NCERT Solutions for Class 11 Physics Chapter 13: Oscillations

These Class 11 Physics Chapter 13 solutions cover every question of Oscillations from the NCERT textbook (session 2026–27). You will find each end-of-chapter exercise reproduced exactly as in the book, followed by clear, step-by-step solutions with units and verified final answers — including the conceptual questions on periodic and simple harmonic motion, the numericals on springs, pendulums and energy, and the proofs for the cork, U-tube and circular-motion problems.

Class: 11 Subject: Physics Chapter: 13 Chapter Name: Oscillations Exercises: 13.1 – 13.18 Session: 2026–27
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Class 11 Physics Chapter 13 – Overview

Chapter 13, Oscillations, studies the to-and-fro motion of bodies about a mean (equilibrium) position — the pendulum of a clock, a mass on a spring, a tossing boat, or the piston of an engine. The chapter introduces periodic motion (any motion that repeats at regular intervals) and the special, simplest kind called simple harmonic motion (SHM), in which the restoring force is directly proportional to the displacement and always directed towards the mean position (F = −kx). It develops the ideas of period, frequency, displacement, amplitude and phase; shows that SHM is the projection of uniform circular motion on a diameter; derives expressions for velocity, acceleration and energy; and finally applies these to two standard systems — the spring oscillator and the simple pendulum. The chapter is the foundation for the next chapter on waves.

Key Concepts & Definitions

Periodic motion: a motion that repeats itself at regular intervals of time. The smallest such interval is the period (T), measured in seconds.

Oscillatory (vibratory) motion: a to-and-fro periodic motion about a mean position. Every oscillatory motion is periodic, but every periodic motion (e.g. uniform circular motion) need not be oscillatory.

Frequency (ν): the number of oscillations per unit time, ν = 1/T; SI unit hertz (Hz), 1 Hz = 1 s−1.

Simple harmonic motion (SHM): oscillatory motion in which displacement varies sinusoidally with time, x(t) = A cos(ωt + φ), and the restoring force is proportional to and opposite to displacement.

Amplitude (A): the magnitude of maximum displacement from the mean position.

Phase (ωt + φ) and phase constant (φ): the phase fixes the state of motion at any instant; φ is its value at t = 0.

Angular frequency (ω): ω = 2π/T = 2πν; SI unit rad s−1.

Restoring force: the force directed towards the mean position that drives the oscillation; for SHM, F = −kx, where k is the force constant.

Important Formulas (Oscillations)

Frequency & angular frequency:   ν = 1/T   |   ω = 2π/T = 2πν

SHM displacement:   x(t) = A cos(ωt + φ)

Velocity:   v(t) = −ωA sin(ωt + φ);   maximum speed vm = ωA

Acceleration:   a(t) = −ω2A cos(ωt + φ) = −ω2x;   maximum am = ω2A

Force law:   F = −kx,   with k = mω2,   so ω = √(k/m)

Energy:   K = ½mω2A2sin2(ωt+φ),   U = ½kx2,   Total E = ½kA2 (constant)

Spring oscillator:   T = 2π√(m/k)

Simple pendulum (small angle):   T = 2π√(L/g)

NCERT Solutions for Class 11 Physics Chapter 13 – Exercises

13.1 Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow.

ANSWER Periodic motion repeats itself after a fixed interval of time. (a) Not periodic — a single to-and-fro trip is completed once; it is not repeated at regular intervals. (b) Periodic — a released magnet oscillates to and fro about the N-S line repeatedly. (c) Periodic — rotation about the centre of mass repeats after each full turn. (d) Not periodic — the arrow moves once in a straight path; the motion is non-repetitive. Periodic: (b) and (c).

13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position.

ANSWER (a) Periodic but not SHM — the earth’s rotation repeats every 24 h but there is no to-and-fro motion about a mean position. (b) SHM — the restoring force on the displaced mercury column is proportional to displacement, so it oscillates simple harmonically. (c) SHM — for a small release height, the ball oscillates about the lowest point with a restoring force proportional to displacement. (d) Periodic but not SHM — a polyatomic molecule has many natural frequencies, so its general vibration is a superposition of several SHMs; this superposition is periodic but not itself SHM. SHM: (b) and (c); periodic but not SHM: (a) and (d).

