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(D) The general equation for $SHM$ is $x = A \cos(\omega t + \phi)$.At $t = 0$,the particle is at $x = A/2$ and moving towards the left (towards the m

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The general equation for $SHM$ is $x = A \cos(\omega t + \phi)$.At $t = 0$,the particle is at $x = A/2$ and moving towards the left (towards the mean position $O$).Substituting $t = 0$ and $x = A/2$ into the equation: $A/2 = A \cos(\phi)$,which gives $\cos(\phi) = 1/2$.This implies $\phi = \pm 60^{\circ}$ or $\pm \pi/3$ radians.Since the particle is moving towards the left (towards the negative direction) from a positive displacement,the velocity $v = dx/dt = -A\omega \sin(\omega t + \phi)$ must be negative at $t = 0$.Thus,$-A\omega \sin(\phi)  0$.Therefore,$\phi = +60^{\circ} = \pi/3$.Substituting $\omega = 2\pi/T$ and $\phi = \pi/3$,the equation of motion is $x = A \cos \left( \frac{2\pi}{T}t + \frac{\pi}{3} \right)$.

$A$ particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards the left as shown in the figure. If the time period of $SHM$ executed about $O$ is $T$ and the amplitude is $A$,then the equation of motion of the particle is:

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