Question

A spring with spring constant 14.5 N/m hangs from the ceiling. A 510 g ball is attached to the spring and allowed to come to rest. It is then pulled down 6.10 cm and released.

What is the time constant if the ball's amplitude has decreased to 3.90 cm after 48.0 oscillations?

Answer 1

Given

Mass of the ball , m = 0.51 kg

Spring constant , k = 14.5 N/m

Initial Amplitude , A = 6.10 cm

The frequency of the single oscillation,  f = (1/2π) √k/m

                                                             = (1/2π) √14.5 N/ m  / 0.51 kg

                                                         f = 0.85 Hz

Time period for 1 oscillation is , T = 1/ f = 1/ 0.85 H z

                                                   T = 1.17 s

For 48 oscillations, the time period is , T = 48 *1.17 s = 56.16 s

Amplitude has decreased to 3.90 cm in 56.16 s, the time constant is

calculated using

             x(t)   =    x_m e ^-bt/2m 

       0.039 m  =  (0.061 m) e ^{-b(56.16s) / 2(0.51 kg)}

      e ^{-b(56.16s) / 2(0.51 kg)} =     0.6393

            - 56.16 b / 1.02 = ln (0.6393)

                - 56.16 b = - 0.4473 * 1.02

                             b = 0.00812 kg /s

          Time constant , τ = 2 m / b = 2(0.51 kg) / (0.00812 kg /s)

                                  τ = 125.615 s