1. Set up the motion equation

Let $x$ be the displacement of the block from its initial equilibrium position.

• Applied constant force $= F$
• Spring restoring force $= kx$ (opposite to displacement)

So the net force is

$m\ddot x = F-kx$

or

$m\ddot x + kx = F$

This is SHM about a new equilibrium position.

2. Find the new equilibrium position

At the new equilibrium,

$kx_0 = F \implies x_0 = \frac{F}{k}$

Thus the block performs SHM about $x=\frac{F}{k}$.

3. Use initial conditions

Initially, the block is at rest at the old equilibrium position:

$x=0, \quad v=0$

Relative to the new equilibrium, the initial displacement is

$A = \left|0-\frac{F}{k}\right| = \frac{F}{k}$

Hence amplitude of SHM is

$A=\frac{F}{k}$

Angular frequency is

$\omega = \sqrt{\frac{k}{m}}$

4. Maximum speed in SHM

For SHM,

$v_{\max} = A\omega$

Therefore,

$v_{\max} = \frac{F}{k}\sqrt{\frac{k}{m}} = \frac{F}{\sqrt{mk}}$

5. Check options

$v_{\max} = \frac{F}{\sqrt{mk}}$

So the correct option is:

D

6. Comparison with stored answer

Stored correct answer: D

Derived answer: D

They agree.