Class 11 Physics Chapter 13 Oscillations – Extra Questions with Answers

Simple harmonic motion is defined by one condition, a=−ω²x, and that relationship underlies nearly every question in this set. Others ask about velocity and acceleration at the mean and extreme positions, and the time period of a spring-mass system.

Last Updated: September 23, 2026

Very Short Answer Questions (1 mark)

Q1. Write the differential condition for SHM.
Ans: a=−ω²x (acceleration proportional to and opposite of displacement).

Q2. What is the formula for the time period of a spring-mass system?
Ans: T=2π√(m/k).

Q3. At the mean position, what is the velocity and acceleration in SHM?
Ans: Velocity is maximum; acceleration is zero.

Q4. At extreme position, what is the velocity in SHM?
Ans: Zero.

Q5. Write the total energy formula for a particle in SHM.
Ans: E=½mω²A².

Short Answer Questions (2–3 marks)

Q6. A simple pendulum has length 1 m. Find its time period (g=9.8 m/s²).
Ans: T=2π√(l/g)=2π√(1/9.8)=2π(0.319)≈2 s.

Q7. A particle in SHM has amplitude 0.1 m and angular frequency 5 rad/s. Find its maximum velocity and maximum acceleration.
Ans: v_max=Aω=0.1(5)=0.5 m/s. a_max=Aω²=0.1(25)=2.5 m/s².

Q8. A mass of 0.5 kg is attached to a spring of spring constant 200 N/m. Find the period of oscillation.
Ans: T=2π√(m/k)=2π√(0.5/200)=2π(0.05)≈0.314 s.

Higher-Order Thinking / Application Questions

Q9. A particle executes SHM with amplitude 4 cm and angular frequency 2 rad/s. Find its displacement, velocity, and acceleration at t such that ωt=π/6 (taking φ=0), and explain the relationship between the three at this instant.
Ans: x=A sin(ωt)=4 sin(30°)=4(0.5)=2 cm. v=Aωcos(ωt)=4(2)cos(30°)=8(0.866)=6.93 cm/s. a=−ω²x=−4(2)=−8 cm/s². At this instant, the particle is partway between the mean and extreme positions; its velocity is still fairly large (not yet zero) while its acceleration (which always points back toward the mean position) is negative, correctly opposing the positive displacement, confirming that acceleration in SHM always acts as a restoring influence proportional to displacement.

Q10. Explain why the kinetic energy and potential energy of a particle in SHM continuously interconvert, and describe how their sum behaves throughout one complete oscillation, using energy conservation reasoning.
Ans: In SHM, as a particle moves from the mean position toward an extreme position, its speed decreases (approaching zero at the extreme) while the restoring force (and hence potential energy stored, ½mω²x²) increases, since displacement from the mean position increases. Conversely, as the particle moves back from an extreme position toward the mean position, potential energy converts back into kinetic energy, with speed (and KE) becoming maximum exactly at the mean position where displacement (and PE) is zero. Because SHM is an idealized frictionless oscillation, no energy is lost to non-conservative forces, so by the law of conservation of energy, the total mechanical energy E=KE+PE=½mω²A² remains constant throughout the motion, even though the individual KE and PE values vary periodically and continuously interconvert into one another.

Quick visual: a worked diagram from the full Solutions page, for reference.

Frequently Asked Questions

A simple pendulum has a length of 1 m, how would you find its time period on Earth?
Using T = 2 pi root(L over g) = 2 pi root(1 over 9.8), which is about 2.01 seconds.

How would you find the maximum velocity of a particle in simple harmonic motion with amplitude 0.1 m and angular frequency 5 rad per second?
Maximum velocity = amplitude times angular frequency = 0.1 times 5 = 0.5 m per second.

Written by Satish

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