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

Answer 1

Answer: (A)

Here, $T=30s$
At $t=0$, $OP$ makes an angle of $\frac{\pi }{2}$ with the $x$-axis.
After a time $t$, it covers an angle of $\frac{2\pi }{T}t$ in the clockwise sense, and makes an angle of $(\frac{\pi }{2}-\frac{2\pi }{T}t)$ with the $x$ -axis.
The projection of $OP$ on the $x$ -axis at time $t$ is given by
$x(t)=Bcos(\frac{\pi }{2}-\frac{2\pi }{T}t)$
$=Bsin(\frac{2\pi }{T}t)$
$x(t)=Bsin(\frac{2\pi }{30}t)(\because T=30s)$