Question

(1) A simple pendulum oscillates back and forth with a maximum angular displacement of θMax “ (pi/12) radians. At t = 1.2s, the pendulum is at its maximum displacement. The length of the pendulum is L = 35cm and the mass of the bob attached to the end of it is m = 300g.

(a) What is the angular frequency of this oscillation?

(b) What is the angular position as a function of time?

(c) What is the angular speed as a function of time?

(d) If we introduce damping to this oscillator, do you expect the total energy of the pendulum to increase, decrease, or stay the same?

(e) This simple pendulum is placed next to a physical pendulum that is a rod has a mass of m = 300g and length L = 35cm. Is the physical pendulum’s angular frequency larger, smaller, or the same as the simple pendulum?

Answer 1

a)
Angular frequency, = SQRT[g/L]
= SQRT[9.81/0.35]
= 5.294 rad/s

b)
The general equation for the oscillating amplitude is,
= max sin(t + c)
Where c is the phase shift in radians.

Given that at t = 1.2 s, = max = /12
Substituting,
/12 = /12 x sin(5.294 x 12 + c)
sin(5.294 x 12 + c) = 1
(5.294 x 12 + c) = /2
c = /2 - 5.294 x 12
= - 61.96 rad

So, the general equation is,
= /12 x sin(5.294 t - 61.96)

c)
Angular speed = d/dt
= d/dt[/12 x sin(5.294 t - 61.96)]
= /12 x 5.294 cos(5.294 t - 61.96)
= 1.386 cos(5.294 t - 61.96)

d)
Since damping is dissipative, total energy decreases.

e)
Angular velocity = SQRT[mgL_cm/I]
Where L_cm is the center of mass of the rod, L_cm = L/2 = 0.175 m
I is the moment of inertia of the rod about the pivot, I = 1/3 mL²
= 1/3 x 0.3 x (0.35)²
= 0.1225

Angular velocity = SQRT[0.3 x 9.81 x 0.175 / 0.1225]
= 2.05 rad/s