MHTCET 18 Apr 2026 Shift 2 Paper

Concept:In SHM, the maximum velocity is given by $v_{max} = a \omega$, where $a$ is the amplitude and $\omega = \sqrt{\frac{K}{m}}$ is the angular frequency.Explanation:For particle A, the maximum velocity is $v_{max,A} = a_A \sqrt{\frac{K_A}{m}}$.For particle B, the maximum velocity is $v_{max,B} = a_B \sqrt{\frac{K_B}{m}}$.Since the maximum velocities are equal, we equate them:$a_A \sqrt{\frac{K_A}{m}} = a_B \sqrt{\frac{K_B}{m}}$.The masses are equal, so $m$ cancels out:$a_A \sqrt{K_A} = a_B \sqrt{K_B}$.Rearranging to find the ratio of the amplitude of B to that of A:$\frac{a_B}{a_A} = \sqrt{\frac{K_A}{K_B}}$.Thus, $a_B : a_A = \sqrt{K_A} : \sqrt{K_B}$.Answer:The ratio of the amplitude of B to that of A is $\sqrt{K_A} : \sqrt{K_B}$, so the correct option is B.