📌 Snapshot
- Establishes the language of periodic and oscillatory motion — period T, frequency ν, displacement, amplitude, phase — and singles out Simple Harmonic Motion (SHM) as the simplest oscillation, defined by a sinusoidal displacement x(t) = A cos(ωt + φ).
- Derives the kinematics of SHM (velocity v = −Aω sin(ωt+φ), acceleration a = −ω²x) by treating SHM as the projection of uniform circular motion on a diameter — a powerful geometric trick used throughout CUET problems.
- Builds the dynamics — restoring force F = −kx with ω = √(k/m) — and the energy picture: KE = ½ mω²(A²−x²), PE = ½ kx², total E = ½ kA² = constant.
- Applies SHM to two canonical systems: the spring–mass oscillator (T = 2π√(m/k)) and the simple pendulum at small angular displacement (T = 2π√(L/g)).
- CUET draws heavily here for formula-substitution numericals (T from m, k or L, g; vmax; KE/PE ratios) and for conceptual sieves (periodic vs SHM, phase of v and a, energy at extremes vs mean).
📖 Detailed Notes
2.1 Core concepts
- Motion that repeats itself at regular intervals of time is called periodic motion; to-and-fro motion about a mean (equilibrium) position is called oscillatory motion. Every oscillatory motion is periodic, but every periodic motion need not be oscillatory — uniform circular motion is periodic but not oscillatory (NCERT §13.1–13.2, p. 259–260).
- At the equilibrium position no net external force acts on the body; a small displacement brings a restoring force into play, producing oscillations or vibrations. The terms oscillation and vibration are essentially synonymous; oscillation is used at low frequency (tree branch), vibration at high frequency (musical string) (NCERT §13.2, p. 260).
- The period T is the smallest time interval after which the motion repeats; its SI unit is the second. The frequency ν = 1/T is measured in hertz (1 Hz = 1 oscillation per second = 1 s⁻¹); ν need not be an integer (NCERT §13.2.1, Eqs. 13.1–13.2, p. 260–261).
- In this chapter displacement is generalised: it refers to the change with time of any physical property (position of a block on a spring, angle of a pendulum from the vertical, voltage across a capacitor, pressure in a sound wave). Displacement can be positive or negative (NCERT §13.2.2, p. 261).
- A periodic displacement is conveniently written as f(t) = A cos ωt (or A sin ωt). The period is T = 2π/ω; any linear combination A sin ωt + B cos ωt is also periodic with the same T. By Fourier's theorem any periodic function can be written as a superposition of sines and cosines (NCERT §13.2.2, Eqs. 13.3a–13.3d, p. 261–262).
- Simple Harmonic Motion (SHM) is the oscillation in which the displacement is a sinusoidal function of time: x(t) = A cos(ωt + φ), where A is the amplitude, ω the angular frequency, (ωt+φ) the phase, and φ the phase constant. SHM is not just any periodic motion — it is the sinusoidal one (NCERT §13.3, Eq. 13.4, p. 262).
- The amplitude A is the magnitude of maximum displacement and is taken positive without loss of generality; the cosine varies between +1 and −1, so x varies between +A and −A. The phase (ωt + φ) determines the instantaneous state of motion; at t = 0 the phase equals the phase constant φ (NCERT §13.3, p. 263).
- ω is related to T by ω = 2π/T (Eq. 13.7), and to frequency by ω = 2πν. ω is called the angular frequency and has SI units of rad s⁻¹ (NCERT §13.3, Eqs. 13.5–13.7, p. 264).
- SHM ↔ uniform circular motion: the projection of a particle moving uniformly on a circle of radius A with angular speed ω on a diameter of that circle executes SHM with amplitude A and angular frequency ω. The circle is called the reference circle and the moving point the reference particle (NCERT §13.4, p. 264–265).
- Velocity in SHM: v(t) = −Aω sin(ωt + φ), obtained either by projecting the tangential velocity v = ωA of the reference particle or by differentiating x(t). Hence vmax = Aω, occurring at x = 0 (mean position), and v = 0 at x = ±A. Eliminating time gives v = ±ω√(A² − x²) (NCERT §13.5, Eqs. 13.8–13.10, p. 266).
- Acceleration in SHM: a(t) = −ω²A cos(ωt + φ) = −ω² x(t), obtained from the centripetal acceleration ω²A of the reference particle, or by differentiating v(t). Thus acceleration is proportional to displacement and always directed towards the mean position; |a|max = ω²A at the extremes, a = 0 at x = 0 (NCERT §13.5, Eqs. 13.11–13.12, p. 266–267).
