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# A simple pendulum with a length of 2.23 m and a mass of 6.74 kg is given an initial speed of 2.06 m/s at its equilibrium position. Assume it undergoes simple harmonic motion. Determine (a) its period, (b) its total energy, and (c) its maximum angular displacement.

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We are given following data for a simple pendulum:

$L=2.23\text{ m}$
$m=6.74\text{ kg}$
$v=2.06\frac{\text{ m}}{\text{ s}}$

Simple pendulum equation of motion:

\begin{align*} I\alpha&=-mgL\sin{\theta}\\\\ mL^2\dfrac{d^2\theta}{dt^2}&=-mgL\theta\\\\ \dfrac{d^2\theta}{dt^2}&=-\dfrac{g}{L}\theta\\ \end{align*}

Where:

$\omega=\sqrt{\dfrac{g}{L}}$

Calculating period:

$T=\dfrac{2\pi}{\omega}=2\pi \sqrt{\dfrac{L}{g}}=2\pi \sqrt{\dfrac{2.23}{9.81}}=\boxed{3\text{ s}}$

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