MHTCET 15 Apr 2026 Shift 1 Paper

Concept:In S.H.M., kinetic energy depends on displacement. When the particle starts from the mean position, its displacement is $x = A \sin \omega t$.Explanation:Total energy of the particle is constant:$$E=\frac{1}{2} k A^{2}$$At displacement $x$, potential energy is:$$U=\frac{1}{2} k x^{2}$$Hence, kinetic energy is:$$K = E - U$$So the ratio of kinetic energy to total energy becomes:$$\frac{K}{E} = 1 - \frac{x^{2}}{A^{2}}$$Substitute $x = A \sin \omega t$:$$\frac{K}{E} = 1 - \sin^{2} \omega t = \cos^{2} \omega t$$Given that $K$ is 50% of $E$, we have:$$\frac{K}{E} = \frac{1}{2}$$Therefore:$$\cos^{2} \omega t = \frac{1}{2}$$$$\cos \omega t = \frac{1}{\sqrt{2}}$$Using $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$, we get:$$\omega t = 45^{\circ} = \frac{\pi}{4}$$The time period is $T = 6$ s, so the angular frequency is:$$\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}$$Now solve for time $t$:$$\frac{\pi}{3} t = \frac{\pi}{4}$$$$t = \frac{3}{4} = 0.75 \text{ s}$$Answer:The kinetic energy is 50% of the total energy at $t = 0.75$ s.Therefore, the correct option is A. 0.75 s.