Question

Figure shows the circular motion of a particle. The radius of the circle, the time period $T$, 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. $x\left(t\right)=\text{B sin}\left(\frac{2\pi t}{30}\right)$
• B. $x\left(t\right)=\text{B cos}\left(\frac{\pi t}{15}\right)$
• C. $x\left(t\right)=\text{B sin}(\frac{\pi t}{15}+\frac{\pi }{2})$
• D. $x\left(t\right)=\text{B cos}(\frac{\pi t}{15}+\frac{\pi }{2})$

Answer 1

The correct option is A $x\left(t\right)=\text{B sin}\left(\frac{2\pi t}{30}\right)$

Given, $T=30\text{s},OQ=B$. The projection of the radius vector on the diameter of the circle when a particle is moving with uniform angular velocity $\left(\omega \right)$ on a circle of reference is SHM. Let the particle go from $P$ to $Q$ in time t.


Then, $\angle POQ=\omega t=\angle OQR$
The projection of radius OQ on X - axis will be $OR=x\left(t\right)$ say.
In $\Delta OQR,\text{sin}\omega t=\frac{x\left(t\right)}{B}$
$x\left(t\right)=\text{B sin}\omega t=\text{B sin}\frac{2\pi }{T}t=\text{B sin}\frac{2\pi }{30}t$