Oscillations NEET PYQs (2005–2026) with Solutions

10 solved Oscillations PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumForce & Energy in SHM

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

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (C) Option 3

   Explanation

   For SHM, velocity is: $$v=A\omega\cos(\omega t+\phi)$$ Hence kinetic energy: $$K=\frac12 mv^2$$ $$K=\frac12 mA^2\omega^2\cos^2(\omega t+\phi)$$ Thus, $$K\propto \cos^2(\omega t+\phi)$$ Since $\cos^2\theta$ is always non-negative and has period $T/2$, kinetic energy becomes zero twice in one oscillation and reaches maximum twice. Therefore the correct variation is **Option (C)**.

2. Question 2 (NEET 2026)

   NEET 2026EasySimple 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: (Consider mass of the bob = 20 g)

   1. AOption A: 0.2 m/s
   2. BOption B: 1.41 m/s
   3. COption C: 14.1 m/s
   4. DOption D: 2.0 m/s

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   Correct answer: (B) 1.41 m/s

   Explanation

   Total mechanical energy of the pendulum remains constant: E = KE + PE = 0.02 J At equilibrium position: Potential energy = 0 Therefore: Total energy = Kinetic energy ½mv² = 0.02 Given: m = 20 g = 0.02 kg Substitute: ½ × 0.02 × v² = 0.02 0.01v² = 0.02 v² = 2 v = √2 v ≈ 1.41 m/s Hence option (B) is correct.

3. Question 3 (NEET 2026)

   NEET 2026MediumSHM of a floating body

   A cylindrical cork of uniform density floats in a liquid of density $\rho_1$. 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 $\rho_2$, then the similar oscillation has time period $2T$. The value of $\rho_2 / \rho_1$ is:

   1. AOption A: $4$
   2. BOption B: $2$
   3. COption C: $1/2$
   4. DOption D: $1/4$

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   Correct answer: (D) $1/4$

   Explanation

   If the cork (mass $m$, cross-section $A$) is pushed down by $x$, the extra buoyant force $\rho A g x$ acts upward, so the force constant is $k = \rho A g$. $T = 2\pi\sqrt{\frac{m}{\rho A g}} \Rightarrow T \propto \frac{1}{\sqrt{\rho}}$. $\frac{T_2}{T_1} = \sqrt{\frac{\rho_1}{\rho_2}} = 2 \Rightarrow \frac{\rho_2}{\rho_1} = \frac{1}{4}$, option (D).

4. Question 4 (NEET 2026)

   NEET 2026MediumEnergy in simple harmonic motion

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

   1. AOption A: $k = \frac{1}{m}$
   2. BOption B: $k = m$
   3. COption C: $k = m^2$
   4. DOption D: $k = \sqrt{m}$

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   Correct answer: (B) $k = m$

   Explanation

   Energy conservation: $\frac{1}{2}kx^2 + \frac{1}{2}mv^2 = E$ (constant), i.e. $\frac{x^2}{2E/k} + \frac{v^2}{2E/m} = 1$. This is an ellipse in the $x$–$v$ plane. It is a circle only when the two semi-axes are equal: $\frac{2E}{k} = \frac{2E}{m}$, i.e. $k = m$ (numerically, in SI units). Hence option (B).

5. Question 5 (NEET 2026)

   NEET 2026EasySimple pendulum

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

   1. AOption A: 0.75 m
   2. BOption B: 1.5 m
   3. COption C: 2 m
   4. DOption D: 1 m

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   Correct answer: (D) 1 m

   Explanation

   Time period T=60/30=2 s. Using T=2π√(l/g), l=gT²/(4π²)=1 m. Hence option (D) is correct.

6. Question 6 (NEET 2025)

   NEET 2025HardSimple Harmonic Motion

   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 $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (B) Option (2)

   Explanation

   As sand leaks out of the box, the mass of the oscillating system decreases gradually. Since the angular frequency of a spring mass system is given by $\\omega = \\sqrt{\\frac{k}{m}}$, the frequency increases with time as mass decreases. At the same time, energy is lost from the system due to leakage and damping effects, so the amplitude decreases with time. Therefore, the correct schematic variation is shown in option (B).

7. Question 7 (NEET 2025)

   NEET 2025MediumSimple Harmonic Motion

   Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_1$ and $k_2$, respectively, oscillate vertically. If their maximum speeds are the same, then the ratio $(A_Q/A_P)$ of the amplitude $A_Q$ of the mass Q to the amplitude $A_P$ of mass P is:

   1. AOption A: $\dfrac{k_2}{k_1}$
   2. BOption B: $\dfrac{k_1}{k_2}$
   3. COption C: $\sqrt{\dfrac{k_2}{k_1}}$
   4. DOption D: $\sqrt{\dfrac{k_1}{k_2}}$

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   Correct answer: (D) $\sqrt{\dfrac{k_1}{k_2}}$

   Explanation

   For SHM, maximum speed is: Also, Since the masses are identical and maximum speeds are equal: $$A_P \sqrt{k_1} = A_Q \sqrt{k_2}$$ Therefore, $$\frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}}$$ Hence, option (D) is correct.

8. Question 8 (NEET 2024)

   NEET 2024EasySimple Harmonic Motion

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

   1. AOption A: 5 cm, 1 s
   2. BOption B: 5 m, 1 s
   3. COption C: 5 cm, 2 s
   4. DOption D: 5 m, 2 s

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   Correct answer: (D) 5 m, 2 s

   Explanation

   Comparing with the standard SHM equation: $$x = A \sin(\omega t + \phi)$$ Amplitude: $$A = 5\,m$$ Angular frequency: $$\omega = \pi\,rad\,s^{-1}$$ Time period: $$T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2\,s$$ Hence, option (D) is correct.

9. Question 9 (NEET 2024)

   NEET 2024MediumSimple Pendulum

   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:

   1. AOption A: 2√3
   2. BOption B: 4
   3. COption C: √3
   4. DOption D: √2

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   Correct answer: (D) √2

   Explanation

   Time period of a simple pendulum: $$T = 2\pi \sqrt{\frac{L}{g}}$$ It is independent of mass. New length: $$L' = \frac{L}{2}$$ Therefore: $$T' = 2\pi \sqrt{\frac{L/2}{g}}$$ $$T' = \frac{1}{\sqrt{2}}T$$ Given: $$T' = \frac{x}{2}T$$ Thus: $$\frac{x}{2} = \frac{1}{\sqrt{2}}$$ $$x = \sqrt{2}$$ Hence, option (D) is correct.

10. Question 10 (NEET 2023)

    NEET 2023EasyForce & Energy in SHM

    The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be :

    1. AOption A: 2U
    2. BOption B: 4U
    3. COption C: 8U
    4. DOption D: 16U

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    Correct answer: (D) 16U

    Explanation

    Potential energy stored in a spring is proportional to x². Stretching changes from 2 cm to 8 cm, i.e. 4 times. Hence energy becomes 4² = 16 times.