Lecture 3 [24 March, 2020] PHY 115

PHY 115 

Wave and Oscillation

    ♦ Simple Pendulum:

    ♦ Compound Pendulum: any shape rigid body...

  

    ♦ Differential Equation for Simple Harmonic Motion:

  (1) Force along the string =  $Mg\cos \theta$

    (2) Force perpendiculer to the string = $Mg\sin\theta$

The component $Mg\cos\theta$ balance the tension.

  $$T = Mg \cos\theta$$  

The Force active on the oscillating particle is,

  $$\begin{align*} F = & - Mg\cos\theta \\ \textrm{when,} & \theta \rightarrow 0, ~\textrm{ the}~ \sin\theta = \theta \\ F = & - Mg\theta \tag{1}\label{eq:1} \end{align*}$$   

We know,  $F = ma \tag{2}\label{eq:2}$  

The linear displacement,  $$\begin{align*} y = & l\theta \\ \implies \frac{dy}{dt} = & \ \frac{d\theta}{dt} \\ \implies \frac{d^2 y}{d t^2} = & a = \ \frac{d^2 \theta}{d t^2} \tag{3}\label{eq:3} \end{align*}$$  

Comparing equation (1) and (2),

  $$F = Ml \frac{d^2 \theta}{d t^2}$$  

Now using equation (1) 

  $$\begin{align*} & Ml \frac{d^2 \theta}{d t^2} = - Mg\theta \\ \implies & \frac{d^2 \theta}{d t ^2} + \frac{g}{l}\theta = 0 \end{align*}$$  

comparing with differential equation of SHM,

  $$\begin{align*} & \omega^2 = \frac{g}{l} \\ \implies & \left(\frac{2\pi}{T } \right)^2 = \frac{g}{l} \\ \implies & t = 2 \pi \sqrt{\frac{l}{g}} \end{align*}$$  

HW: Find out T for compound pendulum.