A spring of force constant 400 N/m is loaded with a mass 0.25 kg. The amplitude of oscillations is 4 cm. When mass comes to equilibrium position, its velocity is?

The force constant of the spring is $k = 400 \text{ N/m}$.
The mass is $m = 0.25 \text{ kg}$.
The amplitude of oscillations is $A = 4 \text{ cm}$. Convert this to meters:

$$A = 4 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 0.04 \text{ m}$$

For a mass-spring system, the angular frequency is given by the formula:

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

Substitute the given values:

$$\omega = \sqrt{\frac{400 \text{ N/m}}{0.25 \text{ kg}}}$$ $$\omega = \sqrt{1600 \text{ s}^{-2}}$$ $$\omega = 40 \text{ rad/s}$$

The velocity at the equilibrium position is the maximum velocity ($v_{max}$) in Simple Harmonic Motion (SHM). The formula for maximum velocity is:

$$v_{max} = A\omega$$

Substitute the values for amplitude and angular frequency:

$$v_{max} = (0.04 \text{ m}) \times (40 \text{ rad/s})$$ $$v_{max} = 1.6 \text{ m/s}$$

The final answer is $\text{1.6 m/s}$.