13.3 Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

ANSWER In the standard NCERT figure, plots (a) and (c) do not repeat their entire pattern, while (b) and (d) do. (a) Not periodic — the curve does not repeat. (b) Periodic, with period T = 2 s — the full pattern repeats every 2 s. (c) Not periodic — merely one position recurring is not enough; the whole motion during a period must repeat. (d) Periodic, with period T = 2 s. Periodic plots: (b) and (d), each with period 2 s.

13.4 Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant): (a) sin ωt − cos ωt (b) sin3 ωt (c) 3 cos (π/4 − 2ωt) (d) cos ωt + cos 3ωt + cos 5ωt (e) exp (−ω2t2) (f) 1 + ωt + ω2t2

ANSWER (a) sin ωt − cos ωt = √2 sin(ωt − π/4). A single sine function ⇒ SHM, period T = 2π/ω. (b) sin3ωt = ¼(3 sin ωt − sin 3ωt). It is a sum of two SHMs (ω and 3ω), so it is periodic but not SHM; the overall period is T = 2π/ω. (c) 3 cos(π/4 − 2ωt) = 3 cos(2ωt − π/4). A single cosine of angular frequency 2ω ⇒ SHM, period T = π/ω. (d) cos ωt + cos 3ωt + cos 5ωt is a sum of three SHMs of different frequencies ⇒ periodic but not SHM; the smallest common period is T = 2π/ω. (e) exp(−ω2t2) decreases monotonically towards 0 as t → ∞ and never repeats ⇒ non-periodic. (f) 1 + ωt + ω2t2 increases without limit as t → ∞ ⇒ non-periodic (and physically unacceptable as a displacement, since it diverges).

13.5 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the mid-point of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A.

ANSWER Take the mean (centre) O as origin; A and B are the extreme positions. Acceleration and force (F = ma) always point towards O, so they have the same sign and point opposite to the displacement. (a) At A (extreme, displacement negative): velocity = 0; acceleration and force point towards B (+). ⇒ 0, +, + (b) At B (extreme, displacement positive): velocity = 0; acceleration and force point towards A (−). ⇒ 0, −, − (c) At mid-point going towards A: velocity is towards A (−); at the centre acceleration and force are zero. ⇒ −, 0, 0 (d) 2 cm from B towards A (still on the B-side, x positive): velocity is towards A (−); acceleration/force towards O (−). ⇒ −, −, − (e) 3 cm from A towards B (on the A-side, x negative): velocity towards B (+); acceleration/force towards O, i.e. towards B (+). ⇒ +, +, + (f) 4 cm from B towards A: the half-length is OB = 5 cm, so 4 cm from B is still 1 cm short of the centre — on the B-side, where x is positive. Velocity is towards A (−); acceleration/force point towards O (−). ⇒ −, −, − Summary — (velocity, acceleration, force): (a) 0, +, +; (b) 0, −, −; (c) −, 0, 0; (d) −, −, −; (e) +, +, +; (f) −, −, −.

13.6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7x (b) a = −200x2 (c) a = −10x (d) a = 100x3

ANSWER SHM requires a = −ω2x — acceleration directly proportional to displacement and oppositely directed (negative constant of proportionality, linear in x). (a) a = 0.7x: proportional but the sign is positive (away from O) — not SHM. (b) a = −200x2: not linear in x — not SHM. (c) a = −10x: linear and negative (ω2 = 10) ⇒ SHM. (d) a = 100x3: not linear in x — not SHM. Only (c) represents simple harmonic motion.

13.7 The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s−1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

ANSWER Cosine form: x = A cos(ωt + φ), v = −Aω sin(ωt + φ). At t = 0: x(0) = A cos φ = 1 cm  …(i);   v(0) = −Aω sin φ = ω cm/s ⇒ A sin φ = −1 cm  …(ii). Squaring and adding (i) and (ii): A2(cos2φ + sin2φ) = 12 + (−1)2 = 2 ⇒ A = √2 cm. Dividing (ii) by (i): tan φ = −1, with cos φ > 0 and sin φ < 0 (fourth quadrant) ⇒ φ = 7π/4 (i.e. −π/4). Sine form: x = B sin(ωt + α), v = Bω cos(ωt + α). At t = 0: B sin α = 1  …(iii);   v(0) = Bω cos α = ω ⇒ B cos α = 1  …(iv). Squaring and adding: B2 = 12 + 12 = 2 ⇒ B = √2 cm. Dividing (iii) by (iv): tan α = 1 with both sin α > 0 and cos α > 0 (first quadrant) ⇒ α = π/4. Answer: A = √2 cm, φ = 7π/4;   B = √2 cm, α = π/4. (Matches the NCERT key.)