- Phase relations: with respect to displacement, velocity has a phase lead of π/2 and acceleration has a phase difference of π (i.e., is in anti-phase). All three quantities have the same period T (NCERT §13.5, Fig. 13.13, p. 267).
- Force law for SHM: by Newton's second law F = ma = −mω² x = −kx, where k = mω² (so ω = √(k/m)). The restoring force is linearly proportional to displacement and directed towards the mean position; the system is a linear harmonic oscillator. Forces with extra x², x³ terms give non-linear oscillators (NCERT §13.6, Eqs. 13.13–13.14, p. 267).
- Energy in SHM: kinetic energy K = ½ mv² = ½ mω²A² sin²(ωt+φ) = ½ k A² sin²(ωt+φ); potential energy U = ½ k x² = ½ k A² cos²(ωt+φ). Both are periodic with period T/2 (NCERT §13.7, Eqs. 13.15–13.17, p. 268).
- The total mechanical energy of the harmonic oscillator is E = K + U = ½ k A² = ½ mω²A² — constant in time, independent of where the particle is. At x = 0 the energy is entirely kinetic; at x = ±A it is entirely potential; in between K and U exchange (NCERT §13.7, Eq. 13.18 and Fig. 13.16, p. 269).
- Spring–mass oscillator: F = −kx gives ω = √(k/m) and T = 2π√(m/k) (NCERT §13.6 and Summary point 8, Eqs. 13.13–13.14, p. 267 and p. 272).
- Simple pendulum: for a bob of mass m on an inextensible massless string of length L, the tangential restoring torque is τ = −L (mg sinθ). Newton's law of rotation gives α = −(mgL/I) sinθ. For small θ, sinθ ≈ θ (Table 13.1 shows sinθ ≈ θ even up to ~20°), so α = −(mgL/I) θ — SHM in angular displacement (NCERT §13.8, Eqs. 13.19–13.24, p. 270–271).
- With I = mL², the angular frequency is ω = √(g/L) and the period is T = 2π√(L/g), independent of mass and amplitude (NCERT §13.8, Eqs. 13.25–13.26, p. 271).
- Two-spring effective stiffness (NCERT Example 13.6, p. 267–268): when a mass is connected to two identical springs of constant k on opposite sides, both contribute restoring force, so the effective k is 2k and T = 2π√(m/2k). The same logic gives k_series = k₁k₂/(k₁ + k₂) and k_parallel = k₁ + k₂ in standard combinations encountered in NCERT exercises.
- Energy interconversion is continuous in SHM: at any displacement x, K = ½ mω²(A² − x²) and U = ½ mω² x²; sum K + U = ½ mω² A² is conserved. Plots of K and U against time are sin² and cos² curves of period T/2, mirror images of each other around the half-energy line (NCERT §13.7, Fig. 13.16, p. 268–269).
- SHM as projection of uniform circular motion explains why the kinematic equations of SHM have the same form as the components of uniform circular motion: x = A cos θ, vₓ = −Aω sin θ, aₓ = −Aω² cos θ, with θ = ωt + φ — this geometric mapping is what NCERT exploits to derive every SHM relation without calculus (NCERT §13.4, p. 264–266).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Periodic motion | Motion that repeats itself at regular intervals of time | 260 |
| Oscillatory motion | To-and-fro motion about a mean (equilibrium) position | 259–260 |
| Period (T) | The smallest interval of time after which the motion is repeated; SI unit second | 260 |
| Frequency (ν) | Number of repetitions per unit time, ν = 1/T; SI unit hertz (Hz) = 1 s⁻¹ | 261 |
| Displacement | Change with time of any physical property under consideration (position, angle, voltage, pressure, etc.) | 261 |
| Amplitude (A) | Magnitude of the maximum displacement of the particle; taken positive | 263 |
| Phase | The time-dependent quantity (ωt + φ) in x(t) = A cos(ωt+φ), determining the state of motion at time t | 263 |
| Phase constant (φ) | Value of phase at t = 0; the phase angle | 263 |
| Angular frequency (ω) | ω = 2π/T = 2πν; SI unit rad s⁻¹ | 264 |
| Simple harmonic motion (SHM) | Oscillation in which displacement is a sinusoidal function of time: x(t) = A cos(ωt+φ) | 262 |
| Restoring force (in SHM) | Force F = −kx, linearly proportional to displacement and directed towards the mean position | 267 |
| Linear harmonic oscillator | System obeying F = −kx exactly; if extra x², x³ terms appear it becomes non-linear | 267 |
| Reference circle / reference particle | The circle (and uniformly revolving point on it) whose projection on a diameter gives the SHM | 265 |
| Equilibrium (mean) position | Point of zero net force where the displacement x is measured from; in SHM, x = 0 here | 260 |
| Extreme position | Points x = ±A where v = 0 and | a |
| Maximum speed | v_max = Aω (at the mean position) | 266 |
| Maximum acceleration | a | |
| Effective spring constant (series) | k_s = k₁k₂/(k₁+k₂) | 267 (implied) |
| Effective spring constant (parallel) | k_p = k₁ + k₂ | 267 (implied) |
| Period of spring–mass system | T = 2π√(m/k) | 267 |
| Period of simple pendulum | T = 2π√(L/g), small θ | 271 |
| Small-angle approximation | sin θ ≈ θ (rad), valid up to ~20° | 271 |
2.3 Diagrams / processes to remember
- Fig. 13.1 (p. 260) — three examples of periodic motion: insect climbing a ramp; child climbing a step; a ball bouncing (parabolic arcs).
