Class 11 · Physics · Chapter 13
Oscillations
This chapter builds the mathematics of periodic motion — simple harmonic motion in particular — from the swinging pendulum and the block on a spring, and links it to circular motion, energy, damping and resonance. It has no board exam at Class 11, but it is the direct foundation for waves in Class 11 and for AC circuits, electromagnetic waves and modern physics oscillators in Class 12, and it is a favourite topic in JEE and NEET.
Before you startWhat the board asks from this chapter
No CBSE board paper at Class 11 (no board exam this class), but it carries real weight in the school final/annual exam and is tested heavily and repeatedly in JEE Main, JEE Advanced and NEET, usually through numericals on time period, energy and phase.
| Question type | Marks | Usually asked |
|---|---|---|
| MCQ | 1 | Reading x–t, v–t or a–t graphs; identifying which quantity is SHM; ratio/graph questions on energy versus displacement. |
| Assertion–Reason | 1 | Tests whether every periodic motion is oscillatory, whether every oscillatory motion is SHM, and the phase relation between x, v and a. |
| Very short answer | 2 | State the differential equation of SHM; define angular frequency; why circular motion by itself is not SHM. |
| Short answer | 3 | Derive T for a spring–mass system or write the energy expressions for SHM at a given displacement. |
| Long answer | 5 | Full derivation: v(t) and a(t) from x(t), or the simple pendulum time period derivation with the small-angle approximation. |
| Numerical/HOTS | 3 | Given amplitude and time period, find velocity/acceleration/KE/PE at a stated displacement or time; combine two springs in series/parallel. |
Hot topics, in order
- Displacement, velocity and acceleration equations of SHM and their phase relationships
- Time period of a spring–mass system (series and parallel combinations)
- Time period of a simple pendulum and the small-angle approximation
- Energy in SHM: KE, PE and total energy versus displacement and time
- Difference between periodic, oscillatory and simple harmonic motion
- SHM as the projection of uniform circular motion
- Damped oscillations and the amplitude decay formula
- Resonance and forced oscillations, with everyday examples
The whole chapter, in orderNotes
01 13.1–13.2 Periodic and Oscillatory Motions
A motion that repeats itself after equal intervals of time is periodic motion; the smallest such interval is the time period T. The number of repetitions per second is the frequency ν = 1/T, measured in hertz (Hz).
A periodic motion that moves back and forth about a fixed mean (equilibrium) position is called oscillatory or vibratory motion — a swinging pendulum, a block on a spring, the atoms in a molecule. Every oscillatory motion is periodic, but the reverse is not always true: uniform circular motion is periodic (it repeats every T) but it is not oscillatory because the particle does not move to and fro about one point.
At the mean position, the restoring force and hence the acceleration are zero and speed is maximum; at the extreme positions (turning points) velocity is zero and the restoring force/acceleration is maximum, directed back towards the mean position. This restoring force is what makes an oscillation possible — remove it and the motion stops repeating.
02 13.3 Simple Harmonic Motion (SHM) PYQ CBSE 2020
Simple harmonic motion is the special periodic motion in which the restoring force (and acceleration) is directly proportional to the displacement from the mean position and is always directed towards it: F = −kx. This single condition is enough to derive the sinusoidal shape of the motion.
The general solution is x(t) = A cos(ωt + φ), where A is the amplitude (maximum displacement from mean position), ω is the angular frequency, and (ωt + φ) is the phase at time t; φ is the phase constant, fixed by the position and velocity at t = 0. Since cosine repeats every 2π, the motion repeats when ωt increases by 2π, giving T = 2π/ω.
SHM can equally be written using sine — x(t) = A sin(ωt + φ) — the two differ only by a phase of π/2 and by the choice of where t = 0 is taken. Both x = A cos ωt (starting at extreme position) and x = A sin ωt (starting at mean position) are common starting conditions in numericals.
03 13.4 SHM and Uniform Circular Motion
SHM can be visualised as the projection, on any diameter, of a particle moving with uniform speed on a reference circle of radius A. If the particle P moves on the circle with constant angular speed ω, its foot of perpendicular on the x-axis (or y-axis) executes SHM of amplitude A and angular frequency ω — this is exactly why ω in SHM is called angular frequency even though nothing is physically rotating.
