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

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$x (t) = rcos (ωt)$

$x (t) = rcos (ωt - \frac{π}{6} ω)$

$x (t) = rcos (ωt + \frac{π}{6})$

$x (t) = rsin (ωt + \frac{π}{6})$

answer is C.

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Detailed Solution

By using concept of phasor diagram. Particle starts from $π$ /6.

$x = Acos (ωt + ϕ)$
$x = r \cos (ωt + π / 6)$

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