The spring -block system will perform SHM about the mean position with an amplitude $5\mathrm{cm}$.



Given, spring constant $\mathrm{k}=50 \mathrm{N}/\mathrm{m}$

m = mass attached = 2 kg

$\therefore \text{Angular frequency}\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5\text{rad/s}$

Assuming the displacement function

y(t) = A sin (ωt + ϕ)

where, ϕ = initial phase

But given at t = 0, y(t) = +A

∴               y(0) = + A = A sin (ω × 0 + ϕ)

or                $\text{sin}\varphi =1\Rightarrow \varphi =\frac{\pi }{2}$

&the desired equation is

$y\left(t\right)=A\text{sin}\left(\omega t+\frac{\pi }{2}\right)$

= A cos ωt

Putting A = 5 cm, ω = 5 rad/s

We get,                  y(t) = 5 sin (5t + π/2)

where, t is in second and y is in centimetre.