The figure shows the circular motion of a par | vedclass

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(D) From the figure,the radius of the circle is $A = 2 	ext{ cm}$.The period of revolution is $T = 1 	ext{ s}$.The angular frequency is $\omega = \f

Vedclass · Accepted Answer

$x = 2\,\cos \left( {2\pi t + \frac{\pi }{3}} \right)$

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(D) From the figure,the radius of the circle is $A = 2 	ext{ cm}$.The period of revolution is $T = 1 	ext{ s}$.The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{1} = 2\pi 	ext{ rad/s}$.At $t = 0$,the particle $P$ is at an angle of $60^\circ$ or $\frac{\pi}{3}$ radians with the positive $x$-axis.The $x$-projection of the position vector is given by $x(t) = A \cos(\omega t + \phi)$.At $t = 0$,$x(0) = A \cos(\phi) = 2 \cos(60^\circ) = 2 	imes 0.5 = 1 	ext{ cm}$.Substituting the values,$x(t) = 2 \cos(2\pi t + \frac{\pi}{3})$.Thus,the correct option is $D$.

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 $P$ is

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