This picture is useful for reading off phase quickly: at t = 0 the reference particle is at angle φ from the axis of projection, and as it goes around, the projection point sweeps between +A and −A, matching x(t) = A cos(ωt + φ).
Do not confuse this with real circular motion — the connection is only geometrical, used as a tool to visualise phase and to derive v(t) and a(t) using simple circular-motion relations (v = Aω, centripetal acceleration = Aω²).
04 13.5 Velocity and Acceleration in SHM PYQ CBSE 2019
Differentiating x(t) = A cos(ωt + φ) gives velocity and acceleration.
v(t) = dx/dt = −Aω sin(ωt + φ). Velocity is maximum (= Aω) at the mean position (x = 0) and zero at the extreme positions (x = ±A). Velocity is 90° (π/2) out of phase with displacement.
a(t) = dv/dt = −Aω²cos(ωt + φ) = −ω²x. Acceleration is directly proportional to displacement and always opposite in sign — it always points towards the mean position. Acceleration is maximum (= Aω²) at the extremes and zero at the mean position, and it is 180° out of phase with displacement (exactly opposite) but 90° out of phase with velocity.
The relation v = ω√(A² − x²) (obtained by eliminating t between x and v) is very useful for numericals that give displacement and ask for speed, or vice versa, without needing time.
05 13.6 Force Law for SHM
From Newton's second law, F = ma = −mω²x = −kx, where k = mω² is called the force constant (or spring constant when the system is literally a spring). This is Hooke's-law-like behaviour: the restoring force is proportional to displacement and opposite to it.
Conversely, whenever the net force (or torque, for angular oscillations) on a system can be written as F = −kx for small displacements from equilibrium, the resulting motion is guaranteed to be SHM with ω = √(k/m). This is the standard method used in the chapter to prove that a spring–mass system, a simple pendulum (for small angles) and a floating cylinder all execute SHM — write down the restoring force/torque, show it is linear in displacement, and read off ω.
Note k has units N/m and depends on the physical system (stiffness of spring, weight and length for a pendulum, etc.), while ω = √(k/m) combines k with the inertia (mass) of the oscillating body.
06 13.7 Energy in Simple Harmonic Motion PYQ CBSE 2018
SHM conserves total mechanical energy because the restoring force is conservative. At displacement x, kinetic energy K = ½mv² = ½mω²(A² − x²), and potential energy U = ½kx² = ½mω²x² (taking U = 0 at the mean position). Both K and U oscillate with time but their sum stays constant.
Total energy E = K + U = ½mω²A² = ½kA², independent of x and of time — it depends only on amplitude, mass and ω (or k). This is one of the most tested numerical ideas: E is fixed by A and k alone; at any displacement x, K = E − U so you can find one from the other without extra data.
K is maximum (= E) at the mean position where v is maximum; U is maximum (= E) at the extreme positions where x = ±A. Both K and U vary as cos²/sin² of (ωt + φ), so each oscillates at twice the frequency of the displacement itself, always staying non-negative.
07 13.8 Some Systems Executing SHM PYQ CBSE 2023 · CBSE 2017
Spring–mass system: A block of mass m on a frictionless surface attached to a spring of force constant k, displaced by x, feels restoring force F = −kx, so ω = √(k/m) and T = 2π√(m/k). This holds whether the spring is horizontal or vertical (gravity only shifts the mean position, it does not change T).
For springs in series, 1/keff = 1/k1 + 1/k2 (effective spring is softer, larger T); in parallel, keff = k1 + k2 (stiffer, smaller T).
Simple pendulum: a point mass m on a light, inextensible string of length L. For angular displacement θ, restoring torque about the support is τ = −mgL sin θ; for small θ, sin θ ≈ θ, giving τ = −mgLθ, i.e. angular SHM with ω² = g/L (moment of inertia I = mL², and mgLθ = Iα gives α = −(g/L)θ). Hence T = 2π√(L/g), independent of amplitude (for small θ) and of the mass of the bob — this is exactly why it is used to measure g experimentally.
08 13.9 Damped Simple Harmonic Motion
Real oscillators lose energy to friction or a resistive medium, so amplitude decreases with time — this is damped oscillation. For a block on a spring in a viscous medium, besides the spring force −kx there is a damping force proportional to velocity, −bv (b = damping constant), so the equation of motion is m(d²x/dt²) = −kx − b(dx/dt).
