Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the 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
Diffcult
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Let angular velocity of the particle executing motion is $\omega$ and when it is at $Q$ makeks and angle $0$ as shown in the diagram.
Clearly, $0=\omega t$
Now, we can write $O R=O Q \cos (90-0)$
$=O Q \sin 0=O Q \sin \omega t$
$=r \sin \omega t|\because O Q=r|$
$\Rightarrow x=r \sin \omega t=B \sin \omega t[\because r=B]$
$B \sin \cdot \frac{2 \pi}{T} t=B \sin \left(\frac{2 \pi}{30} t\right)$
Clearly, this equation represents $SHM.$
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