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?

Step 1: Identify the given values and convert units. 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: $A = 4 cm \times \frac{1 m}{100 cm} = 0.04 \text{ m}$

Step 2: Calculate the angular frequency ($\omega$) of the oscillation. For a mass-spring system, the angular frequency is given by the formula: $\omega = \sqrt{\frac{k}{m}}$ Substitute the given values: $\omega = \sqrt{\frac{400 N/m}{0.25 kg}}$ $\omega = \sqrt{1600 s^{-2}}$ $\omega = 40 \text{ rad/s}$

Step 3: Calculate the velocity when the mass comes to the equilibrium position. 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: $v_{max} = (0.04 m) \times (40 rad/s)$ $v_{max} = 1.6 \text{ m/s}$

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