Question

Show that for small oscillations the motion of a simple pendulum is simple harmonic. Derive an expression for its time period. Does it depend on the mass of the bob?

Answer 1

By drawing a free body diagram as shown in the image (please find attached) we can set up the following differential equation -

$\frac{d^2\theta}{dt^2}=\frac{-m}{g}\sin\theta$

The general solution to this differential equation is very complicated, hence we use a small angle approximation - sin(x) = x

Therefore the differnential equation becomes:

$\theta''=-\omega^2\theta$

$\omega=\sqrt{\frac{m}{g}}$

This is a linear differential equation which solved for -

$\theta(t)=\cos(\omega t)+A\sin(\omega t)$

The time period can be solved for by considering the first equation -

$\theta''=-\omega^2\theta$

The second derivative of theta is given as the angular acceleration -

$\alpha=-\omega^2\theta$

Our acceleration can also be defined using newton's second law as -

$\alpha = \frac{F_e}{m}=-\frac{g}{l}\theta$

Comparing the equations, we obtain -

$\omega=\sqrt{\frac{m}{g}}$

The time period, as the trjaectory is linear combination of sines and cosines is given as -

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

As there is no mass terms in this expression, we can conclude the it does not depend on the mass of the bob.

edit : My answer may not be satisfactory, therefore I recommend checking out the video on pendulums made by Physics With Elliot.