Class 11 Physics NCERT Solutions

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Oscillations Class 11 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 13, Oscillations — 18 questions from 13.1 to 13.18, each worked through step by step in the CBSE marking pattern. Simple harmonic motion, spring and pendulum time periods and the energy in SHM.

Class:11Subject:PhysicsChapter:13

4 Key Formulas

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 11 Physics Chapter 13?

Chapter 13 carries 18 exercise questions, numbered 13.1 to 13.18. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents19 sections

Revision & Practice

Chapter Overview

Oscillations covers the sign vocabulary of SHM, the equations x = A cos(ωt + φ), the reference-circle construction, and the spring and pendulum periods. Boards test the standard forms hard — recognising SHM from a = −ω²x, extracting amplitude and phase from initial conditions, and converting between sine and cosine descriptions. Every question below is from the NCERT Class 11 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

SHM means a = −ω²x: test every candidate against that one relation. When matching to x = A cos(ωt + φ), fix A and φ with the initial position and velocity, and remember sin ↔ cos shifts of π/2. The standard periods — T = 2π√(m/k), T = 2π√(l/g) — handle most numeric parts; quote how the effective g changes (moon, moving car) before substituting.

$Ex 13.1 — Which motions are periodic$
$Ex 13.2 — Which motions are (nearly) simple harmonic$
$Ex 13.3 — Reading x-t plots: periodic motion and its period$
$Ex 13.4 — Classifying functions of time (SHM / periodic / non-periodic)$
$Ex 13.5 — Signs of velocity, acceleration and force in SHM$
$Ex 13.6 — Which a(x) relations represent SHM$
$Ex 13.7 — Amplitude and phase from initial conditions$
$Ex 13.8 — Weight of a body oscillating on a spring balance$
$Ex 13.9 — Frequency, maximum acceleration and speed of a spring$
$Ex 13.10 — Displacement functions for three starting positions$
$Ex 13.11 — SHM from the x-projection of two circular motions$
$Ex 13.12 — Reference circles for four SHM equations$
$Ex 13.13 — Spring extension and periods for one-mass and two-mass setups$
$Ex 13.14 — Maximum speed of a locomotive piston$
$Ex 13.15 — Period of a pendulum on the moon$
$Ex 13.16 — Period of a pendulum in a car on a circular track$
$Ex 13.17 — Floating cork oscillating simple harmonically$
$Ex 13.18 — Mercury column in a U-tube executing SHM$

NCERT Exercise 13.1 — Which Motions Are Periodic

1Exercise question

Step-by-step solution

1(a) The swimmer's trips need not take equal times — no definite period, so not periodic.
2(b) The released magnet oscillates about its equilibrium direction with a definite period — periodic.
3(c) The rotating hydrogen molecule returns to the same state after each full rotation — periodic.
4(d) The arrow moves only forward, never returning — not periodic.

Final answer

(b) and (c) represent periodic motion.

NCERT Exercise 13.2 — Which Motions Are (Nearly) Simple Harmonic

1Exercise question

Step-by-step solution

1(a) The earth's rotation repeats with a fixed period but is not an oscillation about a mean — periodic, not SHM.
2(b) The mercury column oscillates to and fro about a mean level — SHM.
3(c) For small releases the ball's displacement from the lowest point is proportional to its acceleration — SHM.
4(d) A polyatomic molecule vibrates as a superposition of many modes — periodic, not SHM.

Final answer

SHM: (b) and (c); periodic but not SHM: (a) and (d).

NCERT Exercise 13.3 — Reading x-t Plots: Periodic Motion and Its Period

1Exercise question

Step-by-step solution

1(a) Monotonic, unidirectional linear motion — the motion never repeats, so not periodic.
2(b) The pattern repeats every 2 s — periodic motion with T = 2 s.
3(c) The particle repeats over a part, but the full curve does not repeat in equal time intervals — not periodic.
4(d) The pattern repeats every 2 s — periodic motion with T = 2 s.

Final answer

Plots (b) and (d) are periodic, each with period 2 s.

NCERT Exercise 13.4 — Classifying Functions of Time (SHM / Periodic / Non-Periodic)

1Exercise question

Step-by-step solution

1(a) sin ωt − cos ωt = √2 sin(ωt − π/4) — SHM, period 2π/ω.
2(b) sin³ ωt = (3 sin ωt − sin 3ωt)/4 — sum of two SHMs of different frequencies — periodic but not SHM, period 2π/ω.
3(c) 3 cos(π/4 − 2ωt) = 3 cos(2ωt − π/4) — SHM, period 2π/2ω = π/ω.
4(d) Sum of cos ωt, cos 3ωt, cos 5ωt — periodic but not SHM, fundamental period 2π/ω.
5(e) exp(−ω²t²) decays and never repeats — non-periodic.
6(f) 1 + ωt + ω²t² grows without repeating — non-periodic.

Final answer

SHM: (a) 2π/ω, (c) π/ω; periodic not SHM: (b) 2π/ω, (d) 2π/ω; non-periodic: (e), (f).

