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Class 11 · CBSE

Oscillations — NCERT Class 11 Physics

Complete NCERT Solutions for CBSE Class 11 Physics Chapter 14 — all intext and chapter-end exercise questions on SHM equations, spring-mass, pendulum, energy in SHM, damped/forced oscillations answered clearly.

Updated 11 June 202611 min read

How to Use This Page

Work through each NCERT exercise question yourself before reading the solution below. For numericals, follow every step: identify the correct formula, substitute values with proper units, compute, and state the answer with units. Pay attention to angular frequency (ω) versus frequency (ν) versus time period (T) — they are related by ω = 2πν = 2π/T.

Key Concepts (Quick Recap)

  • Displacement in SHM: x = A cos(ωt + φ)
  • Velocity: v = −Aω sin(ωt + φ); |v|_max = Aω
  • Acceleration: a = −Aω² cos(ωt + φ) = −ω²x; |a|_max = Aω²
  • Angular frequency: ω = 2π/T = 2πν
  • Spring-mass system: T = 2π√(m/k), ω = √(k/m)
  • Simple pendulum: T = 2π√(l/g)
  • Total energy: E = ½mω²A² = ½kA² (constant)
  • KE: ½mω²(A² − x²); PE: ½mω²x²
  • Damped oscillations: x = A e^(−bt/2m) cos(ω't + φ), where ω' = √(ω₀² − b²/4m²)
  • Resonance: amplitude is maximum when driving frequency equals natural frequency

Intext Questions

Page 345

Q1. Which of the following are SHM? (a) x = A sin ωt (b) x = A sin² ωt (c) x = A cos ωt + B sin ωt
Answer

(a) Yes — single sinusoid is SHM.

(b) No — sin²ωt = ½(1 − cos 2ωt). This has a constant term (non-zero mean) and the angular frequency is 2ω. Not SHM.

(c) Yes — a linear combination of sin and cos with the same ω is SHM. It can be written as C sin(ωt + δ).

Page 350

Q2. A spring of force constant k is cut into two equal lengths. What is the force constant of each half?
For a spring: k ∝ 1/L (longer spring = smaller k) Original: k, L Half length: k' = 2k (each half is twice as stiff)
Answer

Force constant of each half = 2k

Q3. A simple pendulum has a time period of 2 s. What is its length? g = 9.8 m/s².
T = 2π√(l/g) → T² = 4π²l/g l = gT²/(4π²) = 9.8 × 4 / (4 × 9.87) = 39.2/39.48 = 0.993 m ≈ 1 m
Answer

Length ≈ 1 m

Page 357

Q4. At what displacement is the KE equal to PE in SHM?
KE = PE → ½mω²(A² − x²) = ½mω²x² A² − x² = x² → 2x² = A² → x = A/√2
Answer

At x = A/√2, kinetic energy equals potential energy.

Chapter-End Exercise

Q1. Which of the following examples represent periodic motion? (a) A swimmer completing one return trip. (b) A freely suspended bar magnet displaced from N-S direction. (c) Rotation of Earth about its axis.
Answer

(a) Not periodic (unless repeated) — a single return trip is not necessarily periodic.

(b) Yes — the magnet oscillates about the N-S direction with a definite period (SHM for small displacements).

(c) Yes — Earth rotates with a definite period of 24 hours.

Q2. Which of the following examples represent (nearly) SHM? (a) Motion of Earth about the Sun. (b) Motion of an oscillating mercury column in a U-tube. (c) Motion of a ball in a smooth bowl.
Answer

(a) Not SHM — the orbit is nearly circular, not a restoring force proportional to displacement.

(b) Yes — the mercury column executes SHM (restoring force proportional to displacement from equilibrium).

(c) Yes — for a smooth spherical bowl and small displacements, the ball executes SHM.

Q3. Figure depicts x-t plots for four particles. Which are in SHM?
Answer

SHM plots are sinusoidal (smooth, symmetric sine/cosine curves). Any plot that shows non-sinusoidal, triangular, or asymmetric oscillation is not SHM.

