The displacement-time relation for a particle can be expressed as $\mathrm{y}=0.5\left[{\cos }^2\left(\mathrm{n\pi t}\right)-{\sin }^2\left(\mathrm{n\pi t}\right)\right]$
This relation shows that

$\begin{array}{l} \mathrm{y}=0.5\left[{\cos }^2\left(\mathrm{n\pi t}\right)-{\sin }^2\left(\mathrm{n\pi t}\right)\right]=0.5\cos 2\mathrm{n\pi t}\\ \frac{\mathrm{dy}}{\mathrm{dt}}=-0.5\times 2\mathrm{n\pi }\times \sin 2\mathrm{n\pi t}\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}=-(0.5)(2\mathrm{n\pi }{)}^2\cos 2\mathrm{n\pi t}=-4{\mathrm{n}}^2{\mathrm{\pi }}^2\mathrm{y} ........\left(\mathrm{i}\right)\\ \mathrm{i}.\mathrm{e}., \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}\propto -\mathrm{y}\end{array}$

i.e., particle is executing SHM with amplitude 0.5 m, i.e., choice (a) is correct.
As standard equation of SHM is

     $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}=-{\mathrm{\omega }}^2\mathrm{y}$ …….(ii)

Hence,  $\mathrm{\omega }=2\mathrm{n\pi }$

For a second's pendulum, T = 2 s

Hence, ${\mathrm{\omega }}^{\mathrm{\prime }}=\frac{2\mathrm{\pi }}{\mathrm{T}}=\mathrm{\pi }=2\mathrm{n}$ times that of a second's pendulum.

i.e., choice (b) is not correct.

    ${\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}_{max}=\mathrm{n\pi }$

i.e., choice (c) is also correct.