The solution (for small damping) is x(t) = A e−bt/2m cos(ω′t + φ), where the amplitude A e−bt/2m decays exponentially with time, and the (reduced) angular frequency is ω′ = √(k/m − b²/4m²), slightly less than the undamped ω0 = √(k/m).
Mechanical energy is not conserved in damped SHM: E(t) = ½kA²e−bt/m, decaying exponentially. Larger b means faster decay; if damping is too large, the system does not oscillate at all (overdamped) and just returns slowly to equilibrium — car shock absorbers are deliberately designed this way (critically/heavily damped).
09 13.10 Forced Oscillations and Resonance PYQ CBSE 2022
When an external periodic force F(t) = F0 cos(ωdt) is applied to a damped oscillator (natural frequency ω0), after the initial transients die out the system settles into steady oscillations at the driving frequency ωd, not its own natural frequency — this is forced (driven) oscillation.
The steady-state amplitude depends on how close ωd is to ω0 and on the damping: it is largest when ωd is close to ω0. When ωd = ω0 (or very close to it, for small damping), the amplitude becomes maximum — this condition is called resonance. With very small damping the resonance amplitude can become very large (in the ideal undamped case it would grow without bound); real damping always keeps it finite.
Everyday examples: pushing a swing at its own natural rhythm builds up a large amplitude; soldiers break step crossing a bridge to avoid matching the bridge's natural frequency; a radio tuning circuit is adjusted (LC resonance) to match the frequency of the desired broadcast station.
VocabularyKey terms
Keep this on one pageFormula sheet
| Name | Expression | Notes |
|---|---|---|
| Frequency–period relation | ν = 1/T | ν in hertz, T in seconds |
| Displacement in SHM | x(t) = A cos(ωt + φ) | general equation of SHM |
| Time period from ω | T = 2π/ω | ω in rad/s |
| Velocity in SHM | v(t) = −Aω sin(ωt + φ); v = ω√(A2 − x2) | maximum value Aω at x = 0 |
| Acceleration in SHM | a(t) = −ω2x | maximum value Aω2 at x = ±A; defining relation of SHM |
| Force law for SHM | F = −kx, ω = √(k/m) | k = mω2 is the force constant |
| Time period, spring–mass system | T = 2π√(m/k) | independent of amplitude; horizontal or vertical spring |
| Time period, simple pendulum | T = 2π√(L/g) | valid for small angular amplitude, independent of mass and amplitude |
| Kinetic energy in SHM | K = ½mω2(A2 − x2) | maximum at mean position |
| Potential energy in SHM | U = ½kx2 = ½mω2x2 | maximum at extreme positions |
| Total energy in SHM | E = ½kA2 = ½mω2A2 | constant, independent of x and t |
| Damped oscillation amplitude | x(t) = A e−bt/2m cos(ω′t + φ) | ω′ = √(k/m − b2/4m2) |
| Springs in series/parallel | 1/kseries = 1/k1 + 1/k2; kparallel = k1 + k2 | used to find effective T |
Practise with a pencilDiagrams that earn marks
x–t, v–t and a–t graphs of SHM (x = A cos ωt)
Stack three sinusoidal curves on a shared time axis: x(t) as a cosine from +A to −A; v(t) as a negative sine, zero at x's extremes and maximum where x = 0; a(t) as a negative cosine (mirror of x), maximum where x is extreme, zero where x = 0. Label A, Aω, Aω2 and mark T.
SHM as projection of uniform circular motion
Draw a circle of radius A about O with particle P on it, radius OP at angle (ωt + φ) to the x-axis. Drop a perpendicular from P to the x-axis at N; as P circles at speed ω, N oscillates between +A and −A. Show two or three positions of P with matching N points.
Energy (K, U, E) versus displacement x in SHM
Plot x from −A to +A on the horizontal axis, energy on the vertical. Draw U = ½kx2 as an upward parabola, zero at x = 0, rising to E at x = ±A; draw K as the mirror downward parabola. Draw E as a horizontal line at ½kA2, and shade to show K + U = E at every x.
Simple pendulum forces at angular displacement θ
Draw a bob of mass m on a string of length L, displaced through small angle θ from vertical. Resolve mg into mg cos θ along the string (balanced by tension) and mg sin θ perpendicular to it, the restoring force pulling the bob back; mark arc length x = Lθ.
