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Question

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 mL=2.23\text{ m}
m=6.74 kgm=6.74\text{ kg}
v=2.06 m sv=2.06\frac{\text{ m}}{\text{ s}}

Simple pendulum equation of motion:

Iα=mgLsinθmL2d2θdt2=mgLθd2θdt2=gLθ\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:

ω=gL\omega=\sqrt{\dfrac{g}{L}}

Calculating period:

T=2πω=2πLg=2π2.239.81=3 sT=\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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