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(C) Let the particle move on a circle of radius $R$ with constant angular speed $\omega$. At $t = 0$,the particle is at a position $P_0$. At time $t$,

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(C) Let the particle move on a circle of radius $R$ with constant angular speed $\omega$. At $t = 0$,the particle is at a position $P_0$. At time $t$,the particle is at position $P$,such that the angle $\angle P_0OP = 	heta = \omega t$.The displacement $S$ is the chord length $P_0P$. In the isosceles triangle $	riangle P_0OP$,the length of the chord is given by $S = 2R \sin(\frac{	heta}{2}) = 2R \sin(\frac{\omega t}{2})$.This function represents a sine curve with a period $T = \frac{2\pi}{\omega}$.At $t = 0$,$S = 0$. At $t = \frac{\pi}{\omega}$,$S = 2R \sin(\frac{\pi}{2}) = 2R$ (maximum displacement). At $t = \frac{2\pi}{\omega}$,$S = 2R \sin(\pi) = 0$.Thus,the graph is a sine curve starting from $0$,reaching a maximum at $t = \frac{\pi}{\omega}$,and returning to $0$ at $t = \frac{2\pi}{\omega}$. This corresponds to the graph shown in option $C$.

$A$ particle moves along a circle with a constant angular speed $\omega$. Its displacement,with respect to the initial position of the particle at time $t = 0$,is plotted against time. The graph would look like:

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