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NEET Oscillations and Waves MCQs with answers

429 questions on Oscillations and Waves for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easyNEET 2022

1. Time period of simple pendulum depends on:

  1. Mass of bob
  2. Length and gravity
  3. Amplitude
  4. All of above
Show answer and explanation

Answer: (B) Length and gravity

T = 2π√(L/g). Independent of mass and amplitude (small oscillations).

Quick tip: T = 2π√(L/g). On moon: T increases (g smaller).

NCERT Class 11, Chapter 14, page 340

easy

2. Two points on a progressive wave have a path difference of λ/6. The phase difference between the oscillations of these two points is:

  1. π/3 radians
  2. π/6 radians
  3. 2π/3 radians
  4. π/4 radians
Show answer and explanation

Answer: (A) π/3 radians

Phase difference Δφ is related to path difference Δx by Δφ = (2π/λ) × Δx. Substituting Δx = λ/6 yields Δφ = (2π/λ) × (λ/6) = 2π/6 = π/3 radians.

medium

3. A transverse harmonic wave travels along a string in the +x-direction. At a certain instant t, a particle at x = x₀ has a positive displacement y > 0 and its velocity is directed downwards (in the -y direction). What can be deduced about the slope (∂y/∂x) of the string profile at x = x₀ at this instant?

  1. The slope ∂y/∂x must be zero.
  2. The slope ∂y/∂x must be negative.
  3. The slope ∂y/∂x must be positive.
  4. The slope can be positive or negative depending on the amplitude.
Show answer and explanation

Answer: (C) The slope ∂y/∂x must be positive.

For a wave traveling in the positive x-direction, the particle velocity v_p and the wave speed v satisfy the relation v_p = -v (∂y/∂x). Since the wave travels in the +x direction, v > 0. Given that v_p < 0 (moving downwards), we have -|v_p| = -v (∂y/∂x), which gives ∂y/∂x = |v_p| / v > 0. Therefore, the slope of the string profile must be positive.

hard

4. Tuning fork A has frequency 384 Hz. Sounding A with fork B gives 6 beats per second. When B is loaded with a little wax, the number of beats decreases. The original frequency of B is:

  1. 390 Hz
  2. 378 Hz
  3. 372 Hz
  4. 396 Hz
Show answer and explanation

Answer: (A) 390 Hz

B is 384 ± 6 Hz. Loading lowers B's frequency; beats decrease only if B was above A, so B = 384 + 6 = 390 Hz.

medium

5. A transverse progressive wave on a wire is given by y = 0.02 sin(30t - 2x + π/3), where x, y are in meters and t in seconds. The distance between two particles which have a phase difference of π/2 at any given instant is:

  1. π m
  2. π/2 m
  3. π/4 m
  4. 2π m
Show answer and explanation

Answer: (C) π/4 m

The phase difference between two points separated by distance Δx at a fixed instant is Δϕ = k Δx. From the equation, k = 2 rad/m. Given Δϕ = π/2 rad, we have 2 Δx = π/2, which gives Δx = π/4 m.

easy

6. A simple pendulum has length 1.69 m. Taking g = π² m/s², its time period is:

  1. 5.2 s
  2. 1.3 s
  3. 2.6 s
  4. 3.38 s
Show answer and explanation

Answer: (C) 2.6 s

For a simple pendulum T = 2π√(L/g). With g = π², T = 2π√(1.69/π²) = 2√1.69 = 2.6 s.

hard

7. Tuning fork A has frequency 320 Hz. Sounding A with fork B gives 8 beats per second. When B is loaded with a little wax, the number of beats decreases. The original frequency of B is:

  1. 336 Hz
  2. 304 Hz
  3. 312 Hz
  4. 328 Hz
Show answer and explanation

Answer: (D) 328 Hz

B is 320 ± 8 Hz. Loading lowers B's frequency; beats decrease only if B was above A, so B = 320 + 8 = 328 Hz.

hard

8. Tuning fork A has frequency 400 Hz. Sounding A with fork B gives 2 beats per second. When B is loaded with a little wax, the number of beats decreases. The original frequency of B is:

  1. 396 Hz
  2. 404 Hz
  3. 398 Hz
  4. 402 Hz
Show answer and explanation

Answer: (D) 402 Hz

B is 400 ± 2 Hz. Loading lowers B's frequency; beats decrease only if B was above A, so B = 400 + 2 = 402 Hz.

