For particle P revolving round the centre O with radius of circular path $\mathrm{r}$ and angular velocity $\omega$, as shown in below figure, the projection of OP on the $x$-axis at time $t$ is

Question

$x(t)=\operatorname{cos}\left(\omega t-\frac{\pi}{6} \omega\right)$

$x(t)=\operatorname{cos}(\omega t)$

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

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

Solution

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