Oscillations: NEET previous year questions

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).

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)).

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.

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).

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).

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