Damped oscillation amplitude decay envelope
Draw x(t) = A e−bt/2m cos(ω′t + φ) with shrinking peaks over time, and overlay two dashed exponential envelopes, +A e−bt/2m and −A e−bt/2m, touching each peak and trough.
Resonance curve: amplitude versus driving frequency
Plot driving frequency ωd against steady-state amplitude. Draw two or three curves for different damping b, each peaking near ωd = ω0 (dashed vertical line); smallest b gives the tallest, narrowest peak, larger b a shorter, broader one.
TextbookNCERT questions the board keeps asking
13.1 Which of the following examples represent periodic motion? (a) A swimmer completing one length of a pool and swimming 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.2 m
(a) Not periodic — the swimmer's back-and-forth motion does not repeat identically at equal time intervals in the same manner each time (turning at the walls involves change of direction due to a conscious action, not a restoring force repeating exactly).
(b) Periodic — the bar magnet oscillates about the N-S direction under the restoring torque due to Earth's magnetic field, repeating after a fixed period.
(c) Periodic — a rotating molecule returns to the same orientation after every rotation, so its motion repeats after a fixed time (its rotational period).
(d) Not periodic — the arrow moves away from the bow only once, this motion does not repeat.
13.4 Which of the following functions of time represent (a) simple harmonic motion, (b) periodic but not SHM, and (c) non-periodic motion? Give period for each case of periodic motion: (i) sin ωt − cos ωt (ii) sin3ωt (iii) 3 cos(π/4 − 2ωt) (iv) cos ωt + cos 3ωt + cos 5ωt (v) exp(−ω2t2) (vi) 1 + ωt + ω2t25 m
(i) sin ωt − cos ωt = √2 sin(ωt − π/4) — SHM, T = 2π/ω.
(ii) sin3ωt = (3 sin ωt − sin 3ωt)/4 — periodic, not SHM, T = 2π/ω.
(iii) 3 cos(π/4 − 2ωt), form A cos(2ωt − π/4) — SHM, T = π/ω.
(iv) cos ωt + cos 3ωt + cos 5ωt — periodic, not SHM, T = 2π/ω (period of slowest term).
(v) exp(−ω2t2) — decays exponentially, non-periodic.
(vi) 1 + ωt + ω2t2 — grows without bound, non-periodic.
13.7 The equation of motion of a particle is x = a cos(αt)2. Is the motion periodic? If so, what is its period?3 m
Rewrite using cos2(αt) = (1 + cos 2αt)/2, so x = (a/2) + (a/2) cos(2αt). This is a constant plus a cosine term, so the motion is periodic (it repeats), with angular frequency 2α and period T = 2π/(2α) = π/α. Because of the constant offset a/2, the motion oscillates about x = a/2, not about x = 0, so strictly it is not SHM about the origin, but it is periodic with period π/α.
13.9 A spring having a spring constant 1200 N/m is mounted on a horizontal table, and a mass of 3 kg is attached to the free end and pulled sideways to a distance of 2.0 cm and released. Determine (i) frequency of oscillations, (ii) maximum acceleration, (iii) maximum speed of the mass.3 m
Given k = 1200 N/m, m = 3 kg, A = 2.0 cm = 0.02 m.
ω = √(k/m) = √(1200/3) = √400 = 20 rad/s.
(i) ν = ω/2π = 20/(2 × 3.14) = 20/6.28 ≈ 3.18 Hz.
(ii) amax = Aω2 = 0.02 × 400 = 8 m/s2.
(iii) vmax = Aω = 0.02 × 20 = 0.4 m/s.
13.10 In Q13.9, let us take the position of the mass when the spring is unstretched 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, (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?5 m
ω = 20 rad/s, A = 0.02 m in all three cases (from Q13.9).
(a) At mean position, moving positive at t = 0: x = A sin ωt = 0.02 sin 20t m.
(b) At maximum stretched (extreme, +A): x = A cos ωt = 0.02 cos 20t m.
(c) At maximum compressed (extreme, −A): x = −A cos ωt = −0.02 cos 20t m.
Frequency and amplitude are identical in all three — only the initial phase φ differs (equivalent to φ = −π/2, 0, π respectively in cosine form).
