described by the equation:

$$ \theta(t) = \theta_0 \sin (\omega t) $$

where $\theta_0$ is the maximum angular displacement (or amplitude) and $\omega$ represents the angular frequency.

The angular frequency ($\omega$) is directly related to the time period ($T$) of the pendulum's motion:

$$ \omega = \frac{2 \pi}{T} $$

To find the angular velocity, we differentiate the angular position:

$$ \frac{d}{dt}(\theta(t)) = \theta_0 \omega \cos (\omega t) $$

The angular momentum $L(t)$ of the pendulum about its point of suspension is expressed as:

$$ L(t) = I \cdot \theta(t) $$

For a simple pendulum, the moment of inertia ($I$) of the bob with mass $m$ and length $l$ is:

$$ I = m l^2 $$

Thus,

$$ L(t) = m l^2 (-\theta_0 \omega \sin (\omega t)) $$

The amplitude $A$ of the angular momentum is the maximum value of $L(t)$, which occurs when $\sin (\omega t) = \pm 1$:

$$ A = m l^2 \theta_0 \omega $$

Given that the pendulum length $l$ is not fixed, it can be expressed in terms of $T$ using the formula for the pendulum's time period:

$$ T = 2 \pi \sqrt{\frac{l}{g}} $$

Rearranging gives:

$$ l = \frac{T^2}{4 \pi^2} g $$

$$ l^2 = \frac{T^4}{16 \pi^4} g^2 $$

Substituting this into the expression for $A$:

$$ A = m \frac{T^4}{16 \pi^4} g^2 \theta_0 \frac{2 \pi}{T} $$

Simplifying results in:

$$ A \propto T^3 $$

Simple Harmonic Motion

Continue this paper