The mass $m$ attached to spring oscillate SHM with amplitude $5 \mathrm{~cm}$ shown in figure.


As given that spring constant $(k)=50 \mathrm{~N} / \mathrm{m}$ Mass attached $(m)=2 \mathrm{~kg}$
$\therefore$ Angular frequency $(\omega)$
$$
=\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5 \mathrm{rad} / \mathrm{second} \text {. }
$$
Let consider the displacement function is
$$
y(t)=A \sin (\omega t+\phi)
$$
where, $\phi=$ initial phase
But given at $t=0, y(t)=+A, y(0)=+A$
$\Rightarrow A \sin (\omega \times 0+\phi)=+A$ or $\sin (0+\phi)=1$
or $\sin \phi=1$ or $\phi=\frac{\pi}{2}$
$\therefore$ The desired equation become
$$
y(t)=A \sin \left(\omega t+\frac{\pi}{2}\right)=A \cos \omega t
$$
By putting $A=5 \mathrm{~cm}, \omega=5 \mathrm{rad} / \mathrm{sec}$. then
$$
y(t)=5 \sin \left(5 t+\frac{\pi}{2}\right)
$$
where, $t$ is in second and $y$ is in centimetre.