Oscillations Part 2

Concept:The time period $T$ of a simple pendulum depends on its length $L$ as $T = 2\pi\sqrt{L/g}$.Explanation:Squaring the formula gives $T^2 = \frac{4\pi^2 L}{g}$.Rearranging, $\frac{1}{T^2} = \frac{g}{4\pi^2} \cdot \frac{1}{L}$.Since $g$ and $4\pi^2$ are constants, $\frac{1}{T^2}$ is inversely proportional to $L$.Plotting $\frac{1}{T^2}$ (y-axis) versus $L$ (x-axis) gives a rectangular hyperbola.As $L$ increases, $\frac{1}{T^2}$ decreases and approaches zero.As $L$ approaches zero, $\frac{1}{T^2}$ becomes very large.The graph has both axes as asymptotes.Among the given options, only graph A shows this hyperbolic shape.Answer:Option A.