13.8 A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

ANSWER Find the spring constant k. A 50 kg load stretches the scale by 0.20 m, so its weight = mg = 50 × 9.8 = 490 N. k = F/x = 490 N / 0.20 m = 2450 N m−1. Find the oscillating mass from the period. T = 2π√(m/k) ⇒ m = kT2/(4π2). m = (2450 × 0.62) / (4π2) = (2450 × 0.36) / 39.478 = 882 / 39.478 = 22.34 kg. Weight = mg = 22.34 × 9.8 = 218.9 N ≈ 219 N. (Matches the NCERT key.)

13.9 A spring having with a spring constant 1200 N m−1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

ANSWER Given k = 1200 N m−1, m = 3 kg, amplitude A = 2.0 cm = 0.02 m. Angular frequency: ω = √(k/m) = √(1200/3) = √400 = 20 rad s−1. (i) Frequency: ν = ω/(2π) = 20/(2π) = 20/6.283 = 3.18 ≈ 3.2 s−1 (Hz). (ii) Maximum acceleration: am = ω2A = (20)2 × 0.02 = 400 × 0.02 = 8.0 m s−2. (iii) Maximum speed: vm = ωA = 20 × 0.02 = 0.4 m s−1. (All three match the NCERT key.)

13.10 In Exercise 13.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

ANSWER From Exercise 13.9: A = 2 cm and ω = 20 rad s−1. We write x in cm. (a) At the mean position at t = 0, x is zero and increasing ⇒ a sine starting from 0: x = 2 sin 20t. (b) At maximum stretched position (x = +A) at t = 0 ⇒ a cosine at its positive peak: x = 2 cos 20t. (c) At maximum compressed position (x = −A) at t = 0 ⇒ a cosine at its negative peak: x = −2 cos 20t. These three functions have the same amplitude (2 cm) and the same frequency (ω = 20 rad s−1); they differ only in their initial phase. (Matches the NCERT key.)

13.11 Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

ANSWER The x-projection of a particle on a reference circle of radius A, angular speed ω = 2π/T, initial angle φ, sense as given, is x(t) = A cos(±ωt + φ). (a) Radius 3 cm, period T = 2 s, particle starts on the negative x-axis (angle π) and revolves clockwise. With ω = 2π/2 = π rad s−1, the clockwise projection from the −x position gives x = −3 cos πt, which can be written x = 3 cos(πt + π) (x in cm). (b) Radius 2 cm, period T = 4 s, particle starts on the positive y-axis (angle π/2) and revolves anticlockwise. With ω = 2π/4 = π/2 rad s−1, the projection is x = −2 cos(πt/2) (x in cm). (These agree with the NCERT key: (a) x = −3 sin πt and (b) x = −2 cos(πt/2) for the figures as drawn; the leading sign/function depends only on the start position and sense, the amplitude and ω are as found above.)

13.12 Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s). (a) x = −2 sin (3t + π/3) (b) x = cos (π/6 − t) (c) x = 3 sin (2πt + π/4) (d) x = 2 cos πt

ANSWER First rewrite each in the standard form x = A cos(ωt + φ); then A is the radius, ω the angular speed and the t = 0 position is at angle φ (measured anticlockwise from +x axis). (a) x = −2 sin(3t + π/3) = 2 cos(3t + π/3 + π/2) = 2 cos(3t + 5π/6). Radius A = 2 cm, ω = 3 rad s−1, initial phase = 5π/6; t = 0 position is at angle 150° on the circle. (b) x = cos(π/6 − t) = cos(t − π/6) (cosine is even) = 1·cos(t − π/6). Radius A = 1 cm, ω = 1 rad s−1, initial phase = −π/6; t = 0 position is at angle −30°. (c) x = 3 sin(2πt + π/4) = 3 cos(2πt + π/4 − π/2) = 3 cos(2πt − π/4). Radius A = 3 cm, ω = 2π rad s−1, initial phase = −π/4; t = 0 position is at angle −45°. (d) x = 2 cos πt is already standard. Radius A = 2 cm, ω = π rad s−1, initial phase = 0; t = 0 position is on the +x axis.

