Question

A mass of 1.62 kg stretches a vertical spring 0.318 m. If the spring is stretched an additional 0.137 m and released, how long does it take to reach the (new) equilibrium position again? ______s

Answer 1

The spring constant of the spring is,

              k = mg/y

                 = (1.62 kg)(9.8 m/s²) / (0.318 m)

                 = 49.92 N/m

The angular frequency is,

          w = [k / m]^1/2

             = [(49.92 N/m) / 1.62 kg]^1/2

             = 5.55 rad/s

The period is,

      T = 2(pi)/ w

         = 2(pi) / w

         = 2(pi) / 5.55 rad/s

         = 1.13 s

The required, time period is T/4 = 0.283 s