Choose the correct length (L) versus square of the time period ( T²) graph for a simple pendulum executing simple harmonic motion. (a) (b) (c) (d)

Answer: (d). T = 2π√(L/g) ⇒ T² = (4π²)/(g)L T² ∝ L: a straight line through the origin with positive slope (4π²)/(g). So the answer is (d).

Choose the correct length (L) versus square of the time period ( T²)… | JEE Main 2023 (1 Feb, Shift 2)

Q42JEE Main 2023 · 1 FEB · SHIFT 2Easy

Simple Pendulum

Choose the correct length (L) versus square of the time period ( $T^2$ ) graph for a simple pendulum executing simple harmonic motion.

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Answer: (d)

$T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow T^2 = \frac{4\pi^2}{g}L$
$T^2 \propto L$ : a straight line through the origin with positive slope $\frac{4\pi^2}{g}$ .
So the answer is (d).

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