Oscillations PYQs on the CBT Screen

Question 1 · NEET 2026 · Q5 · +4 / −1

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

Answer and explanation

Answer: (3)

The velocity in simple harmonic motion (SHM) is $v = A\omega\cos(\omega t + \phi)$ , so the kinetic energy $K = \tfrac{1}{2}mA^2\omega^2\cos^2(\omega t + \phi)$ is proportional to $\cos^2(\omega t + \phi)$ . This is always non-negative and repeats with period $T/2$ , producing two identical humps between 0 and $T$ that fall to zero at the extreme positions. Graph (3) shows exactly this, so the answer is (3).

Question 2 · NEET 2026 · Q16 · +4 / −1

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)

Answer and explanation

Answer: (2)

At the equilibrium (lowest) point the bob's energy is entirely kinetic, so $\tfrac{1}{2}mv^2$ equals the total energy 0.02 J. With m = 20 g = 0.02 kg, $v^2 = \dfrac{2(0.02)}{0.02} = 2$ , giving $v = \sqrt{2} \approx 1.41\ \text{m/s}$ (option (2)). Forgetting to divide by the mass gives $\sqrt{2(0.02)} = 0.2\ \text{m/s}$ (option (1)).

Question 3 · NEET 2026 · Q33 · +4 / −1

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 $\pi^2 = 9.8$ , and $g = 9.8$ m/s $^2$ )

Answer and explanation

Answer: (4)

The time period is $T = 60/30 = 2$ s. From $T = 2\pi\sqrt{L/g}$ , the length is $L = \frac{gT^2}{4\pi^2} = \frac{9.8\times4}{4\times9.8} = 1$ m. The other options result from arithmetic or period errors.

Question 4 · Re-NEET 2026 · Q25 · +4 / −1

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:

Answer and explanation

Answer: (4)

Depressing the cork by $x$ creates an extra buoyant force $A\rho g x$ directed back towards equilibrium, so the effective spring constant is $k = A\rho g$ and $T = 2\pi\sqrt{\frac{m}{A\rho g}}$ . With $m$ and $A$ unchanged, $T \propto \rho^{-1/2}$ . Doubling the period therefore requires the density to fall by a factor of four: $\frac{\rho_2}{\rho_1} = \frac{1}{4}$ . NCERT Class 11, Oscillations (simple harmonic motion (SHM) examples).

Question 5 · Re-NEET 2026 · Q28 · +4 / −1

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

Answer and explanation

Answer: (2)

Conservation of energy gives $\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = E$ , which rearranges to $\frac{v^2}{2E/m} + \frac{x^2}{2E/k} = 1$ - an ellipse in the $x\!-\!v$ plane. An ellipse becomes a CIRCLE only when the two semi-axes are equal, i.e. $\frac{2E}{m} = \frac{2E}{k}$ , which requires $k = m$ numerically. NCERT Class 11, Oscillations (energy in simple harmonic motion).

Question 6 · NEET 2020 · Q153 · +4 / −1

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is

Answer and explanation

Answer: (4)

In SHM, a = - $\omega$ $^{2}$ x. Acceleration is therefore exactly opposite to displacement, giving a phase difference of $\pi$ radians. Chapter: Oscillations.

Oscillations, on the exam screen.