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

Q: 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

a 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 $SHM.$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$x\left( t \right) = B\,\sin \,\left( {\frac{{2\pi t}}{{30}} + \frac{\pi }{4}} \right)$
B$x\left( t \right) = B\,\cos \,\left( {\frac{{\pi t}}{{15}}} \right)$
C$x\left( t \right) = B\,\sin \,\left( {\frac{{\pi t}}{{15}} + \frac{\pi }{2}} \right)$
D$x\left( t \right) = B\,\cos \,\left( {\frac{{\pi t}}{{15}} + \frac{\pi }{2}} \right)$

Diffcult

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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