Skip to content
LoginStart free

Class XI · Physics

Oscillations and Waves: NEET MCQ practice

380 practice questions on Oscillations and Waves, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

380

Practice questions

157

Easy

212

Medium

11

Hard

Get it on Google PlayPractise on the web →

What Oscillations and Waves covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Periodic motion- time period, frequency, displacement as a function of time
  • Simple harmonic motion, phase, oscillations of a spring, kinetic/potential energy in SHM
  • Simple pendulum- derivation of expression for time period
  • Wave motion, longitudinal and transverse waves, speed of travelling wave
  • Principle of superposition of waves, reflection of waves
  • Standing waves in strings and organ pipes, fundamental mode and harmonics, beats

Sample questions on Oscillations and Waves

Try each one, then open the answer and explanation.

Question 1 · Practice question

The displacement of a progressive wave is represented by y = A sin(ωt - kx) where x is distance and t is time. The dimensions of ω/k are same as those of the

  1. A.velocity
  2. B.wave number
  3. C.wavelength
  4. D.frequency
Show answer

Answer: A. velocity

ω has dimension T⁻¹ (angular frequency) and k has dimension L⁻¹ (angular wave number). Dividing ω/k gives dimension LT⁻¹, which is that of velocity. Option (B) wave number has dimension L⁻¹, (C) wavelength has dimension L, and (D) frequency has dimension T⁻¹, so only velocity matches.

Question 2 · Practice question

Two point masses of 3 Kg and 2 Kg are attached at the two ends of a horizontal spring with spring constant k=200N/m. Find the natural frequency of vibration of the system.

  1. A.2.05 Hz
  2. B.2.17 Hz
  3. C.3.15 Hz
  4. D.1.9 Hz
Show answer

Answer: A. 2.05 Hz

The system is equivalent to a single mass equal to the reduced mass μ = m₁m₂/(m₁ + m₂) = 1.2 kg oscillating on the same spring. The natural angular frequency ω = √(k/μ) ≈ √(200/1.2) ≈ 12.91 rad/s, giving f = ω/2π ≈ 2.05 Hz. The other options result from common mistakes — forgetting reduced mass (using total mass gives ~1.9 Hz) or arithmetic errors in conversion.

Question 3 · Practice question

The instantaneous displacement of a simple harmonic oscillator is given by y = A cos (ωt + n/4). Its speed will be maximum at the time:). Its speed will be maximum at the time:

  1. A.n/4ω
  2. B.ω/n
  3. C.ω/2n
  4. D.2n/ω
Show answer

Answer: A. n/4ω

In SHM, speed is maximum when displacement is zero. Setting y = A cos(ωt + n/4) = 0 gives ωt + n/4 = π/2 (taking n as π), leading to t = π/(4ω) = n/4ω. Option (a) is correct. Distractor (d) 2n/ω might arise from incorrectly thinking speed is max at extreme displacement, but that’s when speed is zero.

Question 4 · Practice question

Due to what force a simple pendulum remains in simple harmonic motion?

  1. A.y, displacement
  2. B.mg sinθ, component of weight due to gravitational force
  3. C.a, acceleration
  4. D.mg, weight
Show answer

Answer: B. mg sinθ, component of weight due to gravitational force

A simple pendulum undergoes SHM because the restoring force is mg sinθ, the tangential component of its weight. This force is directly proportional to displacement for small angles. The total weight mg acts vertically, not along the arc, so option (d) is incorrect; options (a) and (c) are not forces at all.

Question 5 · Practice question

The motion of simple pendulum is said to be S.H.M when its angle θ¸ through which bob is displaced from its equilibrium position is

  1. A.θ is very small
  2. B.θ is zero
  3. C.θ is very large
  4. D.θ = cosθ
Show answer

Answer: A. θ is very small

A simple pendulum executes SHM only when the angular displacement θ is very small, because only then does sinθ ≈ θ hold, making the restoring force proportional to the displacement. Option (b) θ is zero would mean no oscillation, while large angles (c) lead to non-linear motion that is not SHM.

More Class XI Physics chapters

All Physics chapters →