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Simple Harmonic Motion question

2023 · 1 Feb · Shift 2 · Q53

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Simple Harmonic Motion
2023 · 1 Feb · Shift 2 · Q53

Simple Harmonic Motion question

2023 · 1 Feb · Shift 2 · Q53

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1

Choose the correct length (L) versus square of the time period (

\mathrm{T}^{2}

) graph for a simple pendulum executing simple harmonic motion.

A
B
C
D

View written solutionFree

Correct answer: D

For a simple pendulum undergoing small oscillations, the time period is

$T = 2\pi \sqrt{\frac{L}{g}}$

Squaring both sides,

$T^2 = 4\pi^2 \frac{L}{g}$

Rearranging for $L$ ,

$L = \frac{g}{4\pi^2} T^2$

This shows that $L$ is directly proportional to $T^2$ :

$L \propto T^2$

So the graph of $L$ versus $T^2$ is a straight line passing through the origin with positive slope.

Therefore, the correct option is the one showing a straight line through the origin.

Hence, the correct answer is D.

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