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Oscillations – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 13. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Oscillations are repeated to-and-fro motions. Simple harmonic motion (SHM) has a restoring force proportional to displacement and is the model for springs and pendulums.

SHM

x = A sin(ωt + φ); restoring force F = −kx; a = −ω²x.
v = ω√(A² − x²); v_max = Aω at the mean position; a_max = Aω² at the extremes.
Total energy = ½kA² (constant); KE and PE interchange.

Systems

Spring–mass: T = 2π√(m/k).
Simple pendulum: T = 2π√(L/g), independent of mass (small amplitude).
Damped oscillations lose energy; forced oscillations show resonance when driving frequency = natural frequency.

Key terms

SHM: periodic motion with acceleration proportional to displacement and directed towards the mean position
Resonance: large amplitude when driving frequency equals natural frequency
Phase: the quantity (ωt + φ) that tells the state of motion (position and direction) of an oscillating particle at time t

Formulas

T = 2π√(m/k)
T = 2π√(L/g)
E = ½kA²

Important questions with answers

Why does a pendulum clock lose time in summer?2 marks

The pendulum rod expands in heat, increasing its length. Since T = 2π√(L/g), the period increases, the clock ticks slower and loses time.

Show that for a particle in SHM, the total energy is constant and equals ½kA².3 marks

x = A sin ωt, v = Aω cos ωt. KE = ½mA²ω² cos²ωt and PE = ½kx² = ½mω²A² sin²ωt (k = mω²). Total = ½mω²A²(sin² + cos²) = ½kA², constant.

Derive the time period of a simple pendulum for small oscillations. Find the length of a seconds pendulum (period 2 s) on Earth (g = 9.8 m/s²) and its period on the Moon (g = g_E/6).5 marks

The bob of mass m on a string of length L displaced by angle θ feels a restoring force −mg sin θ ≈ −mgθ = −mg(x/L) for small θ. So a = −(g/L)x, which is SHM with ω² = g/L and T = 2π√(L/g). Seconds pendulum: L = gT²/4π² = 9.8 × 4/39.48 ≈ 0.993 m (about 1 m). On the Moon, T = 2π√(L/(g/6)) = √6 × 2 ≈ 4.9 s.

Quick revision

Speed is maximum at the mean position.
Acceleration is maximum at the extremes.
Pendulum period is independent of mass.

Exam tips

Pendulum length 4× ⇒ period 2×.
In SHM, x, v and a are related: v leads x by π/2 and a is opposite to x (a = −ω²x).

Practice MCQs

In SHM, acceleration is proportional to
1. velocity
2. displacement
3. time
4. mass
Show answer
Answer: (b) displacement. a = −ω²x.
Period of a simple pendulum depends on
1. length and g
2. mass
3. amplitude only
4. colour
Show answer
Answer: (a) length and g. 2π√(L/g).
If the length of a pendulum becomes 4 times, its period becomes
1. 4 times
2. 2 times
3. ½
4. 16 times
Show answer
Answer: (b) 2 times. T ∝ √L.
Speed of a particle in SHM is maximum at
1. mean position
2. extreme position
3. halfway
4. everywhere same
Show answer
Answer: (a) mean position. All KE.
A particle in SHM has amplitude 5 cm and ω = 4 rad/s. Its maximum speed is
1. 1.25 cm/s
2. 20 cm/s
3. 80 cm/s
4. 9 cm/s
Show answer
Answer: (b) 20 cm/s. Aω.
A 1 kg mass on a spring with k = 100 N/m has period
1. π s
2. 2π s
3. 10 s
4. 2π/10 s
Show answer
Answer: (d) 2π/10 s. 2π√(1/100).
Total energy of a SHM with amplitude A is proportional to
1. A
2. √A
3. 1/A
4. A²
Show answer
Answer: (d) A². ½kA².
Two springs of constant k each are joined in series and a mass m is hung. The time period becomes
1. 2π√(2m/k)
2. 2π√(m/k)
3. 2π√(m/2k)
4. π√(m/k)
Show answer
Answer: (a) 2π√(2m/k). k_series = k/2.

All chapters of Class 11 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.