A spring executes S.H.M. with mass 1 kg attached to it. The force constant of the spring is $4 \mathrm{~N} / \mathrm{m}$. If at any instant its velocity is $20 \mathrm{~cm} / \mathrm{s}$, the displacement at that instant is (Amplitude of S.H.M. is 0.4 m )

The displacement $x$ in simple harmonic motion relates to velocity $v$, angular frequency $\omega$, and amplitude $A$ through the equation $v = \omega \sqrt{A^2 - x^2}$.

Given parameters: $m = 1\, \text{kg}$, $k = 4\, \text{N/m}$, $v = 0.2\, \text{m/s}$, and $A = 0.4\, \text{m}$.

The angular frequency is $\omega = \sqrt{k/m} = \sqrt{4} = 2\, \text{rad/s}$.

Substituting into the velocity equation: $0.2 = 2 \sqrt{0.4^2 - x^2}$.

Simplifying yields $0.1 = \sqrt{0.16 - x^2}$.

Squaring both sides: $0.01 = 0.16 - x^2$.

Solving gives $x^2 = 0.15$, so $x = \sqrt{0.15}\, \text{m}$.

The displacement corresponds to option B.