hard

9. A progressive wave moves along the +x direction. At an instant t = t₀, the particle at position x₁ has a displacement y = +A/2 and its particle velocity is directed along the negative y-direction. If the wave is given by y(x, t) = A sin(ωt - kx + φ), the slope (∂y/∂x) of the string at x₁ at that instant is:

  1. Zero
  2. Negative
  3. Positive
  4. Indeterminate without knowing φ
Show answer and explanation

Answer: (C) Positive

For a progressive wave traveling in the positive x-direction, particle velocity v_p and wave profile slope are related by v_p = -v(∂y/∂x), where v > 0 is the wave speed. Since the particle velocity is along the negative y-direction (v_p < 0), we have -v(∂y/∂x) < 0, which implies ∂y/∂x > 0 (positive).

medium

10. A 0.25 kg mass attached to a spring of force constant 160 N/m oscillates horizontally. Its time period is:

  1. 0.248 s
  2. 159 s
  3. 0.124 s
  4. 0.0395 s
Show answer and explanation

Answer: (A) 0.248 s

For a spring-mass oscillator T = 2π√(m/k) = 2π√(0.25/160) = 0.248 s.

hard

11. Two particles perform SHM of the same frequency and amplitude along the same straight line about the same mean position. They pass each other while moving in opposite directions at x = +A/2. The phase difference between them is:

  1. π/3 rad
  2. π/2 rad
  3. 2π/3 rad
  4. 5π/6 rad
Show answer and explanation

Answer: (C) 2π/3 rad

Let the positions be x₁ = A sin(ϕ₁) and x₂ = A sin(ϕ₂). Since x₁ = x₂ = A/2, we have sin(ϕ₁) = sin(ϕ₂) = 1/2. Because they are moving in opposite directions, one has v₁ = Aω cos(ϕ₁) > 0 and the other has v₂ = Aω cos(ϕ₂) < 0. Thus, ϕ₁ = π/6 rad and ϕ₂ = 5π/6 rad. The phase difference between them is Δϕ = ϕ₂ - ϕ₁ = 5π/6 - π/6 = 4π/6 = 2π/3 rad.

hard

12. Two particles A and B execute SHM of period T along the same line about the same mean position with equal amplitude. Particle A leads particle B by a phase angle of π/3. The minimum time interval after passing through the mean position by particle A until particle B passes through the mean position in the same direction is:

  1. T/6
  2. T/3
  3. T/12
  4. T/4
Show answer and explanation

Answer: (A) T/6

Since particle A leads particle B by a phase difference of Δφ = π/3, particle B reaches the exact same state of motion as particle A after an additional phase of Δφ is covered. The time delay is Δt = (Δφ / 2π) × T = (π/3 / 2π) × T = T/6.

easy

13. A wave has frequency 500 Hz and wavelength 0.68 m. Its speed is:

  1. 500 m/s
  2. 340 m/s
  3. 735 m/s
  4. 170 m/s
Show answer and explanation

Answer: (B) 340 m/s

Wave speed v = νλ = 500 × 0.68 = 340 m/s.

NCERT Class 11

medium

14. Consider the following two statements regarding a plane progressive sound wave propagating along the +x-direction in air: Statement I: The pressure variation is in phase with the displacement of the medium particles. Statement II: At the position of maximum displacement of particles from their mean positions, the pressure is equal to the normal atmospheric pressure.

  1. Statement I is correct and Statement II is incorrect.
  2. Statement I is incorrect and Statement II is correct.
  3. Both Statement I and Statement II are correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (B) Statement I is incorrect and Statement II is correct.

For a progressive sound wave y = A sin(kx - ωt), the excess pressure wave is ΔP = -B (∂y/∂x) = -B k A cos(kx - ωt) = ΔP₀ sin(kx - ωt - π/2). Thus, pressure variation is out of phase by π/2 with displacement, making Statement I incorrect. When displacement is maximum, the spatial gradient ∂y/∂x = 0, so the excess pressure ΔP = 0, meaning the pressure is equal to the normal undisturbed atmospheric pressure, which makes Statement II correct.

easy

15. The average potential energy of a particle of mass m executing simple harmonic motion of amplitude A and angular frequency ω over one complete cycle of motion is:

  1. (1/4)mω²A²
  2. (1/2)mω²A²
  3. (1/8)mω²A²
  4. zero
Show answer and explanation

Answer: (A) (1/4)mω²A²

For a harmonic oscillator with zero potential energy at the mean position, U(t) = (1/2)kA²sin²(ωt). The average value of sin²(ωt) over one full time period is 1/2. Therefore, the time-averaged potential energy is (1/2)kA² × (1/2) = (1/4)kA² = (1/4)mω²A².

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