Question

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

Answer 1

Answer: (A)

Given, $T=30 s , O Q=B$. The projection of the radius vector on the diameter of the circle when a particle is moving with uniform angular velocity $(\omega)$ on a circle of reference is $SHM$. Let the particle go from $P$ to $Q$ in time $t$. Then $\angle POQ =\omega t=\angle OQP$. The projection of radius $OQ$ on $x$ -axis will be $O R=x(t)$ say.

$\ln \Delta O Q R, \sin \omega t=\frac{x(t)}{B}$
or $x(t)=B \sin \omega t=B \sin \frac{2 \pi}{T} t=B \sin \frac{2 \pi}{30} t$