### Related Formula x(t) = x_0 sin^2left(fract2 ight) = x_0 left(frac1 - cos t2 ight)$x(t) = x_0 \sin^2\left(\frac{t}{2} ight) = x_0 \left(\frac{1 - \cos t}{2} ight)$ ### Core Logic Given x_0 = 1\ mathrmm$x_0 = 1\ \mathrm{m}$, the position simplifies to: x = frac1 - cos t2 implies 2x - 1 = -cos t implies cos t = 1 - 2x$x = \frac{1 - \cos t}{2} \implies 2x - 1 = -\cos t \implies \cos t = 1 - 2x$ Differentiating x(t)$x(t)$ to find velocity v$v$: v = fracdxdt = frac12sin t$v = \frac{dx}{dt} = \frac{1}{2}\sin t$ Kinetic energy K$K$ is proportional to v^2$v^2$: K = frac12mv^2 = frac12m left(frac14sin^2 t ight) = fracm8(1 - cos^2 t)$K = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{1}{4}\sin^2 t ight) = \frac{m}{8}(1 - \cos^2 t)$ Substituting cos t = 1 - 2x$\cos t = 1 - 2x$: K = fracm8left[1 - (1 - 2x)^2 ight] = fracm8left[1 - (1 - 4x + 4x^2) ight] = fracm8(4x - 4x^2) = fracm2(x - x^2)$K = \frac{m}{8}\left[1 - (1 - 2x)^2 ight] = \frac{m}{8}\left[1 - (1 - 4x + 4x^2) ight] = \frac{m}{8}(4x - 4x^2) = \frac{m}{2}(x - x^2)$ This is an inverted parabola (K propto x - x^2$K \propto x - x^2$) passing through x=0$x=0$ and x=1$x=1$, reaching a peak at x = 1/2$x = 1/2$. This matches Graph (1). ### Pattern Recognition The position oscillates purely within the boundary interval [0, 1]$[0, 1]$. Kinetic energy peaks perfectly at the equilibrium midpoint x = 1/2$x = 1/2$ where speed is maximum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations