JEE Main 2025 Jan 24 Shift 2, Physics Q19: Simple Harmonic Motion And Oscillation

JEE Main2025Jan 24, Shift 2Physics

A particle oscillates along the x-axis according to the law, $x (t) = x_{0} sin^{2} (\frac{t}{2})$ where $x_{0} = 1 m$ . The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph.

Solution

Use the identity $sin^{2} (t /2) = \frac{1 - cos t}{2}$ :

$x (t) = \frac{x _{0}}{2} (1 - cos t) \Rightarrow x = \frac{1 - cos t}{2} (with x_{0} = 1)$

So $cos t = 1 - 2 x$ .

The motion is simple harmonic about the midpoint $x = 1/2$ with amplitude $1/2$ — the particle oscillates between $x = 0$ and $x = 1$ .

Velocity: $v = \frac{d x}{d t} = \frac{x _{0}}{2} sin t = \frac{1}{2} sin t$ .

Kinetic energy: $K = \frac{1}{2} m v^{2} = \frac{m}{8} sin^{2} t = \frac{m}{8} (1 - cos^{2} t)$

Substituting $cos t = 1 - 2 x$ :

$K = \frac{m}{8} [1 - (1 - 2 x)^{2}] = \frac{m}{8} [4 x - 4 x^{2}] = \frac{m}{2} x (1 - x)$

This is an inverted parabola in $x$ , zero at $x = 0$ and $x = 1$ , maximum at $x = 1/2$ .

Correct option: (1)

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JEE Main 2025 Jan 24 Shift 2, Physics Q19: Simple Harmonic Motion And Oscillation

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