- Fig. 13.2 (p. 261) — (a) block on a spring, the prototype linear SHM system; (b) simple pendulum with angular displacement θ.
- Fig. 13.3 / 13.4 / 13.5 (p. 262–263) — particle vibrating between +A and −A; snapshots at t = 0, T/4, T/2, 3T/4, T; continuous x-vs-t cosine curve.
- Fig. 13.6 (p. 263) — table mapping the symbols A, ω, ωt+φ, φ to their meanings in x(t) = A cos(ωt+φ).
- Fig. 13.7 (p. 263) — two SHMs with same ω, φ but different A (a); same A, ω but different φ (b).
- Fig. 13.8 (p. 264) — two SHMs with same A, φ but different periods T (and hence different ω).
- Fig. 13.9 / 13.10 (p. 265) — ball on a string in horizontal circular motion; its projection on a wall is SHM. Reference-circle construction.
- Fig. 13.11 / 13.12 (p. 266) — projecting the tangential velocity v = ωA of reference particle gives v(t) = −ωA sin(ωt+φ); projecting the centripetal acceleration ω²A gives a(t) = −ω²A cos(ωt+φ) = −ω² x(t).
- Fig. 13.13 (p. 267) — stacked plots of x(t), v(t), a(t): same period T but velocity leads displacement by π/2, acceleration is anti-phase to displacement.
- Fig. 13.14 / 13.15 (p. 268) — Example 13.6: a mass between two identical springs k each gives effective force −2kx and T = 2π√(m/2k).
- Fig. 13.16 (p. 269) — K, U, and total E plotted vs time and vs displacement: K and U both repeat with period T/2; total E = ½kA² is a horizontal line.
- Fig. 13.17 (p. 270) — simple pendulum free-body diagram: T along string; mg cosθ (radial) and mg sinθ (tangential, restoring).
- Table 13.1 (p. 271) — values of θ (degrees, radians) vs sinθ, showing sinθ ≈ θ to good accuracy up to ~20°.
2.4 Common confusions / NTA trap points
- Periodic vs SHM. Every SHM is periodic, but the reverse is false. Earth's rotation is periodic, not SHM. NTA distractors will offer a periodic non-sinusoidal motion (e.g., a bouncing ball, planetary orbit) as "SHM" (NCERT §13.2 and Points to Ponder #2, p. 273).
- Phase of v and a. Students often write "velocity is in phase with displacement". It isn't — v leads x by π/2 and a is in anti-phase (phase difference π), not π/2 (NCERT §13.5, Fig. 13.13, p. 267).
- vmax and amax locations. vmax = Aω occurs at the mean position (x = 0), not at the extremes; amax = ω²A occurs at the extremes (x = ±A), not at the mean. NTA loves swapping these (NCERT §13.5, p. 266–267).
- KE/PE at mean and extreme. At x = 0 energy is entirely kinetic; at x = ±A entirely potential. KE and PE individually have period T/2, but total E is constant — not periodic in any non-trivial sense (NCERT §13.7, Fig. 13.16, p. 268–269).
