A particle oscillates along the x -axis according to the law, x(t) = x₀sin ²(t/2) where x₀ = 1 m. The kinetic energy (K) of the particle as a function of x is correctly represented by the graph (a) (b) (c) (d)

Answer: (d). Kx01/21K max (mean) x = x₀sin ²t/2 = 1/2 - 1/2cos t (with x₀ = 1 m): SHM about x = 1/2 with amplitude 1/2 and ω = 1. K = 1/2mω²[A² - (x - 1/2)²], an inverted parabola in x. K = 0 at the extremes x = 0 and x = 1; K is maximum at the mean position x = 1/2. So the answer is (d).

A particle oscillates along the x -axis according to the law, x(t) =… | JEE Main 2025 (24 Jan, Shift 2)

Q25JEE Main 2025 · 24 JAN · SHIFT 2Medium

Energy in SHM

A particle oscillates along the $x$ -axis according to the law, $x(t) = x_0\sin^2\left(\frac{t}{2}\right)$ where $x_0 = 1\text{ m}$ . The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph

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Answer: (d)

$x = x_0\sin^2\frac{t}{2} = \frac{1}{2} - \frac{1}{2}\cos t$ (with $x_0 = 1$ m): SHM about $x = \frac{1}{2}$ with amplitude $\frac{1}{2}$ and $\omega = 1$ .
$K = \frac{1}{2}m\omega^2\left[A^2 - \left(x - \frac{1}{2}\right)^2\right]$ , an inverted parabola in $x$ .
$K = 0$ at the extremes $x = 0$ and $x = 1$ ; $K$ is maximum at the mean position $x = \frac{1}{2}$ .
So the answer is (d).

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