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

Step-by-step solution

The displacement x in simple harmonic motion relates to velocity v , angular frequency , and amplitude A through the equation v = A^2 - x^2 . Given parameters: m = 1 , kg , k = 4 , N/m , v = 0.2 , m/s , and A = 0.4 , m . The angular frequency is = k/m = 4 = 2 , rad/s . Substituting into the velocity equation: 0.2 = 2 0.4^2 - x^2 . Simplifying yields 0.1 = 0.16 - x^2 . Squaring both sides: 0.01 = 0.16 - x^2 . Solving gives x^2 = 0.15 , so x = 0.15 , m . The displacement corresponds to option B.