A spring executes S.H.M. with mass 10 kg attached to it. The force constant of spring is 10 N/m. If at any instant its velocity is 40 cm/s, the displacement at that instant is (Amplitude of S.H.M. is 0.5 m)

Step-by-step solution

Given m = 10 kg, k = 10 N/m, v = 40 cm/s = 0.4 m/s, and A = 0.5 m. The angular frequency of the spring-mass system is given by = k m = 10 10 = 1 rad/s. The velocity of a particle in S.H.M. is v = A^2 - x^2 . Substituting the given values: 0.4 = 1 (0.5)^2 - x^2 Squaring both sides: 0.16 = 0.25 - x^2 x^2 = 0.25 - 0.16 = 0.09 x = 0.3 m. Answer: 0.3 m