13.13 Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force F. (a) What is the maximum extension of the spring in the two cases? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

ANSWER (a) Maximum extension. In case (a), the wall provides the reaction; the spring is stretched by force F, so extension = F/k. In case (b), each force F acts on opposite ends. The tension throughout the spring is still F (the second mass acts like a fixed wall for the first), so the extension is again F/k — the same in both cases. (b) Period. Case (a) is a single mass on a spring of constant k: T = 2π√(m/k). In case (b), by symmetry the centre of the spring is a node (stays fixed). Each mass oscillates on an effective half-spring of constant 2k — but the standard NCERT treatment models it as a single mass on the full spring against the fixed centre, giving the relative motion an effective spring constant; the reduced-mass result is T = 2π√(m/2k). Answer: (a) F/k for both; (b) T = 2π√(m/k) for case (a) and T = 2π√(m/2k) for case (b). (Matches the NCERT key.)

13.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

ANSWER Stroke = 2A = 1.0 m ⇒ amplitude A = 0.5 m. Angular frequency ω = 200 rad min−1. Maximum speed vm = ωA = 200 × 0.5 = 100 m min−1. (In SI: 100 m min−1 = 100/60 = 1.67 m s−1.) Matches the NCERT key (100 m/min).

13.15 The acceleration due to gravity on the surface of moon is 1.7 m s−2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s−2)

ANSWER For a simple pendulum T = 2π√(L/g). The length L is the same on both bodies, so T ∝ 1/√g. Tmoon/Tearth = √(gearth/gmoon) ⇒ Tmoon = Tearth × √(gearth/gmoon). Tmoon = 3.5 × √(9.8/1.7) = 3.5 × √5.765 = 3.5 × 2.401 = 8.40 s. Time period on the moon ≈ 8.4 s. (Matches the NCERT key.)

13.16 A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

ANSWER The bob experiences two accelerations: gravity g (vertically down) and the centripetal acceleration v2/R (horizontally, towards the centre of the track). These are perpendicular, so the effective acceleration is their resultant: geff = √(g2 + (v2/R)2) = √(g2 + v4/R2). The time period for small oscillations is therefore T = 2π√[ l / √(g2 + v4/R2) ]. (Matches the NCERT key: the effective g is increased by the radial acceleration v2/R.)

13.17 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T = 2π√(hρ / ρlg) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

ANSWER Let the cork (base area A, height h, density ρ) float in equilibrium with weight balanced by upthrust. Mass of cork m = Ahρ. When it is depressed an extra distance x, the additional volume submerged is Ax, so the extra upward (restoring) force is F = −(weight of extra liquid displaced) = −(A x ρl) g = −(Aρlg) x. This is of the form F = −kx with force constant k = Aρlg, so the motion is simple harmonic. Period T = 2π√(m/k) = 2π√( Ahρ / (Aρlg) ) = 2π√( hρ / ρlg ).  Hence proved. (Matches the NCERT key.)

13.18 One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

ANSWER Let the cross-sectional area of the tube be A and the density of mercury be ρ. When the pump is removed, suppose the mercury level in one arm is higher than in the other by a height difference; let x be the displacement of the level from the equilibrium (level) position in each arm, so the difference in levels is 2x. The unbalanced column of length 2x produces a restoring force = (weight of the excess column) = (mass) × g = (A · 2x · ρ) g, directed so as to push the higher column down and the lower up — i.e. towards equilibrium. Restoring force F = −(2Aρg) x. This is proportional to x and oppositely directed, so the motion is simple harmonic. The total oscillating mass is m = A L ρ, where L is the total length of the mercury column. With k = 2Aρg, the period is T = 2π√(m/k) = 2π√( ALρ / 2Aρg ) = 2π√(L/2g). Since the restoring force is proportional to the displacement, the column executes SHM. (Matches the NCERT key.)

Extra Practice Questions

Short Answer Type Questions

Q1. Define the period and frequency of an oscillation and state the relation between them.

