Question

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

(a) What's the angular frequency of this oscillation?
(b) What's the angular position as a function of time?
(c) What's the angular speed as a function of time?
(d) If damping is introduced to this oscillator, would 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's a rod with 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) The angular frequency of oscillation for a simple pendulum is given by,

(b) Now the angular position of the pendulum is given by,

The angular position is 0.26sin(5.29t) rad.

(c) The angular speed is given by,

Therefore angular speed is 1.37cos(5.29t) rad/s.

(d) The energy will remain contant (Conservation of energy) but the dissipative energy increases due to addition of damper, which means that the maximum displacement will reduce.

(e) For a rod, we have to consider the center of mass, for finding the oscillation frequency, as you know that the angular frequency is given by,

From the equation we can say that the angular frequency is inversely proportional to the length L/ distance of center of mass.

For a rod of length 35 cm, the center of mass will be at 17.5 cm, which leads to decrease in L. Therefore the frequency of physical pendulum will be more than the simple pendulum