Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is

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$x (t) = B \sin (\frac{2 π t}{30})$

$x (t) = B \cos (\frac{π t}{15})$

$x (t) = B \sin (\frac{π t}{15} + \frac{π}{2})$

$x (t) = B \cos (\frac{π t}{15} + \frac{π}{2})$

answer is A.

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Detailed Solution

From figure $\sin (ω t) = \frac{x}{B} \Rightarrow x = B \sin (ω t)$

$x = B s i n (\frac{2 π}{T} t) \Rightarrow x = B \sin (\frac{2 π}{30} t)$

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