Q 11-13-090JEE MainJEE Main 2023 (1 Feb, Shift 2)Easy
Choose the correct length ($L$) versus square of time period ($T^2$) graph for a simple pendulum executing simple harmonic motion.
Answer: (C) see figure
$T^2=\dfrac{4\pi^2}{g}L$, so $T^2\propto L$: a straight line through the origin, graph (3).
Solution by Sreeraj P, M.Sc Physics