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

Answer 1

Answer: (A)

x(t) = 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 $\left( \frac{\pi}{2} - \frac{2\pi}{T}t \right)$ with the x axis

The projection of OP on the axis at time t is given by

$x(t) = B\cos\left( \frac{\pi}{2} - \frac{2\pi}{T}t \right) = B\sin\left( \frac{2\pi}{T}t \right)$

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