$A$ mass of $2 \, kg$ is attached to a spring with a spring constant of $50 \, N/m$. The block is pulled to a distance of $5 \, cm$ from its equilibrium position at $x = 0$ on a horizontal frictionless surface and released from rest at $t = 0$. Write the expression for its displacement at any time $t$.

(N/A) The spring-block system executes simple harmonic motion $(SHM)$ with an amplitude $A = 5 \, cm$ from the mean position.
Given:
Spring constant $k = 50 \, N/m$
Amplitude $A = 5 \, cm = 0.05 \, m$
Mass $m = 2 \, kg$
The angular frequency $\omega$ is given by:
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5 \, rad/s$
The general equation for displacement in $SHM$ is:
$x(t) = A \sin(\omega t + \phi)$
At $t = 0$,the block is at its maximum displacement $x = A$ (since it is pulled to $5 \, cm$ and released from rest).
$x(0) = A \sin(\phi) = A$
$\sin(\phi) = 1 \implies \phi = \frac{\pi}{2} \, rad$
Substituting the values into the general equation:
$x(t) = 5 \sin(5t + \frac{\pi}{2})$
Since $\sin(\theta + \frac{\pi}{2}) = \cos(\theta)$,
$x(t) = 5 \cos(5t)$
where $x$ is in $cm$ and $t$ is in seconds.