NCERT Exercise 13.5 — Signs of Velocity, Acceleration and Force in SHM

1Exercise question

Step-by-step solution

1(a) At the extreme A the particle is momentarily at rest: velocity 0; acceleration and force point towards the mean (B side) — both positive.
2(b) At B: velocity 0; acceleration and force point towards A — both negative.
3(c) At the mid-point moving left: velocity negative; acceleration and force at the mean position are zero.
4(d) Between B and the midpoint moving left: velocity negative; displacement on the B side makes acceleration and force negative.
5(e) Between A and the midpoint moving right (A → B): velocity positive; displacement on the A side makes acceleration and force positive.
6(f) Near B moving left: velocity negative, and (like (d)) acceleration and force negative.

Final answer

(a) 0, +, + (b) 0, −, − (c) −, 0, 0 (d) −, −, − (e) +, +, + (f) −, −, − (velocity, acceleration, force).

NCERT Exercise 13.6 — Which a(x) Relations Represent SHM

1Exercise question

Step-by-step solution

1SHM requires the force law a = −(k/m)x, i.e. a proportional to x with a negative constant.
2(a) a = +0.7x: wrong sign — not SHM.
3(b) a = −200x²: proportional to x² — not SHM.
4(c) a = −10x: matches a = −ω²x with ω² = 10 — SHM.
5(d) a = 100x³: proportional to x³ — not SHM.

Final answer

Only (c) a = −10x represents simple harmonic motion.

NCERT Exercise 13.7 — Amplitude and Phase From Initial Conditions

1Exercise question

Step-by-step solution

1At t = 0: A cos φ = 1 and (differentiating) v₀ = −Aω sin φ = ω, so A sin φ = −1.
2Adding squares: A²(cos²φ + sin²φ) = 1² + 1² = 2, so A = √2 cm.
3tan φ = (A sin φ)/(A cos φ) = −1 with cos φ > 0, sin φ < 0, so φ = −π/4 (7π/4).
4For x = B sin(ωt + α): B sin α = 1 and Bω cos α = ω, so B cos α = 1.
5B² = 2 → B = √2 cm; tan α = 1 with both components positive, so α = π/4.

Final answer

Cosine form: A = √2 cm, φ = −π/4; sine form: B = √2 cm, α = π/4.

NCERT Exercise 13.8 — Weight of a Body Oscillating on a Spring Balance

1Exercise question

Step-by-step solution

1Full scale: 50 kg stretches the spring 0.20 m, so k = mg/x = 50 × 9.8/0.20 = 2450 N m⁻¹.
2T = 2π√(m/k) with T = 0.6 s: m = kT²/4π² = 2450 × 0.36/39.5 = 22.4 kg.
3Weight = mg = 22.4 × 9.8 ≈ 219 N.

Final answer

The body weighs about 219 N (mass ≈ 22.4 kg).

NCERT Exercise 13.9 — Frequency, Maximum Acceleration and Speed of a Spring

1Exercise question

Step-by-step solution

1Angular frequency ω = √(k/m) = √(1200/3) = 20 rad s⁻¹; A = 0.02 m.
2(i) f = ω/2π = 20/2π = 3.18 Hz.
3(ii) a_max = ω²A = 400 × 0.02 = 8 m s⁻².
4(iii) v_max = ωA = 20 × 0.02 = 0.4 m s⁻¹.

Final answer

(i) 3.18 Hz (ii) 8 m s⁻² (iii) 0.4 m s⁻¹.

NCERT Exercise 13.10 — Displacement Functions for Three Starting Positions

1Exercise question

Step-by-step solution

1From Ex 13.9, A = 2 cm and ω = 20 rad s⁻¹.
2(a) Starting at the mean moving right: x = A sin ωt = 2 sin 20t.
3(b) Starting at maximum stretch: x = A sin(ωt + π/2) = 2 cos 20t.
4(c) Starting at maximum compression: x = A sin(ωt + 3π/2) = −2 cos 20t.
5The three functions have the same frequency (20/2π Hz) and amplitude (2 cm); only the initial phase differs (0, π/2, 3π/2).

Final answer

(a) x = 2 sin 20t (b) x = 2 cos 20t (c) x = −2 cos 20t; same amplitude and frequency, different initial phase.

NCERT Exercise 13.11 — SHM From the x-Projection of Two Circular Motions

1Exercise question

Step-by-step solution

1First circular motion: T = 2 s, A = 3 cm, and at t = 0 the radius OP makes +π/2 with the +x-axis.
2x = A cos(2πt/T + φ) = 3 cos(πt + π/2) = −3 sin πt cm.
3Second circular motion: T = 4 s, A = 2 m, and at t = 0 the radius OP is along the negative x-axis (φ = π).
4x = 2 cos(2πt/4 + π) = −2 cos(πt/2) m.

Final answer

First: x = −3 sin πt cm; second: x = −2 cos(πt/2) m.