- Pendulum period independent of mass and amplitude. T = 2π√(L/g) — students sometimes write it as depending on mass m or on initial amplitude θ₀. Both are wrong for small oscillations (NCERT §13.8 and Points to Ponder #7, Eq. 13.26, p. 271 and p. 273).
- Two-spring trap (Example 13.6). With identical springs k on either side of a mass, both pull it back: F = −2kx, so ω = √(2k/m), T = 2π√(m/2k) — not T = 2π√(m/k). Students forget to add the spring constants (NCERT §13.6, Example 13.6, p. 267–268).
- Small-angle approximation for pendulum. SHM holds only for small θ where sinθ ≈ θ. For large amplitudes the pendulum is still periodic, but not simple harmonic (NCERT §13.8 and Points to Ponder #8, p. 271, 273).
- Confusing T (period) with t (time). T is the smallest interval for one full repetition; t is the running time. The phase ωt + φ uses the running t, not T.
- Energy quadratic in amplitude. Doubling A multiplies E by 4 (not 2), because E ∝ A². Many students miss the square.
- Pendulum on the Moon. Since T ∝ 1/√g and g_Moon ≈ g_Earth/6, the same pendulum has period √6 ≈ 2.45 times longer on the Moon — NTA loves this swap.
- k for a vertically hanging spring. The natural length shifts to a new equilibrium under gravity; the SHM about this new equilibrium still has T = 2π√(m/k) — gravity does not enter the period.
2.5 Key formulas table
| Quantity | Symbol / Formula | NCERT reference |
|---|---|---|
| Period and frequency | T = 1/ν; ν = 1/T | §13.2.1, Eqs. 13.1–13.2, p. 261 |
| Angular frequency | ω = 2π/T = 2πν | §13.3, Eqs. 13.5–13.6, p. 264 |
| SHM displacement | x(t) = A cos(ωt + φ) | §13.3, Eq. 13.4, p. 262 |
| SHM velocity | v(t) = −Aω sin(ωt + φ) | §13.5, Eq. 13.9, p. 266 |
| SHM acceleration | a(t) = −ω² x(t) | §13.5, Eq. 13.11, p. 266 |
| Maximum speed | v_max = Aω at x = 0 | §13.5, p. 266 |
| Maximum acceleration | a | |
| Velocity–displacement | v = ±ω √(A² − x²) | §13.5, Eq. 13.10, p. 266 |
| Restoring force | F = −k x; k = m ω² | §13.6, Eq. 13.13, p. 267 |
| Period (spring–mass) | T = 2π √(m/k) | §13.6, Eq. 13.14, p. 267 |
| Period (two equal springs both sides) | T = 2π √(m/2k) | Ex. 13.6, p. 268 |
| Kinetic energy in SHM | K = ½ mω²(A² − x²) = ½ k(A² − x²) | §13.7, Eq. 13.15, p. 268 |
| Potential energy in SHM | U = ½ k x² | §13.7, Eq. 13.16, p. 268 |
| Total energy | E = ½ k A² = ½ m ω² A² | §13.7, Eq. 13.18, p. 269 |
| Period of K and U | T/2 | §13.7, p. 268 |
| Period (simple pendulum) | T = 2π √(L/g) | §13.8, Eq. 13.26, p. 271 |
| ω (simple pendulum) | ω = √(g/L) | §13.8, p. 271 |
| Phase lead of v over x | π/2 | §13.5, p. 267 |
| Phase of a relative to x | π (anti-phase) | §13.5, p. 267 |
| Small-angle validity | sin θ ≈ θ for θ < ~0.35 rad (~20°) | Table 13.1, p. 271 |
🎯 Practice MCQs
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Q1. Which one of the following statements is correct?
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Answer: B
Circular motion is a periodic but non-oscillatory example. (A) is false (only sinusoidal periodic motion is SHM), (C) reverses the truth, (D) is meaningless — SHM is independent of amplitude.
Q2. The displacement of a particle executing SHM is given by x(t) = A cos(ωt + φ). The quantity (ωt + φ) is called the
▸ Show answer & explanation
Answer: C
The textbook explicitly names (ωt + φ) the phase; its value at t = 0, namely φ alone, is the phase constant. A and B are unrelated to that argument.
Q3. A particle in SHM has amplitude A and angular frequency ω. Its maximum speed is
▸ Show answer & explanation
Answer: B
Since v(t) = −Aω sin(ωt+φ), the maximum value of |v| is Aω, occurring when sin(…)= ±1, i.e., at the mean position. Aω² is the maximum acceleration, not speed.
