Class XI · Chapter 13
Oscillations — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering simple harmonic motion, energy in SHM, and pendulums.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why is the motion of a simple pendulum only approximately, not exactly, simple harmonic?
Answer
The true restoring force is proportional to sinθ, not θ itself. Only for small angles does sinθ ≈ θ (in radians), making the motion approximately SHM. For larger swing angles, this approximation breaks down.
The true restoring force is proportional to sinθ, not θ itself. Only for small angles does sinθ ≈ θ (in radians), making the motion approximately SHM. For larger swing angles, this approximation breaks down.
Derive the expression for the time period of a mass-spring system executing simple harmonic motion.
Answer
Restoring force: F = −kx. By Newton’s second law: ma = −kx → a = −(k/m)x.
Restoring force: F = −kx. By Newton’s second law: ma = −kx → a = −(k/m)x.
Comparing to the defining SHM relation a = −ω²x: ω² = k/m → ω = √(k/m).
Since T = 2π/ω: T = 2π√(m/k).
Q3. (a) Derive expressions for the velocity and acceleration of a particle in SHM, as functions of displacement. (b) A particle executes SHM with amplitude 4 cm and period 2 s. Find its maximum velocity, and its velocity at x = 2 cm.
Answer
(a) From x = A cos(ωt), v = dx/dt = −Aω sin(ωt). Using sin²+cos²=1: v = ±ω√(A²−x²), maximum |v| = Aω at x=0.
Acceleration: a = dv/dt = −Aω²cos(ωt) = −ω²x.
(a) From x = A cos(ωt), v = dx/dt = −Aω sin(ωt). Using sin²+cos²=1: v = ±ω√(A²−x²), maximum |v| = Aω at x=0.
Acceleration: a = dv/dt = −Aω²cos(ωt) = −ω²x.
(b) ω = 2π/T = π rad/s. vmax = Aω = 0.04×π ≈ 0.126 m/s.
At x=0.02m: v = ω√(A²−x²) = π×√(0.0016−0.0004) = π×0.0346 ≈ 0.109 m/s.
2NEET Style (Single-Correct MCQ)
Q4. In SHM, the total mechanical energy is proportional to:
Solution
E = ½kA² = ½mω²A² — total energy scales with A², so doubling the amplitude quadruples the energy.
E = ½kA² = ½mω²A² — total energy scales with A², so doubling the amplitude quadruples the energy.
Q5. At the extreme position in SHM, the kinetic energy of the particle is:
Solution
At the extremes, velocity is momentarily zero (the turning point), so KE = 0 and all energy is potential.
At the extremes, velocity is momentarily zero (the turning point), so KE = 0 and all energy is potential.
Q6. The time period of a simple pendulum is independent of:
Solution
T = 2π√(L/g) — mass doesn’t appear at all, echoing the same mass-independence seen in free fall.
T = 2π√(L/g) — mass doesn’t appear at all, echoing the same mass-independence seen in free fall.
Q7. The phase difference between the displacement and acceleration of a particle in SHM is:
Solution
Since a = −ω²x, acceleration is always exactly opposite in sign to displacement — a phase difference of π.
Since a = −ω²x, acceleration is always exactly opposite in sign to displacement — a phase difference of π.
3JEE Main Style
Q8. A mass of 0.5 kg is attached to a spring of spring constant 200 N/m. Find the frequency of its oscillation.
Solution
ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s.
ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s.
f = ω/2π = 20/6.283 ≈ 3.18 Hz.
Q9. At resonance, the amplitude of a forced oscillation becomes:
Solution
When the driving frequency matches the system’s natural frequency, energy transfer from the driver to the oscillator is most efficient, producing maximum amplitude.
When the driving frequency matches the system’s natural frequency, energy transfer from the driver to the oscillator is most efficient, producing maximum amplitude.
Q10. The length of a simple pendulum is increased by 44%. Find the percentage change in its time period.
Solution
T ∝ √L. New T′ = T√(1.44) = T × 1.2.
T ∝ √L. New T′ = T√(1.44) = T × 1.2.
Percentage increase = 20%.
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