The figure shows the circular motion of a par | vedclass

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(C) The particle moves in a circle of radius $B$ with angular velocity $\omega = \frac{2\pi}{T}$.At $t=0$,the particle is on the positive $y$-axis. As

Vedclass · Accepted Answer

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

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(C) The particle moves in a circle of radius $B$ with angular velocity $\omega = \frac{2\pi}{T}$.At $t=0$,the particle is on the positive $y$-axis. As it moves clockwise,the angle $	heta$ with the positive $y$-axis at time $t$ is $	heta = \omega t = \frac{2\pi t}{T}$.The angle $\phi$ that the radius vector makes with the positive $x$-axis is $\phi = \frac{\pi}{2} - 	heta = \frac{\pi}{2} - \omega t$.The $x$-projection of the radius vector is $x(t) = B \cos(\phi) = B \cos\left( \frac{\pi}{2} - \omega t ight)$.Using the trigonometric identity $\cos(\frac{\pi}{2} - 	heta) = \sin(	heta)$,we get $x(t) = B \sin(\omega t) = B \sin\left( \frac{2\pi t}{T} ight)$.

The figure shows the circular motion of a particle. The radius of the circle is $B$. The particle starts at $t=0$ from the positive $y$-axis and moves clockwise. The simple harmonic motion of the $x$-projection of the radius vector of the rotating particle is given by:

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