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

Answer 1

Answer: (A)

$x\left(t\right) = B \,sin \left(\frac{2\pi}{30}t\right)$

Answer 2

Here, $T = 30 \,s$ 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 $x(t) = B\,cos (\frac{\pi}{2} - \frac{2\pi}{T} t )$ $= B \,sin (\frac{2\pi}{T}t)$ $x(t) = B \,sin (\frac{2\pi}{30} t) \quad (\because T = 30\,s)$