Period $(T)=2 \,s$
$\omega=\frac{2 \pi}{2}=\pi \,rad / s$
When block just represent from a piston, maximum acceleration must be equal to $g$.
$g=-\omega^2 x$
Acceleration is maximum when $x=A$
$g=-\omega^2 A$
or $A=\frac{9.8}{\pi^2}$
Maximum velocity $=A \omega$
$=\frac{9.8}{\pi^2} \times \pi$
$=\frac{9.8}{\pi} \,m / s$
$=3.119 \,m / s =3.12 \,m / s$
Statement $I :$ A second's pendulum has a time period of $1$ second.
Statement $II :$ It takes precisely one second to move between the two extreme positions.
In the light of the above statements, choose the correct answer from the options given below:


$ x = 2 \sin \omega t \,;$ $ y = 2 \sin \left( {\omega t + \frac{\pi }{4}} \right)$
The path of the particle will be :