MHTCET 19 Apr 2026 Shift 1 Paper

Concept:For SHM starting from the mean position, displacement is written as $x = A \sin(\omega t)$.Explanation:The time period is given as $T = 8$ s, so the angular frequency is$$\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ \text{rad s}^{-1}$$At $t = 0$, the particle is at the mean position, so the displacement is$$x = A \sin\left(\frac{\pi}{4}t\right)$$Distance travelled in the 1st second is the change in position from $t = 0$ to $t = 1$ s.At $t = 0$: $x_0 = A \sin 0 = 0$At $t = 1$ s: $x_1 = A \sin\left(\frac{\pi}{4}\right) = A \sin 45^\circ = \frac{A}{\sqrt{2}}$So, $s_1 = \frac{A}{\sqrt{2}} - 0 = \frac{A}{\sqrt{2}}$Distance travelled in the 2nd second is the change in position from $t = 1$ s to $t = 2$ s.At $t = 2$ s: $x_2 = A \sin\left(\frac{\pi}{2}\right) = A \sin 90^\circ = A$So, $s_2 = A - \frac{A}{\sqrt{2}}$The required ratio is$$\frac{s_2}{s_1} = \frac{A - \frac{A}{\sqrt{2}}}{\frac{A}{\sqrt{2}}}$$Simplifying,$$\frac{s_2}{s_1} = \sqrt{2}\left(1 - \frac{1}{\sqrt{2}}\right) = \sqrt{2} - 1$$Answer:The ratio of distances travelled in the 2nd and 1st seconds is $\sqrt{2} - 1$, so the correct option is C.