NCERT Exercise 13.12 — Reference Circles for Four SHM Equations

1Exercise question

Step-by-step solution

1(a) −2 sin(3t + π/3) = 2 cos(3t + 5π/6): radius 2 cm, initial phase 5π/6 (150°), angular speed 3 rad s⁻¹.
2(b) cos(π/6 − t) = cos(t − π/6): radius 1 cm, initial phase −π/6 (−30°), angular speed 1 rad s⁻¹.
3(c) 3 sin(2πt + π/4) = −3 cos(2πt + 3π/4): radius 3 cm, initial phase 3π/4 (135°), angular speed 2π rad s⁻¹.
4(d) 2 cos πt: radius 2 cm, initial phase 0, angular speed π rad s⁻¹.
5For each, mark the radius vector at angle φ anticlockwise from the +x-axis at t = 0.

Final answer

(a) A = 2 cm, φ = 5π/6, ω = 3 (b) A = 1 cm, φ = −π/6, ω = 1 (c) A = 3 cm, φ = 3π/4, ω = 2π (d) A = 2 cm, φ = 0, ω = π.

NCERT Exercise 13.13 — Spring Extension and Periods for One-Mass and Two-Mass Setups

1Exercise question

Step-by-step solution

1(a) One-mass setup: force F at the free end gives extension l = F/k.
2(a) Two-mass setup: each end pulled by F displaces its end x = l/2; the net force 2kx equals F, so l = F/k again — the extensions are the same.
3(b) One mass: mx'' = −kx, so ω = √(k/m) and T = 2π√(m/k).
4(b) Two masses: the centre of the spring is fixed; each half has effective force constant 2k, so T = 2π√(m/2k).

Final answer

(a) Extension is F/k in both cases (b) T₁ = 2π√(m/k), T₂ = 2π√(m/2k).

NCERT Exercise 13.14 — Maximum Speed of a Locomotive Piston

1Exercise question

Step-by-step solution

1Amplitude A = stroke/2 = 0.5 m.
2ω = 200 rad min⁻¹ = 200/60 = 3.33 rad s⁻¹.
3v_max = ωA = (200/60) × 0.5 = 1.67 m s⁻¹ (≏ 100 m min⁻¹).

Final answer

Maximum speed = 1.67 m s⁻¹ ≈ 100 m min⁻¹.

NCERT Exercise 13.15 — Period of a Pendulum on the Moon

1Exercise question

Step-by-step solution

1Tₑ = 2π√(l/gₑ), so l = gₑTₑ²/4π².
2On the moon, T_m = 2π√(l/g_m) = Tₑ √(gₑ/g_m).
3T_m = 3.5 × √(9.8/1.7) = 3.5 × 2.40.
4T_m = 8.4 s.

Final answer

The period on the moon is 8.4 s.

NCERT Exercise 13.16 — Period of a Pendulum in a Car on a Circular Track

1Exercise question

Step-by-step solution

1The bob feels gravity g and the centripetal acceleration v²/R (at right angles).
2Effective acceleration a_eff = √(g² + (v²/R)²).
3T = 2π√(l/a_eff) = 2π√(l/√(g² + v⁴/R²)).

Final answer

T = 2π√(l/√(g² + v⁴/R²)).

NCERT Exercise 13.17 — Floating Cork Oscillating Simple Harmonically

1Exercise question

Step-by-step solution

1At equilibrium the weight of the cork is balanced by the upthrust; depressing it by x displaces an extra volume Ax.
2Extra upthrust (restoring force) F = −A x ρ_l g, directed upward.
3Since F ∝ −x, the motion is SHM with force constant k = Aρ_l g.
4Mass of cork m = A h ρ.
5T = 2π√(m/k) = 2π√(A h ρ/A ρ_l g) = 2π√(hρ/ρ_l g).

Final answer

Restoring force = −Aρ_l g x gives SHM with T = 2π√(hρ/ρ_l g).

NCERT Exercise 13.18 — Mercury Column in a U-Tube Executing SHM

1Exercise question

Step-by-step solution

1If one level is displaced by h above the other, the restoring force is the weight of the unbalanced mercury column of height 2h.
2F = −(A × 2h × ρ)g = −2Aρg h.
3F ∝ −h, so the motion is SHM with k = 2Aρg.
4Mass of mercury m = A l ρ (l = total length of the mercury column).
5T = 2π√(m/k) = 2π√(A l ρ/2Aρg) = 2π√(l/2g).

Final answer

Restoring force = −2Aρg h gives SHM with period T = 2π√(l/2g).

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Equation of SHM

a = - ω^{2} x

Simple pendulum

T = 2 π \frac{L}{g}

Spring time period

T = 2 π \frac{m}{k}

Total energy in SHM

E = \frac{1}{2} m ω^{2} A^{2}

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

A simple pendulum's time period is independent of the mass and, at small angles, of the amplitude too.
Energy in SHM is entirely kinetic at the mean position and entirely potential at the extremes.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Physics Chapter 13 (Oscillations)?

There are 18 exercise questions in this chapter, numbered 13.1 to 13.18. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Oscillations Class 11 Physics?

The formulas this chapter's questions actually turn on are: Equation of SHM, Simple pendulum, Spring time period, Total energy in SHM. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Oscillations important for JEE Main and NEET?

Essential for both — SHM and the pendulum and spring time periods sit at the junction of mechanics and waves and appear in JEE Main and NEET every year.

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