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. \(y(t)=3 \cos \left(\frac{\pi \mathrm{t}}{2}\right)\), where \(y\) in m
2. \(y(t)=-3 \cos 2 \pi t\) , where \(y\) in m
3. \(y(t)=4 \sin \left(\frac{\pi t}{2}\right)\), where \(y\) in m
4. \(y(t)=3 \cos \left(\frac{3 \pi \mathrm{t}}{2}\right) \),  where \(y\) in m

Subtopic:  Phasor Diagram |
 74%
From NCERT
NEET - 2019
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NEET 2023 - Target Batch - Aryan Raj Singh
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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(t)=B sin2πt30

2. x(t)=B cosπt15

3. x(t)=B sinπt15+π2

4. x(t)=B cosπt15+π2

Subtopic:  Simple Harmonic Motion | Phasor Diagram |
 52%
From NCERT
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The displacement \( x\) of a particle varies with time \(t\) as \(x = A sin\left (\frac{2\pi t}{T} +\frac{\pi}{3} \right)\)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 |
From NCERT
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