JEE Main 2026 — Oscillations Question with Solution

The time period $T$ of a simple pendulum of length $L$ is given by the formula:
$T=2\pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get:
$T^2=4{\pi }^2\frac{L}{g}$
Rearranging the equation to find the relationship between $\frac{1}{T^2}$ and $L$:
$\frac{1}{T^2}=\frac{g}{4{\pi }^{2}L}$
Let $y=\frac{1}{T^2}$ and $x=L$. The equation becomes:
$y=\frac{k}{x}$ where $k=\frac{g}{4{\pi }^2}$ is a constant.
This is the equation of a rectangular hyperbola in the first quadrant.
As $L$ increases, $\frac{1}{T^2}$ decreases non-linearly, approaching the axes asymptotically.
Comparing this with the given plots, option (2) correctly represents a hyperbolic relationship between $\frac{1}{T^2}$ and $L$.