Question 113 of 194

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

ABCD

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Answer: (C)

A simple pendulum consists of a point mass attached to an inextensible string and suspended from a support.
The formula to calculate the time period $\left(T\right)$ of a simple pendulum is given by
$T=2\pi \sqrt{\dfrac{l}{g}}.......................(1)$
where, $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
Simplify equation (1) to obtain the relation between the time period and the length.
$\begin{matrix}{T}^{2} & = & \dfrac{4{\pi }^{2}}{g}\times 𝓁 \\ {T}^{2} & \propto & l............................(2)\end{matrix}$
Equation (2) suggests that the graph between the square of the time period versus the length of the pendulum is a straight line passing through the origin.

JEE Main Physics practice question on Oscillations (Time Period and Frequency), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.