Question

A simple pendulum of length `40 cm` oscillates with an angular amplitude of `0.04` rad. Find (a) the time period, (b) the linear amplitude of the bob, (c) the speed of the bob when the spring makes `0.02` rad with the vertical and (d) the angular acceleration when the bob is in momentary rest. Take `g = 10 m//s^(2)`

Answer 1

(a) The angular frequency is `omega = sqrt(g//l) = sqrt((10m//s^(2))/(0.4m)) = 5s^(-1)`
the time period is `(2pi)/(omega) = (2pi)/(5s^(-1)) = 1.26 s`.
(b) Linear amplitude `= 40 cm xx 0.04 = 1.6 cm`
(c) Angular speed at displacement `0.02` rad is
`omega = (5c^(-1)) sqrt((0.04)^(2) - (0.02)^(2)) rad = 0.17 rad//s`.
where speed of the bob at this instant
`= (40 cm) xx 0.175^(-1) = 6.8 sm//s`.
(d) At momentary rest, the bob is in extreme position.
Thus, the angular accelertion
`alpha = (0.4 rad)(25 s^(-2)) = 1 red//s^(2)`.