MHTCET 20 Apr 2026 Shift 1 Paper

Concept:When a mass is connected to two springs on opposite sides, both springs provide restoring forces in the same direction when the mass is displaced. Hence, the equivalent spring constant is the sum of the individual spring constants.Explanation:Let the mass $m$ be displaced slightly by a distance $x$ from its equilibrium position.One spring gets stretched by $x$, while the other spring gets compressed by $x$.In both cases, the spring forces act towards the equilibrium position, opposing the displacement.The restoring force due to the spring with constant $2k$ is:$F_1 = -2k x$The restoring force due to the spring with constant $k$ is:$F_2 = -k x$Both forces act in the same direction, so the total restoring force is:$F = F_1 + F_2 = -2k x - k x = -3k x$Comparing with $F = -k_{\text{eq}} x$, we get:$k_{\text{eq}} = 3k$For a mass-spring system, the frequency of oscillation is:$f = \frac{1}{2\pi} \sqrt{\frac{k_{\text{eq}}}{m}} = \frac{1}{2\pi} \sqrt{\frac{3k}{m}}$Answer:$f = \frac{1}{2 \pi} \sqrt{\frac{3 k}{m}}$Therefore, the correct option is D.