13.13 (a) A spring with spring constant 1200 N/m is mounted on a horizontal table and a mass of 3 kg attached to the free end. The mass is displaced from mean position by 2.0 cm and released. What is (i) the maximum speed, (ii) maximum acceleration of the mass, (iii) the total energy of oscillation? (b) If the same experiment is performed with the spring mounted vertically instead of horizontally, will any of the answers change?5 m
(a) k = 1200 N/m, m = 3 kg, A = 0.02 m; ω = √(k/m) = 20 rad/s.
(i) vmax = Aω = 0.02 × 20 = 0.4 m/s.
(ii) amax = Aω2 = 0.02 × 400 = 8 m/s2.
(iii) Total energy E = ½kA2 = ½ × 1200 × (0.02)2 = ½ × 1200 × 4 × 10−4 = 0.24 J.
(b) No, none of the answers change. Mounting the spring vertically only shifts the mean (equilibrium) position downward due to gravity (to where mg equals the spring's initial extension force); it does not change k, m, or A about the new mean position, so ω, vmax, amax and E all stay the same.
13.15 A body oscillates with SHM according to the equation x = 5 cos[2πt + π/4], where x is in metres and t is in seconds. At t = 1.5 s, calculate (a) the displacement, (b) speed, (c) acceleration of the body.5 m
x = 5 cos(2πt + π/4), so A = 5 m, ω = 2π rad/s.
At t = 1.5 s: phase = 2π(1.5) + π/4 = 3π + π/4 = 13π/4.
(a) x = 5 cos(13π/4) = 5 cos(13π/4 − 2π) = 5 cos(5π/4) = 5 × (−1/√2) ≈ −3.54 m.
(b) v = −Aω sin(13π/4) = −5(2π) sin(5π/4) = −10π × (−1/√2) = 10π/√2 ≈ 22.2 m/s.
(c) a = −ω2x = −(2π)2(−3.54) = 4π2 × 3.54 ≈ 139.9 m/s2, directed towards the mean position (positive, since x was negative).
13.17 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?5 m
When the car moves in a circle of radius R at speed v, the bob (in the car's non-inertial frame) experiences an effective gravity that is the resultant of the real gravity g (downward) and the centrifugal acceleration v2/R (horizontal, radially outward). The effective acceleration is geff = √(g2 + (v2/R)2).
The pendulum's time period is then T = 2π√(l/geff) = 2π√[l/√(g2 + v4/R2)].
13.18 A cylindrical piece of cork 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.)5 m
At equilibrium, weight of cork = weight of liquid displaced. If pushed down by small x, the extra buoyant (restoring) force is F = −A xρlg, since the cork's own weight is unchanged.
So F = −kx with k = Aρlg — hence SHM. Mass of cork m = Ahρ, so ω2 = k/m = ρlg/(hρ).
T = 2π/ω = 2π√(hρ/ρlg), as required.
13.21 You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k, and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.5 m
(a) For one wheel supporting mass m = 750 kg, sag x = 15 cm = 0.15 m under weight mg.
k = mg/x = (750 × 9.8)/0.15 = 7350/0.15 = 49000 N/m (≈ 4.9 × 104 N/m).
(b) ω ≈ √(k/m) = √(49000/750) = √65.3 ≈ 8.08 rad/s, so T = 2π/ω ≈ 0.777 s.
Amplitude decays as A e−bt/2m; amplitude halves in one period T, so A e−bT/2m = A/2, giving e−bT/2m = 0.5, i.e. bT/2m = ln 2 = 0.693.
b = 2m ln2/T = (2 × 750 × 0.693)/0.777 ≈ 1039.5/0.777 ≈ 1338 kg/s (≈ 1.33 × 103 kg s−1).
13.5 A particle executes SHM with amplitude A. At what value of displacement does its kinetic energy become equal to its potential energy?2 m
K = U means ½mω2(A2 − x2) = ½mω2x2, so A2 − x2 = x2, giving 2x2 = A2, so x = ±A/√2 ≈ ±0.707A. At this displacement, both K and U equal E/2.
