Question

Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e., clockwise or anti-clockwise) are indicated on each figure

Obtain the corresponding simple harmonic motions of the $x-\text{projection}$ of the radius vector of the revolving particle $P$, in $A$.

$A$)
$B$)

Answer 1

$A$)
Time Period, $T=2s$
Amplitude, $A=3cm$
At time, $t=0$,
Phase angle $\varphi =\pi$
Hence, the equation of simple harmonic motion for the $x-\text{projection}$ of $OP$, at the time $t$, is given by the displacement equation:

$x=A\sin [\frac{2\pi t}{T}+\varphi ]$

$=3\sin [\left(\frac{2\pi t}{2}\right)+\pi ]=-3\sin \left(\frac{2\pi t}{2}\right)$

Therefore,

$x=-3\sin \left(\pi t\right)cm$

Final Answer: $x=-3\sin \left(\pi t\right)cm$


$B$)
Time Period, $T=4s$
Amplitude, $A=2m$

At time $t=0$, P is at negative exterme. Thus, phase angle $\varphi =\frac{3\pi }{2}$.

Hence, the equation of simple harmonic motion for the x-projection of $OP$, at the time $t$, is given as:

$x=a\sin [\left(\frac{2\pi t}{T}\right)+\varphi ]=2\sin [\left(\frac{2\pi t}{4}\right)+\frac{3\pi }{2}]$

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

Final Answer: $x=-2\cos \left(\frac{\pi }{2}t\right)$