Question

A simple pendulum in a clock is displaced to initial angle of 5.00° and released from rest. When the pendulum passes through its equilibrium position, the bob’s speed is 0.300 m/s. (a) What are the length, angular frequency, and period of the pendulum? (2.2s) (b) If the bob is displaced to initial angle of 10.0° instead, what are the angular frequency and period of the pendulum and the maximum speed of the bob? (c) If the same pendulum clock were on the Moon (g = 51.62 m/s2), what would its period be?

Our teacher gave this for our Homework, but never went over pendulums. thank you for the help!

Answer 1

a) by conservation of energy, KE= mgR[1- cos 5 degree]

                           v^2/2 = gR[1- cos 5 degree]

                           R = 0.5*0.3^2/[9.8*(1- cos 5 degree)]

                           = 1.21 m

period = 2 pi sqrt(1.21/9.8)

            = 2.20 s

angular frequency = 2 pi /2.2

                    = 2.86 rad/s

b) T = 2.2 s, w = 2.86 rad/s,

1.21 = 0.5*v^2/[9.8*(1- cos 10 degree)]

v = sqrt([9.8*(1- cos 10 degree)]*1.21/0.5)

= 0.6 m/s

c) period on moon = 2 pi* sqrt(1.21/51.62)

            = 0.962 s