Previous years and frequently askedBoard questions with model answers
MCQ The motion of a particle executing SHM is given by x = 0.01 sin 100π(t + 0.05), where x is in metres and t in seconds. The time period is:- 0.01 s
- 0.02 s
- 0.1 s
- 0.2 s
Frequently asked1 m
MCQ Which of the following quantities is always zero at the mean position in SHM?- Velocity
- Displacement
- Kinetic energy
- Total energy
Frequently asked1 m
MCQ A particle executes SHM with time period T. The time taken by it to travel from the mean position to half of its amplitude is:- T/12
- T/8
- T/6
- T/4
Frequently asked1 m
MCQ The total energy of a particle executing SHM depends on its:- Displacement only
- Amplitude and frequency
- Velocity only
- Amplitude only
Frequently asked1 m
MCQ For a particle in SHM, the phase difference between displacement and acceleration is:- 0
- π/2
- π
- 2π
Frequently asked1 m
MCQ If a simple pendulum is taken to the Moon, where g is about 1/6th of that on Earth, its time period will:- Decrease
- Increase
- Remain unchanged
- Become zero
Frequently asked1 m
MCQ Two identical springs of spring constant k are connected in series. The effective spring constant of the combination is:- 2k
- k
- k/2
- k2
Frequently asked1 m
Assertion–Reason Assertion (A): Every oscillatory motion is periodic, but every periodic motion need not be oscillatory. Reason (R): Uniform circular motion is periodic but not oscillatory because there is no fixed mean position about which the particle moves to and fro.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): In SHM, the total mechanical energy remains constant at all displacements. Reason (R): The restoring force in SHM is a conservative force.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): The time period of a simple pendulum does not depend on the mass of the bob. Reason (R): Acceleration due to gravity is independent of the mass of the bob.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Very short answer Define angular frequency and write its relation with the time period and frequency of SHM.Frequently asked2 m
Angular frequency ω is 2π times the number of oscillations per second; it measures the rate of change of the phase angle of the oscillating particle and is measured in rad s−1. It is related to time period T and frequency ν by ω = 2π/T = 2πν.
Very short answer A particle executes SHM of amplitude A. At what displacement from the mean position is its speed equal to half its maximum speed?Frequently asked2 m
v = ω√(A2 − x2) and vmax = Aω. Setting v = vmax/2: ω√(A2 − x2) = Aω/2, so A2 − x2 = A2/4, giving x2 = 3A2/4, so x = ±(√3/2)A ≈ ±0.87A.
Very short answer Why does the amplitude of an oscillating simple pendulum go on decreasing, and what type of motion does it eventually execute if left undisturbed for a very long time?Frequently asked2 m
The amplitude decreases because air resistance and friction at the support continuously dissipate the pendulum's mechanical energy — this is damped oscillation. If left undisturbed indefinitely, the amplitude decays to zero and the bob eventually comes to rest at the mean (equilibrium) position.
Very short answer State two differences between simple harmonic motion and uniform circular motion.Frequently asked2 m
(1) In SHM the particle moves to and fro in a straight line about a mean position with varying speed, whereas in uniform circular motion the particle moves on a circular path with constant speed. (2) In SHM, acceleration is variable and always directed towards the mean position (a = −ω2x); in uniform circular motion, the (centripetal) acceleration has constant magnitude and is always directed towards the centre of the circle.
Short answer Derive an expression for the velocity and acceleration of a particle executing SHM given by x = A cos(ωt + φ), and state where each is maximum and minimum.Frequently asked3 m
Given x(t) = A cos(ωt + φ).
Velocity: v = dx/dt = −Aω sin(ωt + φ). Its magnitude is maximum, Aω, when sin(ωt + φ) = ±1, i.e. at the mean position (x = 0); it is zero when sin(ωt + φ) = 0, i.e. at the extreme positions (x = ±A).
Acceleration: a = dv/dt = −Aω2cos(ωt + φ) = −ω2x. Its magnitude is maximum, Aω2, at the extreme positions (x = ±A), and zero at the mean position (x = 0).
Short answer A block of mass 1 kg is attached to a spring of spring constant 100 N/m and set into oscillation with amplitude 0.1 m. Find (i) the angular frequency, (ii) the maximum velocity, and (iii) the total energy of oscillation.Frequently asked3 m
m = 1 kg, k = 100 N/m, A = 0.1 m.
(i) ω = √(k/m) = √(100/1) = 10 rad/s.
(ii) vmax = Aω = 0.1 × 10 = 1 m/s.
(iii) E = ½kA2 = ½ × 100 × (0.1)2 = ½ × 100 × 0.01 = 0.5 J.
