MHTCET 17 Apr 2026 Shift 1 Paper

Concept:In SHM, total mechanical energy is conserved as it continually exchanges between potential energy and kinetic energy. The displacement where potential energy becomes 8 times kinetic energy is found using the standard energy expressions.Explanation:For a particle executing SHM, the potential energy at displacement $x$ is:$$U = \frac{1}{2} k x^2$$The kinetic energy at the same displacement is:$$K = \frac{1}{2} k \left(A^2 - x^2\right)$$According to the given condition, potential energy is 8 times kinetic energy:$$\frac{1}{2} k x^2 = 8 \left[ \frac{1}{2} k \left(A^2 - x^2\right) \right]$$Cancelling the common factor $\frac{1}{2} k$ from both sides:$$x^2 = 8\left(A^2 - x^2\right)$$$$x^2 = 8A^2 - 8x^2$$Bringing the $x^2$ terms together:$$9x^2 = 8A^2$$$$x^2 = \frac{8A^2}{9}$$Taking the square root on both sides:$$|x| = \sqrt{\frac{8A^2}{9}} = \frac{2\sqrt{2}}{3} A$$Thus, the particle can be at either side of the mean position:$$x = \pm \frac{2\sqrt{2}}{3} A$$The magnitude of the required displacement is:$$\frac{2\sqrt{2}}{3} A$$Answer:Option C: $\frac{2\sqrt{2}}{3} A$