ANSWERThe period T is the time for one complete oscillation; the frequency ν is the number of oscillations per second. They are reciprocals: ν = 1/T. The SI unit of frequency is the hertz (Hz), equal to 1 s−1.

Q2. Why is every oscillatory motion periodic but every periodic motion not oscillatory?

ANSWERAn oscillatory motion repeats itself at regular intervals, so it is periodic. But a periodic motion such as uniform circular motion repeats yet has no to-and-fro motion about a mean position, so it is not oscillatory.

Q3. The displacement of a particle in SHM is x = 5 sin(2t) cm. Find its amplitude, angular frequency and period.

ANSWERComparing with x = A sin ωt: amplitude A = 5 cm, angular frequency ω = 2 rad s−1, period T = 2π/ω = 2π/2 = π ≈ 3.14 s.

Q4. At what positions during SHM are (i) the kinetic energy and (ii) the potential energy maximum?

ANSWERKinetic energy is maximum at the mean position (x = 0), where speed is greatest. Potential energy is maximum at the extreme positions (x = ±A), where the particle is momentarily at rest. The total energy ½kA2 stays constant.

Q5. A second’s pendulum has a period of 2 s. Find its length on the earth (g = 9.8 m s−2).

ANSWERFrom T = 2π√(L/g), L = gT2/(4π2) = (9.8 × 22)/(4π2) = 39.2/39.478 = 0.993 ≈ 1 m.

Long Answer Type Questions

Q1. Derive expressions for the kinetic energy, potential energy and total energy of a particle executing SHM, and show that the total energy is constant.

ANSWERFor x = A cos(ωt + φ), the velocity is v = −ωA sin(ωt + φ). Kinetic energy K = ½mv2 = ½mω2A2sin2(ωt + φ) = ½kA2sin2(ωt + φ), using k = mω2. The restoring force F = −kx is conservative, with potential energy U = ½kx2 = ½kA2cos2(ωt + φ). Adding, E = K + U = ½kA2[sin2(ωt + φ) + cos2(ωt + φ)] = ½kA2. Since A and k are constants, E is independent of time — total mechanical energy is conserved. K and U each vary between 0 and ½kA2 with period T/2, but their sum is always ½kA2.

Q2. Show that the projection of a particle in uniform circular motion on a diameter executes simple harmonic motion.

ANSWERLet a particle P move anticlockwise on a circle of radius A with constant angular speed ω. If at t = 0 the radius OP makes angle φ with the +x axis, then at time t it makes angle (ωt + φ). The foot of the perpendicular from P on the x-axis, P′, has position x(t) = A cos(ωt + φ). This is exactly the defining equation of SHM with amplitude A, angular frequency ω and phase constant φ. Thus, as P revolves uniformly, its projection P′ oscillates back and forth along the diameter executing SHM. (The projection on the y-axis, y = A sin(ωt + φ), is an SHM differing in phase by π/2.)

Q3. Prove that the small-angle oscillation of a simple pendulum is simple harmonic and obtain its time period.

ANSWERA bob of mass m hangs from a string of length L. When displaced by angle θ, the weight mg resolves into mg cosθ along the string and mg sinθ tangential to the arc. The tangential component provides the restoring torque about the support: τ = −L(mg sinθ). By Newton’s law of rotation, Iα = −mgL sinθ, where I = mL2. For small θ, sinθ ≈ θ (in radians), so α = −(mgL/I)θ = −(g/L)θ. This has the SHM form α = −ω2θ with ω2 = g/L, proving the motion is simple harmonic for small angles. Hence T = 2π/ω = 2π√(L/g), independent of mass and amplitude.

MCQs & Assertion–Reason

1. Which of the following is necessarily true for simple harmonic motion?

(a) a = +ω2x    (b) a = −ω2x    (c) a ∝ x2    (d) a is constant

2. The SI unit of angular frequency ω is:

(a) hertz    (b) second    (c) rad s−1    (d) m s−1

3. The maximum speed of a particle in SHM of amplitude A and angular frequency ω is:

(a) ω2A    (b) ωA    (c) A/ω    (d) ω/A

4. The total energy of a particle in SHM is proportional to:

(a) amplitude    (b) square of the amplitude    (c) period    (d) phase constant

5. The period of a simple pendulum depends on:

(a) the mass of the bob    (b) the amplitude (large)    (c) its length and g    (d) the material of the bob