Short answer Show that for a particle executing SHM, the average kinetic energy over one complete time period equals the average potential energy, and both equal half the total energy.Frequently asked3 m
K(t) = ½mω2A2sin2(ωt + φ) and U(t) = ½mω2A2cos2(ωt + φ). Over one full period, the average value of sin2 and of cos2 is each 1/2 (standard result for a sinusoid over a full cycle).
So ⟨K⟩ = ½mω2A2 × ½ = ¼mω2A2, and similarly ⟨U⟩ = ¼mω2A2. Since E = ½mω2A2, both ⟨K⟩ and ⟨U⟩ equal E/2, and ⟨K⟩ = ⟨U⟩.
Short answer What is a damped oscillation? Write the expression for the displacement of a damped harmonic oscillator and explain the meaning of each term.Frequently asked3 m
A damped oscillation is one whose amplitude continuously decreases with time because a resistive force (e.g. friction or a viscous medium) dissipates its mechanical energy.
Its displacement is x(t) = A e−bt/2m cos(ω′t + φ), where A is the initial amplitude, m the mass of the oscillator, b the damping constant, ω′ = √(k/m − b2/4m2) the (reduced) angular frequency of the damped motion, and φ the phase constant. The factor A e−bt/2m is the exponentially decaying amplitude envelope.
Long answer Derive the expression for the time period of a simple pendulum performing small oscillations, starting from the restoring torque on the bob. State the assumption used and one application of the result.Frequently asked5 m
A bob of mass m hangs from a light, inextensible string of length L, displaced through small angle θ from the vertical. Weight mg resolves into mg cos θ along the string (balanced by tension) and mg sin θ perpendicular to it, directed back towards equilibrium.
Restoring torque about the suspension point: τ = −mgL sin θ. For small θ, sin θ ≈ θ, so τ = −mgLθ.
With I = mL2 and τ = Iα: −mgLθ = mL2α, so α = −(g/L)θ — angular SHM with ω2 = g/L.
Hence T = 2π/ω = 2π√(L/g).
Assumption: θ small enough that sin θ ≈ θ (valid up to about 10°). Application: since T is independent of mass and amplitude, a pendulum of known L is used to measure g experimentally.
Long answer A block of mass m attached to a spring of force constant k executes SHM on a frictionless surface. Derive expressions for its displacement, velocity and acceleration as functions of time, and show that the total mechanical energy remains constant throughout the motion.Frequently asked5 m
Restoring force F = −kx, so by Newton's second law m(d2x/dt2) = −kx, i.e. d2x/dt2 = −ω2x, where ω = √(k/m). Its general solution is x(t) = A cos(ωt + φ).
Velocity: v(t) = dx/dt = −Aω sin(ωt + φ). Acceleration: a(t) = dv/dt = −Aω2cos(ωt + φ) = −ω2x(t).
K = ½mv2 = ½mω2A2sin2(ωt+φ); U = ½kx2 = ½mω2A2cos2(ωt+φ) (using k = mω2).
E = K + U = ½mω2A2[sin2+cos2] = ½mω2A2 = ½kA2, independent of t — total mechanical energy is conserved.
Long answer Explain, with the help of a labelled diagram, how SHM can be represented as the projection of uniform circular motion on a diameter. Using this representation, derive the expressions for the displacement and velocity of the SHM particle.Frequently asked5 m
A reference particle P moves with uniform angular speed ω on a circle of radius A centred at O. At time t its radius vector makes angle (ωt + φ) with the x-axis (φ being the angle at t = 0). Let N be the foot of the perpendicular from P onto the x-axis.
As P moves round the circle, N oscillates along the x-axis between +A and −A, with position x = A cos(ωt + φ) — the SHM equation. So N, the projection of the uniformly rotating P, executes SHM of amplitude A and angular frequency ω.
P's speed is Aω, tangential to the circle. Resolving this along the x-axis gives the velocity of N: v = −Aω sin(ωt + φ), matching v = dx/dt.
Diagram: circle of radius A about O, radius OP at angle (ωt+φ), perpendicular from P to N on the x-axis, oscillation range −A to +A marked.
