Question

The simple pendulum shown above consists of a 0.8 kg bob suspended from the end of a

l = 1.5 m long chord. While the pendulum is swinging, an obstacle is placed a distance of 3l/4

directly beneath the pivot. Find the period of the pendulum now that this obstacle has been

placed in its path.

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Answer 1

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Period of left portion of pendulum is, t_L = 2 approximately squareroot l/g period for right protion of pendulum is, T_r = 2 approximately squareroot l/4g in both halver, peridutum completes on only half of iB oscillation in one full oscillation. Therefore net time period is, T = T_1/2 + T_r/2 = approximately [squareroot l/g + (squareroot l/g 1/2] or T = 3/2 approximately squareroot l/g = 3/2 approximately squareroot 1.5 m/(9.8m/s^2) or t = 1.84s