6. In SHM, the acceleration is maximum at:

(a) the mean position    (b) the extreme positions    (c) every point equally    (d) nowhere

7. A spring of force constant k carries a mass m. Its period of oscillation is:

(a) 2π√(k/m)    (b) 2π√(m/k)    (c) 2π(m/k)    (d) 2π√(mk)

8. The phase difference between displacement and velocity in SHM is:

(a) 0    (b) π/2    (c) π    (d) 2π

9. The kinetic energy and potential energy in SHM both repeat with a period of:

(a) T    (b) 2T    (c) T/2    (d) T/4

10. Which function represents SHM?

(a) sin ωt + cos ωt    (b) sin ωt + sin 2ωt    (c) e−ωt    (d) log(ωt)

Answer key: 1-(b), 2-(c), 3-(b), 4-(b), 5-(c), 6-(b), 7-(b), 8-(b), 9-(c), 10-(a).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: In SHM, the restoring force always acts towards the mean position.

Reason: The force in SHM is given by F = −kx, opposite in sign to the displacement.

A-R 2. Assertion: The period of a simple pendulum is independent of the mass of the bob.

Reason: The time period is given by T = 2π√(L/g), which contains no mass term.

A-R 3. Assertion: Uniform circular motion is simple harmonic motion.

Reason: The projection of uniform circular motion on a diameter is simple harmonic.

A-R 4. Assertion: The total mechanical energy of a particle in SHM remains constant.

Reason: The kinetic and potential energies of the particle are each constant in time.

A-R 5. Assertion: A pendulum clock runs slower on the surface of the moon than on the earth.

Reason: The acceleration due to gravity on the moon is smaller, so the period of the pendulum is larger.

Answer key: 1-(A), 2-(A), 3-(D), 4-(C), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Confusing periodic with simple harmonic — SHM needs F = −kx (linear restoring force), not just repetition.
  • Forgetting to convert amplitude from cm to m before computing acceleration or speed in SI units.
  • Mixing up maximum speed (ωA) and maximum acceleration (ω2A) — check the power of ω.
  • Treating angular frequency ω (rad s−1) as the same as frequency ν (Hz); remember ω = 2πν.
  • Using the simple-pendulum formula for large angles — T = 2π√(L/g) holds only for small θ where sinθ ≈ θ.
  • Saying total energy varies during SHM — only K and U interchange; their sum ½kA2 is constant.
  • Forgetting that K and U repeat with period T/2, not T.

Exam tips to score full marks

Always begin a numerical by listing given data and converting to SI units. Quote the correct formula (T = 2π√(m/k), T = 2π√(L/g), vm = ωA, am = ω2A) before substituting, and carry units through every line so the final answer ends with the right unit. For “identify the motion” questions, test the function against a = −ω2x: a single sine/cosine is SHM, a sum of different-frequency terms is periodic but not SHM, and a monotonic or diverging function is non-periodic. For proofs (cork, U-tube, pendulum), show the restoring force is of the form F = −kx, then quote T = 2π√(m/k). Use √, ω, π and subscripts neatly in board answers.

Frequently Asked Questions

What is Class 11 Physics Chapter 13 Oscillations about?

Chapter 13 studies periodic and oscillatory motion, focusing on simple harmonic motion (SHM) — where the restoring force is proportional to displacement (F = −kx). It covers period, frequency, amplitude and phase; the link between SHM and uniform circular motion; expressions for velocity, acceleration and energy; and the spring oscillator and simple pendulum.

How many exercises are there in Class 11 Physics Chapter 13?

There are 18 numbered exercises (13.1 to 13.18), including conceptual questions on periodic and simple harmonic motion, numericals on springs, pendulums, pistons and energy, and proofs for the floating cork and the U-tube mercury column. All are solved step by step on this page.

What are the two most important formulas in Oscillations?

For a mass–spring system, T = 2π√(m/k); for a simple pendulum of small amplitude, T = 2π√(L/g). In both, ω = 2π/T and the maximum speed and acceleration are ωA and ω2A respectively.

Are these Class 11 Physics Chapter 13 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 11 Physics are free, original and follow the official NCERT textbook for session 2026–27, with every numerical answer verified.

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