Case-based A student sets up a spring–mass system on a horizontal frictionless table: a block of mass 0.5 kg is attached to a spring of force constant 200 N/m. The block is pulled 4 cm from its mean position and released from rest at t = 0. Study the setup and answer: (i) What is the angular frequency of oscillation? (ii) Write the equation of displacement x(t) for this motion, taking the direction of initial pull as positive. (iii) What is the maximum speed attained by the block, and where does it occur? (iv) If the same block were instead attached to a spring of force constant 800 N/m with the same amplitude, how would the maximum acceleration compare to the original case?Frequently asked4 m
(i) ω = √(k/m) = √(200/0.5) = 20 rad/s.
(ii) Released from rest at the extreme position, so x(t) = A cos ωt = 0.04 cos 20t m.
(iii) vmax = Aω = 0.04 × 20 = 0.8 m/s, at the mean position.
(iv) amax = A(k/m). Original: 0.04 × 400 = 16 m/s2. With k = 800: 0.04 × 1600 = 64 m/s2, i.e. 4 times larger, since amax ∝ k for fixed A, m.
Case-based A swing in a park is pushed periodically by a child's father. He notices that if he pushes at just the right rate, matching the swing's own natural back-and-forth rhythm, the swing goes higher and higher with very little effort; pushing at a very different rate barely raises the amplitude at all. Answer: (i) Name the phenomenon being described. (ii) State the condition (in terms of frequencies) under which this phenomenon occurs. (iii) Why does even a small periodic push build up a large amplitude under this condition? (iv) Give one other real-life example of the same phenomenon.Frequently asked4 m
(i) Resonance in forced (driven) oscillations.
(ii) It occurs when the driving frequency ωd (rate of pushing) equals, or is very close to, the swing's natural frequency ω0.
(iii) Each push then adds energy in step with the swing's own motion, so pushes reinforce cycle after cycle; with only small damping removing energy, amplitude keeps building up.
(iv) A radio tuned so its circuit's natural frequency matches the broadcast frequency; or soldiers breaking step on a bridge to avoid matching its natural frequency.
Test yourselfPractice quiz
Marks lost every yearCommon mistakes
- Treating every periodic motion as SHM — a motion can be periodic without the restoring force being proportional to displacement (e.g. the sum cos ωt + cos 3ωt is periodic but not SHM).
- Writing T = 2π√(L/g) for large-angle pendulum swings — the formula is valid only for small angular amplitude where sin θ ≈ θ.
- Confusing amplitude decay with energy decay in damped oscillations — amplitude falls as e−bt/2m while energy falls as e−bt/m, twice as fast in the exponent.
- Forgetting that mounting a spring vertically instead of horizontally does not change its time period — only the mean position shifts due to gravity, T stays 2π√(m/k).
- Mixing up phase relationships — displacement and acceleration are 180° out of phase, but displacement and velocity are 90° out of phase, not 180°.
- Assuming resonance changes the frequency of oscillation — resonance only maximises amplitude at ωd = ω0; the system still oscillates at the driving frequency, not a new one.
- Using x = A sin ωt and x = A cos ωt interchangeably without checking the actual initial condition (position and direction of velocity at t = 0) given in the problem.
- Forgetting the negative sign in a = −ω²x or F = −kx, which leads to a wrong direction for the restoring force/acceleration in derivation-based answers.
How to prepareStrategy for this chapter
This chapter rewards precision in derivations more than memorisation. Master the differential-calculus route from x(t) = A cos(ωt + φ) to v(t) and a(t) — nearly every long-answer question in Class 12 boards and JEE/NEET numericals reduce to these three equations plus v = ω√(A²−x²).
Learn the two standard time-period derivations cold: spring–mass system (T = 2π√(m/k)) from F = −kx, and simple pendulum (T = 2π√(L/g)) from the restoring torque and the small-angle approximation. Both are asked as full 5-mark derivations, and the pendulum derivation is a favourite because it also tests moment of inertia and torque from the rotational-motion chapter.
For energy questions, remember the single governing fact: E = ½kA² is constant, and K + U = E at every instant and every displacement — most 'find K given U' or 'find x given K = fraction of E' problems fall out of this one equation. Practice the standard fractions: x = A/2 gives U = E/4, K = 3E/4; x = A/√2 gives K = U = E/2.
Keep damped oscillations and resonance conceptual rather than derivation-heavy at this stage — know the formulas, know what each symbol does to the graph (larger b flattens and broadens the resonance curve), and know two or three real-life examples of each.