Correct Answer is Option (A)

$\frac{d^{2}x}{d{t}^2}=-\alpha x=-{\omega }^{2}x$

$\Rightarrow \omega =\sqrt{\alpha }$ $$ or $$ $T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\alpha }}$

Correct Answer is Option (B)

Given that, initial angular velocity = ${\omega }_0$

and at any instant time t, angular velocity = $\omega$

So when displacement is x then the resultant acceleration

f = $\left({\omega }_0^2-{\omega }^2\right)x$

So the external force, F = $m\left({\omega }_0^2-{\omega }^2\right)x$ ............(i)

But given that $F\propto \cos \omega t$

From (i) we get,

$m\left({\omega }_0^2-{\omega }^2\right)x\propto \cos \omega t$ .........(ii)

From equation of SHM we know,

$x=A\sin \left(\omega t+\varphi \right)$

When t = 0 then x = A

$\therefore$ A = $A\sin \left(\varphi \right)$

$\Rightarrow A=\frac{\pi }{2}$

$\therefore$ $x=A\sin \left(\omega t+\frac{\pi }{2}\right)=A\cos \omega t$

Putting value of x in (ii), we get

$m\left({\omega }_0^2-{\omega }^2\right)A\cos \omega t\propto \cos \omega t$

$\Rightarrow A\propto \frac{1}{m\left({\omega }_0^2-{\omega }^2\right)}$

Correct Answer is Option (C)

$x=4\left(\cos \pi t+\sin \pi t\right)$

$=\sqrt{2}\times 4\left(\frac{\sin \pi t}{\sqrt{2}}+\frac{\cos \pi t}{\sqrt{2}}\right)$

$x=4\sqrt{2}\sin \left(\pi t+{45}^{\circ }\right)$