Q4. A particle is in linear SHM with time period 2 s and amplitude 1 cm. What is the displacement at t = 0.5 s, if at t = 0 the displacement is 1 cm?
x = A cos(ωt) (since at t = 0, x = A = 1 cm, φ = 0) ω = 2π/T = 2π/2 = π rad/s x(0.5) = 1 × cos(π × 0.5) = cos(π/2) = 0 cm
Answer

Displacement at t = 0.5 s = 0 cm

Q5. A particle in SHM has displacement 4 cm and velocity 20 cm/s when the phase is π/6. Find the amplitude and period.
x = A cos φ → 4 = A cos(π/6) → A = 4/(√3/2) = 8/√3 = 4.62 cm v = −Aω sin φ → 20 = −(8/√3) × ω × sin(π/6) → 20 = −(8/√3) × ω × 0.5 ω = −20 × 2√3 / 8 = −5√3 (taking magnitude, ω = 5√3 = 8.66 rad/s) T = 2π/ω = 2π/(5√3) = 6.283/8.66 = 0.726 s
Answer

Amplitude ≈ 4.62 cm, Period ≈ 0.73 s

Q6. A spring of constant 50 N/m is hung vertically. A 0.5 kg mass is attached. What is the period of oscillation?
T = 2π√(m/k) = 2π√(0.5/50) = 2π × 0.1 = 0.628 s
Answer

Period = 0.63 s

Q7. The period of oscillation of a mass m attached to a spring is T. If the mass is increased by 2 kg, the period becomes 3T. Find m.
T = 2π√(m/k); 3T = 2π√((m+2)/k) 9T²/T² = (m+2)/m → 9 = (m+2)/m 9m = m + 2 → 8m = 2 → m = 0.25 kg
Answer

m = 0.25 kg

Q8. Two springs of constants k₁ and k₂ are connected (a) in series (b) in parallel. Find the time period of a mass m attached.
Answer

(a) Series: effective k = k₁k₂/(k₁ + k₂)

T = 2π√(m(k₁ + k₂)/(k₁k₂))

(b) Parallel: effective k = k₁ + k₂

T = 2π√(m/(k₁ + k₂))
Q9. A simple pendulum has a time period T on Earth. What is the period on the Moon where g is 1/6th?
T ∝ 1/√g → T_moon = T × √(g_earth/g_moon) = T × √6 ≈ 2.45 T
Answer

Period on Moon = √6 × T ≈ 2.45T

Q10. A simple pendulum of length 1 m has a bob of mass 50 g. The bob is given a charge of 1 μC. A uniform electric field of 10⁵ V/m is applied vertically upward. Find the new time period. g = 10 m/s².
Electric force on bob = qE = 10⁻⁶ × 10⁵ = 0.1 N (upward) Weight = mg = 0.05 × 10 = 0.5 N (downward) Net effective g' = g − qE/m = 10 − 0.1/0.05 = 10 − 2 = 8 m/s² T = 2π√(l/g') = 2π√(1/8) = 2π × 0.354 = 2.22 s
Answer

New time period ≈ 2.22 s

Q11. A pendulum clock gives correct time at 20°C. It loses 5 s per day at 35°C. Find the coefficient of linear expansion of the pendulum material.
ΔT_day = 86400 s Time lost = 5 s Fractional change in T = 5/86400 T ∝ √l → ΔT/T = (1/2)Δl/l = (1/2)αΔT_temp ΔT_temp = 35 − 20 = 15°C 5/86400 = (1/2) × α × 15 α = 2 × 5 / (86400 × 15) = 10/1296000 = 7.72 × 10⁻⁶ °C⁻¹
Answer

α ≈ 7.7 × 10⁻⁶ °C⁻¹

Q12. What is the frequency of oscillation of a mass of 20 kg attached to a spring of constant 80 N/m?
ω = √(k/m) = √(80/20) = √4 = 2 rad/s ν = ω/(2π) = 2/(2π) = 0.318 Hz
Answer

Frequency ≈ 0.32 Hz

Q13. The total energy of a particle in SHM is 3.2 J. What is the kinetic energy when the displacement is half the amplitude?
E = ½mω²A² = 3.2 J At x = A/2: KE = ½mω²(A² − A²/4) = ½mω²(3A²/4) = (3/4) × ½mω²A² KE = (3/4) × 3.2 = 2.4 J
Answer

KE = 2.4 J

Q14. A 2 kg particle is in SHM of amplitude 0.1 m. Its KE is 1 J at x = 0.05 m. Find the total energy and the period.
KE = ½mω²(A² − x²) 1 = ½ × 2 × ω²(0.01 − 0.0025) = ω² × 0.0075 ω² = 1/0.0075 = 133.33 → ω = 11.55 rad/s E = ½mω²A² = ½ × 2 × 133.33 × 0.01 = 1.333 J T = 2π/ω = 2π/11.55 = 0.544 s
Answer

Total energy ≈ 1.33 J, Period ≈ 0.54 s

Q15. What are damped oscillations? Give the expression for displacement.
Answer

Damped oscillations occur when a resistive (damping) force proportional to velocity acts on an oscillating system. The amplitude decreases exponentially with time:

x(t) = A₀ e^(−bt/2m) cos(ω't + φ) where ω' = √(ω₀² − b²/4m²), b is the damping constant

The system oscillates with reduced frequency and diminishing amplitude.

