The length of a simple pendulum is 0.70 m and the mass of the particle (the “bob

Question

The length of a simple pendulum is

0.70 m and the mass of the particle (the “bob”) at the end of the cable is 0.21 kg. The pendulum is pulled away from its equilibrium position by an angle of 7.6° and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What
is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob’s speed as it passes through the lowest point of the swing?

Explanation / Answer

given length of the simple pendulum is l = 0.7 m
mass at the end of of cable, m = 0.21 kg
initial angle from equilibrium be theta0 = 7.6 degree
assuming that the motion is simple harmonic motion, and friction can be neglected
a) Angular frequency of simle pendulum = sqroot(g/L) = sqroot(9.81/0.7) = 3.7435 rad/s
b) as friction is not in play the total mechanical energy of the bob has to remain constant through out
again, KE at lowest point = PE at highest poinnt
so, PE at highest point = mgh
h = l*(1 - cos(thetao))
PE = 0.21*9.81*0.7(1 - cos(7.6)) = 0.012667 J = total energy of the pendulum
c) at the lowest point the speed be v
then 0.5mv^2 = 0.01267
v = 2*0.1266/0.21 = 0.12064 m/s

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