Question

A 0.770-kg mass attached to a vertical spring of force constant 147 N/m oscillates with a maximum speed of 0.322 m/s. Calculate the period related to the motion of the mass. Calculate the amplitude. Calculate the maximum magnitude of the acceleration.

Answer 1

The angular frequency of the oscillation is,

        w = [k / m]^1/2

           = [147 N/m / 0.770 kg]^1/2

           = 13.8 rad/s

Thus, the period of the oscillation is,

               T = 2(pi)/w

                  = 2(pi) / (13.8 rad/s)

                  = 0.4547 s

                  = 0.45 s

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The amplitude of the oscillation is,

              A = v_max/ w

                 = 0.322 m/s / 13.8 rad/s

                 = 0.023 m

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Maximum acceleration is,

               a_max = w²A

                        = (13.8 rad/s)²(0.023 m)

                        = 4.44 m/s²