Question

Figure depicts a circular motion of the particle $P$. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated on the figure. Obtain the SHM motion of the $x$ -projection of the radius vector of the rotating particle $P$

• A. $R\sin \left[\frac{\pi }{2}-\frac{\pi }{15}t\right]$
• B. $R\sin \left[\frac{\pi }{15}t-\frac{\pi }{2}\right]$
• C. $R\cos \left[\pi -\frac{\pi }{15}t\right]$
• D. $R\cos \left[\frac{\pi }{15}t-\frac{\pi }{2}\right]$

Answer 1

Answer: (D)

In this case, at $t=0,$ OP makes an angle of $90^{\circ }$
or $\frac{\pi }{2}$ with the $x$ -axis. After a time $t,$ it covers an angle
$\frac{2\pi }{T}t$ in the clockwise sense and makes an angle of
$\left(\frac{\pi }{2}-\frac{2\pi t}{T}\right)$ with the $x$ -axis. The projection $OP$ on the $x$
axis at time $t$ is given by $x(t)=R\cos \left[\frac{\pi }{2}-\frac{2\pi t}{T}\right]$
For $T=30s,x(t)=R\cos \left[\frac{\pi }{2}-\frac{\pi }{15}t\right]$
$=R\cos \left[\frac{\pi t}{15}-\frac{\pi }{2}\right]$
as $\cos (-\theta )=\cos \theta$