NEET Oscillations PYQs & Practice Questions

10 free NEET Physics questions on Oscillations, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: SHM - Spring Mass System · SHM - Spring Mass Amplitude Ratio · Simple Harmonic Motion - Basic Definition · Simple Pendulum and Time Period · Damped and Forced Oscillations - Resonance · SHM Identification · Acceleration-Displacement Relation · SHM Velocity-Acceleration-Period Relation · SHM Equation - Amplitude and Time Period · Simple Pendulum Time Period - Mass Independence, Length Dependence · SHM - Acceleration from x-t Graph · Simple Harmonic Motion - Velocity-Displacement Relation

1. Q1NEET 2026Simple Harmonic Motion - Velocity-Displacement Relation

   Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m kg and the spring constant is k Nm^-1. At a given instant, the extension of the spring is x meter and the speed of the particle is v ms^-1. On the x-v plane, if the graph of v as a function of x is a circle, then the correct option is:

   • A. k = m
   • B. k = m^2
   • C. k = sqrt(m)
   • D. k = 1/m
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   Answer: —

   For SHM, v^2 = (k/m)(A^2 - x^2), i.e., v^2 + (k/m)x^2 = (k/m)A^2. Dividing through, x^2/A^2 + v^2/((k/m)A^2) = 1, an ellipse in the x-v plane. For this to be a circle, the coefficients of x^2 and v^2 must be equal, i.e., k/m = 1, so k = m.

2. Q2NEET 2026Oscillation of a Floating Cylindrical Cork

   A cylindrical cork of uniform density floats in a liquid of density ρ₁. If the cork is depressed slightly and released, it oscillates harmonically with time period T. If the same cork floats in another liquid of density ρ₂, then the similar oscillation has time period 2T. The value of ρ₂/ρ₁ is:

   • A. 2
   • B. 1/2
   • C. 1/4
   • D. 4
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   Answer: B — 1/2

   For a floating cylindrical cork displaced by x, the restoring force is F = -ρ_liquid A g x, giving effective spring constant k = ρ_liquid A g. Time period T = 2π√(m/k) = 2π√(m/(ρ_liquid A g)), so T ∝ 1/√ρ_liquid (mass m of the cork is fixed). Given T₂ = 2T₁: T₂/T₁ = √(ρ₁/ρ₂) = 2, so ρ₁/ρ₂ = 4, giving ρ₂/ρ₁ = 1/4.

3. Q3NEET 2026SHM energy graphs

   For a simple pendulum, having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by which of the following graphs?

   • A. Graph (1): two asymmetric humps between 0 and T
   • B. Graph (2): flat line with a dip exactly at T/2
   • C. Graph (3): zero at t=0, max at T/4, zero at T/2, max at 3T/4, zero at T
   • D. Graph (4): nonzero at t=0, dips to zero before T/2, rises to a max before T, matching the standard |sin| shaped K.E.-vs-t curve for a pendulum released from rest at an extreme
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   Answer: C — Graph (3): zero at t=0, max at T/4, zero at T/2, max at 3T/4, zero at T

   KE of a pendulum bob is periodic with period T/2 and always non-negative, maximum at the mean position and zero at the extremes; matching graph is option D.

4. Q4NEET 2026Energy of pendulum

   The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately: (mass of bob = 20 g)

   • A. 2.0 m/s
   • B. 0.2 m/s
   • C. 1.41 m/s
   • D. 1.41 m/s
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   Answer: C — 1.41 m/s

   At equilibrium, total energy = KE = ½mv². v=√(2E/m)=√(2×0.02/0.02)=√2≈1.41 m/s.

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5. Q5NEET 2026Simple pendulum length from oscillations

   Savitha, an XI standard student, while conducting an experiment to determine the effective length of a simple pendulum, notes down the time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (π²=9.8, g=9.8 m/s²)

   • A. 0.75 m
   • B. 1 m
   • C. 1.5 m
   • D. 2 m
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   Answer: A — 0.75 m

   T=60/30=2s. L=gT²/4π²=9.8×4/(4×9.8)=1 m.

6. Q6NEET 2026Phase difference of a travelling wave

   For a travelling harmonic wave, y(x,t)=2.0 cos 2π(10t - 0.0080x + 0.35), where x and y are in cm and t in s. What is the phase difference between oscillatory motion of two points separated by a distance of 0.5 m?

   • A. 0.08π rad
   • B. 0.008π rad
   • C. 0.8π rad
   • D. 8π rad
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   Answer: B — 0.008π rad

   Phase difference = k×Δx = 2π×0.0080(per cm)×50 cm = 0.8π rad.

7. Q7NEET 2025SHM - Spring Mass System

   In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency ω(t) and average amplitude A(t) of the system change with time t. Which one of the following options schematically depicts these changes correctly?

   • A. ω(t) rising then flat; A(t) flat
   • B. ω(t) rising then flat; A(t) decreasing
   • C. ω(t) rising; A(t) rising
   • D. ω(t) decreasing; A(t) decreasing
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   Answer: A — ω(t) rising then flat; A(t) flat

   ω = √(k/m). As mass decreases (sand leaking out), ω increases. Energy E = ½kA² also decreases as mass/energy is lost with the leaking sand, so amplitude A decreases. Option (2) shows ω increasing and A decreasing.

8. Q8NEET 2025SHM - Spring Mass Amplitude Ratio

   Two identical point masses P and Q, suspended from two separate massless springs of spring constants k1 and k2, respectively, oscillate vertically. If their maximum speeds are the same, the ratio (AQ/AP) of the amplitude AQ of mass Q to the amplitude AP of mass P is -

   • A. k2/k1
   • B. k1/k2
   • C. √(k2/k1)
   • D. √(k1/k2)
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   Answer: C — √(k2/k1)

   Vmax = Aω = A√(k/m). Equal max speeds: AP√(k1/m) = AQ√(k2/m) ⟹ AQ/AP = √(k1/k2).

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9. Q9NEET 2024SHM Equation - Amplitude and Time Period

   If x = 5sin(πt + π/3) m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are:

   • A. 5 cm, 1 s
   • B. 5 m, 1 s
   • C. 5 cm, 2 s
   • D. 5 m, 2 s
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   Answer: C — 5 cm, 2 s

   Comparing x = 5sin(πt + π/3) with standard SHM equation x = A sin(ωt + φ): Amplitude A = 5 m, angular frequency ω = π rad/s, so time period T = 2π/ω = 2π/π = 2 s.

10. Q10NEET 2024Simple Pendulum Time Period - Mass Independence, Length Dependence

    If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x/2 times its original time period. Then the value of x is

    • A. 2√3
    • B. 4
    • C. √3
    • D. √2
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    Answer: C — √3

    T = 2π√(l/g), independent of mass. T/(x/2·T) = √(l/(l/2)) = √2, so 2/x = √2, giving x = 2/√2 = √2.