Appendix B:      Natural Frequency of Simple Pendulum (Beyond Syllabus)

To derive the pendulum’s period of oscillation, it is best to analyze it as a rotational motion.

We will still be applying N2L, but Instead of force F, inertia m and acceleration a, we have to work with torque \displaystyle \displaystyle \tau , moment of inertia I (a body’s resistance to rotate, which for the pendulum is mL2) and angular acceleration \displaystyle \displaystyle \ddot{\theta }.

\displaystyle \displaystyle \begin{aligned}{{\tau }_{{net}}}&=\mathrm{I}\ddot{\theta }\\mg\sin \theta \times L&=m{{L}^{2}}\ddot{\theta }\\\ddot{\theta }&=\frac{{g\sin \theta }}{L}\end{aligned}

If θ is small, we canmake the \displaystyle \sin \theta \approx \theta approximation. (Yes. This means that our formula is valid only for small amplitude oscillations)

\displaystyle \displaystyle \ddot{\theta }=\frac{g}{L}\theta

Comparing this with the SHM equation \displaystyle a=-{{\omega }^{2}}x (\displaystyle \displaystyle \ddot{\theta }=-{{\omega }^{2}}\theta ), we can deduce that

\displaystyle {{\omega }^{2}}=\frac{g}{L}

Hence, the natural frequency is \displaystyle {{f}_{n}}=\frac{1}{{2\pi }}\sqrt{{\frac{g}{L}}}  and the natural period is \displaystyle {{T}_{n}}=2\pi \sqrt{{\frac{L}{g}}}.

From \displaystyle \displaystyle {{\omega }^{2}}=\frac{{\ddot{\theta }}}{\theta }, we can see that the natural frequency depends on the (angular) acceleration per unit (angular) displacement ratio. A higher g increases the restoring force (torque), whereas a shorter L decreases the (moment of) inertia, both resulting in a higher (angular) acceleration per unit (angular) displacement.

The mass of the pendulum, surprisingly, does not affect the period. This is because the restoring torque \displaystyle (mg\sin \theta ) and moment of inertia (mL2) are both proportional to m. So m cancels itself out when it comes to angular acceleration. (This is similar to how all object free fall at g regardless of mass).

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