For a simple pendulum, having time period T , the variation of...

The displacement of a simple pendulum executing simple harmonic motion is given by $x = A \sin(\omega t + \phi)$.

The velocity of the pendulum is $v = \dfrac{dx}{dt} = A \omega \cos(\omega t + \phi)$.

The kinetic energy is given by $K = \dfrac{1}{2} m v^2 = \dfrac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi)$.

Since kinetic energy is proportional to the square of velocity, it can never be negative. This eliminates the graphs that show negative values.

Also, the kinetic energy of a pendulum is not constant, which eliminates the constant graph.

Using the trigonometric identity $\cos^2 \theta = \dfrac{1 + \cos(2\theta)}{2}$, we get:

$K = \dfrac{1}{4} m A^2 \omega^2 [1 + \cos(2\omega t + 2\phi)]$

This shows that the kinetic energy oscillates with an angular frequency $2\omega$, meaning its time period is half that of the pendulum, i.e., $T' = \dfrac{T}{2}$.

The graph in option (3) correctly represents a non-negative oscillating quantity with a period of $T/2$.

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