The radius of the circle, the period of revolution, initial position and direction of revolution are indicated in the figure.

The $$y$$-projection of the radius vector of rotating particle $$P$$ will be:
1. $$y(t)=3 \cos \left(\dfrac{\pi \mathrm{t}}{2}\right)$$, where $$y$$ in m
2. $$y(t)=-3 \cos 2 \pi t$$ , where $$y$$ in m
3. $$y(t)=4 \sin \left(\dfrac{\pi t}{2}\right)$$, where $$y$$ in m
4. $$y(t)=3 \cos \left(\dfrac{3 \pi \mathrm{t}}{2}\right)$$,  where $$y$$ in m

Subtopic:  Phasor Diagram |
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NEET - 2019
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The figure shows the circular motion of a particle. The radius of the circle, the period, the sense of revolution, and the initial position are indicated in the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle $$P$$ will be:

1. $$x \left( t \right) = B$$ $$\text{sin} \left(\dfrac{2 πt}{30}\right)$$

2. $$x \left( t \right) = B$$ $$\text{cos} \left(\dfrac{πt}{15}\right)$$

3. $$x \left( t \right) = B$$ $$\text{sin} \left(\dfrac{πt}{15} + \dfrac{\pi}{2}\right)$$

4. $$x \left( t \right) = B$$ $$\text{cos} \left(\dfrac{πt}{15} + \dfrac{\pi}{2}\right)$$

Subtopic:  Simple Harmonic Motion | Phasor Diagram |
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The displacement $$x$$ of a particle varies with time $$t$$ as $$x = A sin\left (\frac{2\pi t}{T} +\frac{\pi}{3} \right)$$$\mathrm{}$The time taken by the particle to reach from $$x = \frac{A}{2}$$ to $$x = -\frac{A}{2}$$ will be:

 1 $$\frac{T}{2}$$ 2 $$\frac{T}{3}$$ 3 $$\frac{T}{12}$$ 4 $$\frac{T}{6}$$

Subtopic:  Phasor Diagram |
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