f is the first overtone (third harmonic), not the second overtone. The second overtone in a closed pipe is the fifth harmonic (5f).
In SHM, a particle is at the mean position. At this instant, its:
Answer: C. At mean position x = 0: velocity v = Aω (maximum magnitude) and acceleration a = −ω²x = 0. Velocity is maximum where displacement (and hence restoring force and acceleration) is zero (NCERT Class 11 Physics Chapter 13, page 267).
Why A is wrong: A is wrong because this describes the extreme position (x = ±A), where v = 0 and |a| = ω²A is maximum. At the mean position the situation is reversed.
Why B is wrong: B is wrong because velocity and acceleration reach their maxima at different positions. They are never simultaneously maximum — when v is max (mean position), a is zero, and vice versa.
Why D is wrong: D is wrong because if both were zero simultaneously, the particle would be at rest with no restoring force — it would never move. This does not describe any position in SHM.
A spring of spring constant k = 200 N/m is attached to a block of mass 0.50 kg on a frictionless surface. The block is displaced 0.10 m from equilibrium and released. The period of oscillation is closest to:
Answer: D. T = 2π√(m/k) = 2π√(0.50/200) = 2π√(0.0025) = 2π × 0.050 = 0.314 s ≈ 0.31 s (NCERT Class 11 Physics Chapter 13, page 263). Note: the displacement 0.10 m sets the amplitude but does not affect the period.
Why A is wrong: A is wrong because 0.10 s would require k/m = (2π/0.10)² ≈ 3948 s⁻², i.e., k ≈ 1974 N/m for m = 0.50 kg. The given k = 200 N/m gives a longer period.
Why B is wrong: B is wrong because 1.0 s would require m/k = (1/(2π))² ≈ 0.0253, meaning m ≈ 5.06 kg for k = 200 N/m. The actual mass is 0.50 kg, an order of magnitude smaller.
Why C is wrong: C is wrong because 0.63 s ≈ 2 × 0.314 s — this is double the correct period. A common error is using T = 2π√(m/k) but accidentally doubling, or confusing period with half-period.
A simple pendulum of length 1.0 m has period T. If the length is increased to 4.0 m (same location), the new period is:
Answer: B. T = 2π√(L/g). When L increases by a factor of 4, T increases by √4 = 2. New period = 2T (NCERT Class 11 Physics Chapter 13, page 271).
Why A is wrong: A is wrong because halving the period would require L to decrease to L/4, not increase to 4L. Period increases with the square root of length.
Why C is wrong: C is wrong because this would mean period is independent of length. Period depends on √L — quadrupling L doubles T.
Why D is wrong: D is wrong because this assumes T scales linearly with L (T ∝ L). The correct relation is T ∝ √L, so a 4× increase in length gives a 2× increase in period, not 4×.
An open-open pipe of length L has a fundamental frequency f₀. A closed-end pipe of the same length L has a fundamental frequency f_c. The ratio f₀/f_c is:
Answer: A. Open-open pipe: f₀ = v/(2L). Closed pipe: f_c = v/(4L). Ratio f₀/f_c = [v/(2L)] / [v/(4L)] = 4L/(2L) = 2 (NCERT Class 11 Physics Chapter 13, page 271).
Why B is wrong: B is wrong because this assumes both pipes have the same fundamental. The closed pipe's effective wavelength is 4L (quarter-wave resonator) versus 2L for the open pipe, giving different frequencies.
Why C is wrong: C is wrong because this inverts the ratio. The open pipe has a HIGHER fundamental frequency than the closed pipe of the same length, not lower.
Why D is wrong: D is wrong because a ratio of 4 would require f₀ = v/L and f_c = v/(4L), which would mean the open pipe formula is nv/L with n=1. The correct open-pipe fundamental is v/(2L), not v/L.
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Given
A particle's displacement varies with time as y(t) = 2 sin(πt) + 2 sin(2πt), with y in centimetres and t in seconds.
Required
(a) Whether the motion is periodic. (b) Whether it is simple harmonic. (c) Its period.
Concept
These are two separate questions and the answers can differ. A motion is PERIODIC if it repeats exactly after some fixed interval. It is SIMPLE HARMONIC only if, in addition, the acceleration is proportional to the displacement and directed back towards the mean position — a = −ω²y for one single ω (NCERT Class 11 Physics Chapter 13, page 260).
A sum of two sines is the standard case where the first holds and the second does not. Each term on its own is simple harmonic; added together with different frequencies they give a shape that still repeats, but no longer sinusoidally.
Formula
• Period of sin(ωt): T = 2π/ω• A sum repeats after the shortest interval that is a whole number of periods of EVERY term.• SHM test: a = d²y/dt² must equal −ω²y for a single constant ω.
Substitution
First term: ω₁ = π rad/s, so T₁ = 2π/π = 2 s
Second term: ω₂ = 2π rad/s, so T₂ = 2π/2π = 1 s
Acceleration: a = d²y/dt² = −2π² sin(πt) − 8π² sin(2πt)
Calculation
In 2 s the first term completes exactly 1 cycle and the second exactly 2, so both return together and y repeats: T = 2 s. Checking that nothing shorter works, at t + 1 the first term has advanced half a cycle and reverses sign while the second returns, so y(t + 1) ≠ y(t). The period is therefore 2 s, and f = 1/2 = 0.5 Hz.
For the SHM test, −ω²y would have to be −ω²[2 sin(πt) + 2 sin(2πt)], which multiplies BOTH terms by the same factor. The actual acceleration multiplies them by −π² and −4π². No single ω does that, so the motion is not simple harmonic.
Note on exact values: the coefficients 2, the angular frequencies π and 2π, and the factor 2π in T = 2π/ω are exact, so nothing here limits significant figures.
Final answer
(a) Periodic — yes. (b) Simple harmonic — no. (c) T = 2 s, so f = 0.5 Hz.
Common trap
The reflex is to read "sin" and answer "SHM". Every SHM is periodic; not every periodic motion is SHM. The other half of the trap is taking the period to be the SHORTER of the two, 1 s, because the fastest term repeats then — but the slow term has not, so the sum has not either. The period of a sum is set by the SLOWEST component, not the fastest.
Similar NEET-style question
Which of these is periodic but not simple harmonic: (i) y = sin ωt, (ii) y = sin ωt + cos ωt, (iii) y = sin ωt + sin 3ωt? Approach: (i) is SHM by definition. (ii) LOOKS like a sum but combines to √2 sin(ωt + π/4) — one frequency, so still SHM. (iii) has two different frequencies that cannot be combined into one, so it is periodic with T = 2π/ω and not SHM. Always check whether the terms share a frequency before ruling SHM out.
Periodic motion repeats at regular intervals. Oscillatory motion is periodic motion about an equilibrium position. Time period T is the duration of one cycle; frequency ν = 1/T (Hz).
-- NCERT Class 11 Physics, Ch. 13, p. 260More in Oscillations and Waves: 6 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
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 →