A particle oscillates along the x-axis accord | vedclass

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(A) Given $x(t) = x_0 \sin^2\left(\frac{t}{2}ight)$.Using the trigonometric identity $\sin^2	heta = \frac{1 - \cos(2	heta)}{2}$,we get:$x(t) = x_0

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(A) Given $x(t) = x_0 \sin^2\left(\frac{t}{2}ight)$.Using the trigonometric identity $\sin^2	heta = \frac{1 - \cos(2	heta)}{2}$,we get:$x(t) = x_0 \left(\frac{1 - \cos t}{2}ight) = \frac{x_0}{2} - \frac{x_0}{2} \cos t$.Since $x_0 = 1 	ext{ m}$,$x(t) = \frac{1}{2} - \frac{1}{2} \cos t$.This represents simple harmonic motion about the mean position $x = \frac{1}{2} 	ext{ m}$ with amplitude $A = \frac{1}{2} 	ext{ m}$.The velocity is $v = \frac{dx}{dt} = \frac{1}{2} \sin t$.The kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{1}{4} \sin^2 tight) = \frac{m}{8} \sin^2 t$.From $x - \frac{1}{2} = -\frac{1}{2} \cos t$,we have $\cos^2 t = 4(x - \frac{1}{2})^2$.Since $\sin^2 t = 1 - \cos^2 t$,we get $K = \frac{m}{8} [1 - 4(x - \frac{1}{2})^2]$.This is the equation of a downward parabola with respect to $x$,which is zero at $x = 0$ and $x = 1$,and maximum at $x = \frac{1}{2}$. Thus,graph $(A)$ is correct.

$A$ 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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