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Q4. In SHM the acceleration of the particle at displacement x is
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Answer: B
Equation 13.11 gives a(t) = −ω² x(t), so the acceleration is always directed towards the centre. (D) is wrong because at the extremes |a| is maximum, not zero.
Q5. The total mechanical energy of a particle executing SHM of amplitude A under a spring constant k is
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Answer: B
Equation 13.18 states E = ½ k A², independent of time. At the mean position the energy is all kinetic (½ kA²), not zero — so (D) is a typical NTA trap.
Q6. A spring of force constant k = 200 N m⁻¹ is attached to a block of mass m = 2 kg on a frictionless surface. The period of oscillation is
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Answer: A
T = 2π√(m/k) = 2π√(2/200) = 2π × (1/10) = π/5 s. Distractor (B) is the answer one gets by forgetting the square root.
Q7. The length of a simple pendulum which "ticks seconds" (i.e., period T = 2 s) at a place where g = 9.8 m s⁻² is approximately
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Answer: C
Worked out in Example 13.8 — L = gT²/(4π²) = (9.8 × 4)/(4π²) ≈ 1 m. (D) would arise from forgetting the 4π² factor.
Q8. A particle of mass 1 kg attached to a spring of force constant 50 N m⁻¹ is pulled to x = 10 cm from the mean position and released from rest. When the particle is at x = 5 cm, the ratio KE : PE is
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Answer: B
Total E = ½ k A² = ½ × 50 × (0.1)² = 0.25 J. PE = ½ k x² = ½ × 50 × (0.05)² = 0.0625 J, so KE = E − PE = 0.1875 J. Ratio KE : PE = 0.1875 : 0.0625 = 3 : 1. Example 13.7 calculates these values directly.
Q9. With respect to displacement, the velocity in SHM has a phase difference of ___ and the acceleration has a phase difference of ___ .
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Answer: C
v leads x by a quarter cycle, while a is in anti-phase with x.
Q10. Two identical springs of spring constant k are attached on either side of a block of mass m on a frictionless surface (Fig. 13.14). The period of small oscillations of the block is
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Answer: B
Both springs push/pull towards the mean position, giving F = −2kx; effective spring constant 2k, hence T = 2π√(m/2k). (A) is the single-spring answer, the classic trap.
Q11. **Assertion (A):** In SHM the kinetic energy and potential energy each have period T/2, where T is the period of the displacement. **Reason (R):** Both KE and PE depend on the *square* of sinusoidal functions of time, so they repeat twice in every cycle of x(t).
▸ Show answer & explanation
Answer: A
KE = ½ k A² sin²(ωt+φ) and PE = ½ k A² cos²(ωt+φ); using sin² and cos² (both squared) reduces their period from T to T/2. Reason correctly explains the assertion.
Q12. Match the location of a particle in SHM (List I) with its dynamical quantity (List II): | List I (Location) | List II (Quantity) | |---|---| | (P) Mean position (x = 0) | (1) Speed is maximum | | (Q) Extreme position (x = ±A) | (2) Acceleration is maximum in magnitude | | (R) Mean position (x = 0) | (3) Total energy is entirely kinetic | | (S) Extreme position (x = ±A) | (4) Total energy is entirely potential |
▸ Show answer & explanation
Answer: A
At the mean position v = Aω (maximum) and PE = 0, so E = K. At the extremes v = 0, so K = 0 and E = U = ½kA²; also |a| = ω²A is maximum there. Hence the one-to-one mapping P-1, Q-2, R-3, S-4.
Q13. A simple pendulum of period 2.0 s on the Earth is taken to the Moon where g = g_E/6. Its new period is approximately
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Answer: C
T_Moon = T_Earth × √(g_E/g_Moon) = 2.0 × √6 ≈ 4.9 s. The pendulum runs slower because g is weaker.
Q14. A particle in SHM has amplitude 5 cm and angular frequency 10 rad s⁻¹. Its maximum acceleration is
▸ Show answer & explanation
Answer: B
|a|_max = Aω² = 0.05 × 100 = 5 m s⁻². Distractor (C) drops the conversion of cm to m.
Q15. Assertion (A): The total energy of a simple harmonic oscillator is proportional to the square of its amplitude. Reason (R): At any instant the sum K + U equals ½ k A², independent of x.
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Answer: A
E = ½ k A² follows from adding K = ½ k (A² − x²) and U = ½ k x². E ∝ A² is the direct consequence, and R is the exact reason.
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