What is the Q Factor and Resonance of a Simple Pendulum?

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The Q factor, or quality factor, of a simple pendulum indicates the energy loss per cycle, with a higher Q signifying lower damping. For a pendulum with a length of 0.50 m and a Q of 400, the amplitude decreases by two-thirds over time, which can be calculated using the Q factor. The initial energy loss rate in watts can be determined based on the pendulum's mass of 0.20 kg and the initial amplitude of 3.0 cm. To achieve resonance with a sinusoidal driving force, the driving frequency must be very close to the pendulum's natural frequency, which is derived from its angular frequency. Understanding the relationships between angular frequency, amplitude, and energy is crucial for analyzing the pendulum's motion.
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I have no idea what the Q factor is in the following problem:

Consider a simple pendulum (point mass bob) 0.50 m long with a Q of 400.

How long does it take for the amplitude (assumed small) to decrease by two-thirds?

If the amplitude is 3.0 cm and the bob has mass 0.20 kg, what is the initial energy loss rate of the pendulum in watts? (The answer should have a negtive sign.)

If we are to stimultate resonance with a sinusoidal driving force, how close must the driving frequency be to the natural frequency of the pendulum?
 
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Q is the "quality factor" and is a measure of the energy loss per cycle:

Q = \omega \left| \frac {E}{\Delta E} \right|

Low Q means high damping and high Q means low dampiing. You should be able to take it from there!
 
How is the angular frequency related to amplitude and energy?
 
The angular frequency is related to the length of the pendulum:

\omega = \sqrt \frac {g}{L}

and the amplitude will look something like

x = x_0 \cos \omega t

from which you can calculate the velocity and kinetic energy (and you can also obtain the potential energy).
 
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