MHTCET 11 May 2024 Shift 1 Paper

Concept:In linear S.H.M., acceleration magnitude is $a = \omega^2 x$ and $v^2 = \omega^2 (A^2 - x^2)$, where $x$ is displacement from mean position.Explanation:Let the two positions on the same side of the mean position be $x_1$ and $x_2$.Then,$a_1 = \omega^2 x_1$ and $a_2 = \omega^2 x_2$$$a_1 + a_2 = \omega^2 (x_1 + x_2) \quad \text{...(i)}$$Using the velocity relation,$$u^2 = \omega^2 A^2 - \omega^2 x_1^2$$$$V^2 = \omega^2 A^2 - \omega^2 x_2^2$$Subtracting,$$V^2 - u^2 = \omega^2 (x_1^2 - x_2^2)$$$$V^2 - u^2 = \omega^2 (x_1 - x_2)(x_1 + x_2)$$Using equation (i),$$x_1 - x_2 = \frac{V^2 - u^2}{a_1 + a_2}$$Therefore,$$x_2 - x_1 = \frac{u^2 - V^2}{a_1 + a_2}$$Since $a_2 > a_1$, the particle is farther from the mean position at the second instant, so $x_2 - x_1$ gives the required distance.Answer:$$\frac{u^2 - V^2}{a_1 + a_2}$$Hence, the correct option is D.