Projection of uniform circular motion on an axis

If a particle moves in a circle of radius $r$ with angular velocity $\omega$, then its projection on the $x$-axis executes SHM: $x(t)=r\cos\theta(t)$ where $\theta(t)$ is the angular position measured from the positive $x$-axis.

Initial phase from the figure

From the given options and standard representation, the particle has an initial angular displacement of $\dfrac{\pi}{6}$ from the positive $x$-axis at $t=0$.

Hence, $\theta(t)=\omega t+\frac{\pi}{6}$

Write the projection on the $x$-axis

Therefore, $x(t)=r\cos\left(\omega t+\frac{\pi}{6}\right)$

Check options

• A: $x(t)=\cos\left(\omega t-\frac{\pi}{6}\omega\right)$ This is dimensionally incorrect and also misses factor $r$.
• B: $x(t)=\cos(\omega t)$ Missing amplitude $r$ and no initial phase.
• C: $x(t)=r\cos\left(\omega t+\frac{\pi}{6}\right)$ Correct.
• D: $x(t)=r\sin\left(\omega t+\frac{\pi}{6}\right)$ This would correspond to a different phase convention for projection, not the stated $x$-projection of $OP$.

Final answer

The correct option is: $\boxed{C}$