Find the equation of motion of a simple pendulum (see Chapter 7, Problem 2.13), that is, the differential equation for$\theta$ Vas a function of$t$. Show that, for small,$\theta$ this is approximately a simple harmonic motion equation, and find$\theta$if$\theta ={\theta }_0,d\theta /dt=0$when$t=0$.

A heavy particle is attached by a light string to a fixed point and oscillating under gravity constitutes a simple pendulum.

Force is given by:

$F=ma$

Here, $a$is acceleration of the bob,$m$is mass of bob.

Let$O$be the fixed point and $I$be the length of the string and $A$be the position of the bob initially.

If P be the position of the bob at any time$\text{t}$, such that$\mathrm{arc}AP=s$and $\angle AOP=\theta$then .$=l\theta$

The equation of motion along PT is

$\left[\text{Angle}=\frac{\text{arc}}{\text{radius}}\Rightarrow \theta =\frac{s}{l}\Rightarrow s=l\theta \right]$

The equation of motion along PT is

localid="1664289458208" $\begin{array}{c}F=m\frac{d^{2}s}{d{t}^2}\\ =-mg\sin \theta \end{array}$

Force,$F=ma$

$=m\frac{d^{2}s}{d{t}^2}\{a=\frac{d^{2}s}{d{t}^2}\}$

and $F=mg$here is the acceleration due to gravity this force acts vertically (in the direction of) downward direction and is component of force.

$\begin{array}{c}\frac{d^{2}s}{d{t}^2}=-g\sin \theta \\ \frac{d^2\left(I\theta \right)}{d{t}^2}=-g\sin \theta \\ I\frac{d^2\theta }{d{t}^2}=-g\sin \theta \\ \frac{d^2\theta }{d{t}^2}=\frac{-g}{I}\sin \theta \end{array}$

Which is the equation of motion of a simple pendulum.

The equation is:

$\frac{d^2\theta }{d{t}^2}=\frac{-g}{I}(\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}\dots \dots )$

For small,$\frac{{\theta }^3}{3!}=0,\frac{{\theta }^5}{3!}=0\dots \dots \dots$

$\frac{d^2\theta }{d{t}^2}=\frac{-g}{I}\theta$

$\frac{d^2\theta }{d{t}^2}\alpha (-\theta )$Where$\frac{-g}{I}$is constant,

Which is the motion of a simple harmonic motion.

$\frac{d^2\theta }{d{t}^2}=\frac{-g}{I}\theta$

$(D^2+\frac{-g}{l})\theta =0$Where$D=\frac{d}{dt}.$

$\begin{array}{c}\text{A.E.is}{m}^2+\frac{g}{l}=0\\ m^2=-\frac{g}{l}\\ m=\pm \sqrt{-\frac{g}{l}}\\ m=\pm i\sqrt{\frac{g}{l}}\end{array}$

The angle is given as

$\theta =e^{0.2}[C_1\cos \sqrt{\frac{g}{l}}t+C_2\sin \sqrt{\frac{g}{l}}t]$

$\theta =C_1\cos \sqrt{\frac{g}{l}}t+C_2\sin \sqrt{\frac{g}{l}t}$ ……. (1)

$\frac{d\theta }{dt}=-C_1\sqrt{\frac{g}{l}}\sin \sqrt{\frac{g}{l}}t+C_2\sqrt{\frac{g}{l}}\cos \sqrt{\frac{g}{l}t}$ ……. (2)

Put $\theta ={\theta }_{0}t=0$in equation (1)

$\begin{array}{l}{\theta }_0=C_1\cdot 1+C_2\cdot 0\\ {\theta }_0=C_1\end{array}$

$\text{Put}\frac{d\theta }{dt}=0,t=0$in equation (2),

$\begin{array}{l}0=0+C_2\sqrt{\frac{g}{l}}\\ C_2=0\end{array}$

From equation (1),

$\begin{array}{l}\theta ={\theta }_0\cos \sqrt{\frac{g}{l}}t+0\cdot \sin \sqrt{\frac{g}{l}t}\\ \theta =C_1\cos \sqrt{\frac{g}{l}}t\end{array}$