Q16. What is resonance? Give two examples.
Answer

Resonance occurs when the frequency of an external driving force equals the natural frequency of a system, producing maximum amplitude oscillations.

Examples: (a) A swing pushed at its natural frequency goes higher each time. (b) Soldiers break step while crossing a bridge to avoid resonance with the bridge's natural frequency.

Q17. Show that the motion of a simple pendulum is SHM for small displacements.
Answer

For a pendulum of length l displaced by angle θ:

Restoring force = −mg sin θ ≈ −mgθ (for small θ, sin θ ≈ θ) Since arc length s = lθ: F = −mg(s/l) = −(mg/l)s This is of the form F = −ks, where k = mg/l Hence ω = √(g/l) and T = 2π√(l/g) — confirming SHM
Q18. A body of mass 0.1 kg is attached to a spring of constant 40 N/m. The body is pulled 0.1 m from equilibrium and released. Find (a) amplitude (b) period (c) maximum speed (d) maximum acceleration.
(a) A = 0.1 m (b) T = 2π√(m/k) = 2π√(0.1/40) = 2π × 0.05 = 0.314 s (c) v_max = Aω = A√(k/m) = 0.1 × √(40/0.1) = 0.1 × 20 = 2 m/s (d) a_max = Aω² = A(k/m) = 0.1 × 400 = 40 m/s²
Q19. A body oscillates with SHM described by x = 5 cos(2πt + π/4) cm. Find (a) amplitude (b) angular frequency (c) period (d) initial phase (e) displacement at t = 1 s.
(a) A = 5 cm (b) ω = 2π rad/s (c) T = 2π/ω = 1 s (d) φ = π/4 (e) x(1) = 5 cos(2π × 1 + π/4) = 5 cos(π/4) = 5 × 0.707 = 3.54 cm
Q20. The maximum velocity of a particle in SHM is 0.08 m/s and the maximum acceleration is 0.32 m/s². Find the amplitude and period.
v_max = Aω = 0.08; a_max = Aω² = 0.32 ω = a_max/v_max = 0.32/0.08 = 4 rad/s A = v_max/ω = 0.08/4 = 0.02 m = 2 cm T = 2π/ω = 2π/4 = π/2 = 1.57 s
Answer

Amplitude = 2 cm, Period ≈ 1.57 s

Q21. What is the phase difference between displacement and velocity in SHM? Between displacement and acceleration?
Answer
  • Displacement and velocity: Phase difference = π/2 (velocity leads displacement by 90°)
  • Displacement and acceleration: Phase difference = π (acceleration is exactly opposite to displacement)
Q22. The length of a simple pendulum is increased by 1%. What is the percentage change in its time period?
T ∝ √l → ΔT/T = (1/2)Δl/l = (1/2) × 1% = 0.5%
Answer

Time period increases by 0.5%

Q23. Two identical springs of constant k are attached to a mass m — one on each side. Find the time period.
Both springs exert restoring force. If displaced by x: F = −kx − kx = −2kx T = 2π√(m/2k)
Q24. A tunnel is dug through the centre of the Earth. Show that a body dropped into it executes SHM. Find the time period.
At depth d from surface, distance from centre = r = R − d g' = g(R − d)/R = gr/R Force on body of mass m: F = −mg' = −mg r/R = −(mg/R)r This is SHM with k = mg/R T = 2π√(m/k) = 2π√(R/g) = 2π√(6400000/9.8) ≈ 5070 s ≈ 84.5 min
Answer

Time period ≈ 84.5 minutes

Q25. Define forced oscillations and resonance.
Answer

Forced oscillations: A system oscillates under the influence of an external periodic force (in addition to the restoring force). The system eventually oscillates at the frequency of the driving force, not its natural frequency.

Resonance: When the driving frequency equals the natural frequency of the system, the amplitude of oscillation becomes very large (maximum). This condition of maximum response is resonance.

Internal Links

  • [Chapter 14 Guide — Oscillations](/school/cbse/class/11/physics/cbse-oscillations)
  • [All Class 11 Physics Solutions](/school/cbse/class/11/physics/solutions)
  • [Class 11 Physics Subject Guide](/school/cbse/class/11/physics)

Frequently Asked Questions

SHM is oscillatory motion where the restoring force is directly proportional to the displacement from equilibrium and directed opposite to it: F = −kx. The acceleration is a = −ω²x, and displacement follows x = A cos(ωt + φ).
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