The length of a simple pendulum is 0.79 m and the mass of the particle (the “bob”) at the end of the cable is 0.24 kg.

Question

The length of a simple pendulum is 0.79 m and the mass of

the particle (the “bob”) at the end of the cable is 0.24 kg. The pendulum

is pulled away from its equilibrium position by an angle of 8.50

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?

Answer 1

Let the mass of the bob, m = 0.24kg

the displaced angle, θ = 8.5°

Length of pendulum, L = 0.79m

a) the angular frequency, ω = √g/L = √(9.8m/s²/0.79m) = 3.5rad/s

b) At the lowest point (same as when the bob is at the center position), the velocity become maximum and the height tends to zero. This means that the total mechanical energy, E = KE + PE

where KE = Kinetic Energy and PE = Potential Energy. but PE = mgh = 0 [since height, h = 0 at center position]

==> E = KE = (1/2)mv²

c) Using KE = PE ==> (1/2)mv² = mgh

==> v = √(2gh) = √(2gL(1 - cosθ)) = √(2*9.8*.79(1- cos(8.5°)) = 0.46m/s as the bob's speed