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The figure given below depicts a circular motion. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P
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The correct answer is:
$x\left( t \right)=Bcos\left( \frac{\pi }{15}t-\frac{\pi }{2} \right)$
Correct Option is : (B)
$x\left( t \right)=Bcos\left( \frac{\pi }{15}t-\frac{\pi }{2} \right)$
$x\left( t \right)=Bcos\left( \frac{\pi }{15}t-\frac{\pi }{2} \right)$
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