Forced oscillation & resonance: Steady-state amplitude is maximum at ω_d = ω₀ (less damping ⇒ taller, narrower peak).
Phase relations: v leads x by π/2; a is opposite to x (phase π).
NCERT Exercises — Worked Solutions
Exercise 13.1: Identify periodic motions
Which of the following examples represent periodic motion: (a) a swimmer completing one trip from one bank 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.
(a) Not periodic for a single round trip; would be periodic if repeated indefinitely.
(b) Periodic and SHM for small displacement (magnet oscillates about geographic field direction).
(c) Periodic (rotation), not SHM (no restoring force toward an equilibrium angle).
(d) Not periodic — straight-line projectile motion that ends at the target.
Exercise 13.2: Periodic vs SHM
Which of the following is/are SHM? (a) the rotation of Earth about its own 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 slightly off the bottom (d) general vibration of a polyatomic molecule about its equilibrium position.
(a) Periodic, not SHM (uniform rotation, no restoring force).
(b) SHM (restoring pressure ∝ displacement of column).
(c) SHM for small oscillations near the bottom (parabolic potential).
(d) Periodic combination of multiple SHMs (normal modes), generally not pure SHM but can be decomposed.
Exercise 13.3: Reading SHM from graph
The displacement of a particle is x(t) = 0.05 cos(2πt) m. Find period, frequency, amplitude, maximum speed and maximum acceleration.
A = 0.05 m, ω = 2π rad/s ⇒ T = 1.0 s, ν = 1.0 Hz.
v_max = Aω = 0.05 × 2π = 0.314 m/s; a_max = Aω² = 0.05 × 4π² = 1.97 m/s².
Exercise 13.4: From v(x) to ω, A
An SHM has v = 0.30 m/s at x = 0.04 m, and v = 0.16 m/s at x = 0.05 m. Find ω and A.
A 5 kg block is attached to a horizontal spring with k = 1200 N/m. The spring is compressed by 2.0 cm and released. Find ω, T, ν, total energy, max speed.
A particle in SHM has T = 4 s, A = 0.10 m. At t = 0, x = +0.05 m and moving in the +x direction. Write x(t).
ω = 2π/T = π/2 rad/s. x = A cos(ωt + φ). At t = 0: 0.05 = 0.10 cos φ ⇒ cos φ = 0.5 ⇒ φ = ±π/3. Velocity = −Aω sin φ; at t = 0 velocity is positive, so sin φ < 0 ⇒ φ = −π/3.
x(t) = 0.10 cos(πt/2 − π/3) m.
Exercise 13.9: Damping in a wristwatch
A simple pendulum-style oscillator has m = 0.20 kg, k = 90 N/m, b = 0.040 kg/s. Find (a) ω₀, (b) the time for the amplitude to halve.
Show that the potential energy of an SHM averaged over a complete period equals half the total energy.
U(t) = ½kA² cos²(ωt + φ). Time-average of cos² over a full period = ½. So ⟨U⟩ = ½kA² × ½ = ¼kA². Total energy E = ½kA². Hence ⟨U⟩ = E/2. By symmetry ⟨K⟩ = E/2 too — virial theorem.
Interactive: Universal SHM Solver L3 Apply
Choose oscillator type and parameters; the applet returns ω, T, ν, v_max, a_max and total energy.
0.50
50
0.05
Self-Assessment Activity — Build Your Own SHML6 Create
Challenge: Design an experiment using only household items to measure the local gravitational acceleration g to within 2%.
Use a string + small bob as a pendulum of measured length L (use a metre ruler).
Time 50 oscillations using a stopwatch (mobile phones suffice).
Compute T = (total time)/50 and use T = 2π√(L/g) → g = 4π² L / T².
Repeat with different L; estimate uncertainty.
Typical numbers: L = 1.000 m, 50T = 100.5 s ⇒ T = 2.010 s ⇒ g = 4π² × 1.000 / 4.040 = 9.77 m/s² (within 0.4 % of 9.81). Pendulum was used by Picard (1671) to define the metre as the length of a seconds-pendulum at Paris — historic accuracy. Sources of error: finite swing amplitude (correction ≈ θ_0²/16), air drag, string mass.
Competency-Based Questions
A 0.5 kg block on a horizontal frictionless surface is attached to a spring (k = 200 N/m). It oscillates with amplitude A = 0.1 m. Take g = 9.8 m/s² where needed.
Q1. L1 Remember Define amplitude, period, frequency and angular frequency of an SHM.
Amplitude A = maximum magnitude of displacement. Period T = smallest time after which the motion repeats. Frequency ν = 1/T (Hz). Angular frequency ω = 2πν = 2π/T (rad/s).
Q2. L2 Understand Why does a bouncing ball not represent SHM even though it is roughly periodic?
In SHM the restoring force is linear in displacement and acts continuously. A bouncing ball experiences gravity (constant downward force, not linear in position) plus brief impulsive contacts with the floor. Neither force is of the form −kx, so motion is not SHM — and energy is lost on each bounce.
Q3. L3 Apply Find ω, T, total E and v_max of the spring-block system.
Q4. L4 Analyse The spring is now mounted vertically and the block hangs from it. Does the period change? What about the equilibrium position?
Period is unchanged: gravity merely shifts the equilibrium downward by mg/k (= 0.5 × 9.8 / 200 = 0.0245 m), but the restoring force about that new equilibrium is still −k(displacement). The vertical SHM has T = 2π√(m/k) = 0.314 s, identical to the horizontal case.
Q5. L5 Evaluate Why is a steel guitar string a better-quality oscillator (sharper resonance) than a string of soft rubber?
Steel has very low internal damping (b small), so the resonance Q-factor (= ω₀ × τ = m ω₀ / b) is very high — sharp peak in amplitude vs ω_d, long ring-time, pure tone. Rubber has high internal friction; b is large, so oscillations decay quickly, the resonance peak is broad, and the tone is muffled. This is why violins, pianos, guitars use steel/nylon strings, not rubber.
Assertion-Reason Questions
Assertion (A): A simple pendulum kept inside an artificial satellite orbiting Earth does not oscillate.
Reason (R): Inside the orbiting satellite, gravity is exactly balanced by centrifugal force, giving an effective g = 0.
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is NOT the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.
Answer: A. In a freely-falling/orbiting frame, effective gravity vanishes, T → ∞: no restoring force, no oscillation.
Assertion (A): Two SHMs of equal amplitude and frequency may have the same energy regardless of phase.
Reason (R): Total energy of an SHM depends only on m, ω, and A — not on phase.
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is NOT the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.
Answer: A. E = ½ m ω² A². Phase only shifts where on the cycle the motion starts; it does not affect the total energy.
Assertion (A): Resonance can cause a wine glass to shatter when a singer holds the right note.
Reason (R): When the driving sound frequency matches the glass's natural frequency, the glass's vibration amplitude grows until material failure.
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is NOT the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.
Answer: A. Tested experimentally — narrow resonance peak, low damping in the glass, sufficient sound pressure ⇒ amplitude exceeds glass's elastic limit, and the glass shatters.
Did You Know?
Frequently Asked Questions - NCERT Exercises and Solutions: Oscillations
What are the key NCERT exercise types in Chapter 13 Oscillations?
NCERT Class 11 Physics Chapter 13 Oscillations exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Oscillations?
For numerical problems in NCERT Class 11 Physics Chapter 13 Oscillations: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 13?
From NCERT Class 11 Physics Chapter 13 (Oscillations), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 13 Oscillations problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 13 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 13 Oscillations exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 13 Oscillations solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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