$\frac{d^2 x}{d t^2} =-c x = -\omega^2 x \quad(c>0)$

where; $\omega$ = angular frequency

Solution

$x(t)=A \cos \omega t+B \sin \omega t$

Constants A and B are determined by two conditions

(i) x(0), v(0)

(ii) $x(t_1), x(t_2)$

Physically : A spring mass system (Spring follow Hooke's law) performs SHM.

k = Spring Constant

m = mass of the black

g = gravitational acceleration

spring is stretches by l

What happens if we pull / push the mass by a distance y and release it.

$F_{spring} = -k(l-y)$

$F_m$ = mg

Displacement is downwards and the force is upwards.

Net force = $k(l-y)-mg$ = -ky upwards

or F= -ky downwards

Now F = mg

$m \frac{d^2 y}{dt^2}=-ky \text{or} \frac{d^2 g}{d t^2}=-\frac{k}{m} g$

$\omega^2=\frac{k}{m}$

We are using examples to show you how different systems perform simple harmonic motion.

In this system, take a string which is pulled so that it has a tension T.

$\frac{x}{l} \ll 1$

$F_{\text {net }} =T \sin \theta+T \sin \theta=2 T \sin \theta$

Since $\quad x \ll l$

$\sin \theta=\tan \theta \simeq\theta \simeq\frac{x}{l}$

$F_{\text {net }}=-2 T \frac{x}{l}$

Equation of motion: $m \ddot{x}=-\frac{2 T x}{l}$

$\ddot{x}=-\left(\frac{2 T}{l m}\right) x$

Angular frequency $\omega=\sqrt{\frac{2 T}{l m}}$

Time period $2 \pi \sqrt{\frac{\mathrm{lm}}{2 T}}$

$y(t)=A \cos \sqrt{\frac{2 T}{l m}} t+B \sin \sqrt{\frac{2 T}{l m}} t$

A wooden block floating in liquid.

Uniform surface area: A

m = mass of the block

$\rho$ = Density of liquid

l = Depth to which the block is submerged

g = Gravitational acceleration

Archimedes principle: A $\rho$ gl = mg

A $\rho$ l = m

$F_{\text{buoyancy}} = A (l+y)\rho$g, buoyancy force is always upwards.

$F_{gr}$ = - mg, gravitation force is always downwards.

$F_{net}$ = Ay$\rho$g, upwards.

depth = (l-y)

$F_{buoyancy}$ = A$\rho$g(l-y)

$F_{gr}$ = - mg

$F_{net}$ = - Ag$\rho$y

Equation of motion = m$\frac{a^2y}{at^2} = - Ag\rho$y

Mass of the liquid = (Al)$\rho$

Equation of motion = $\frac{md^2y}{d} = -2A$\rho$y

Al$\rho \frac{d^2y}{dt^2}$ = - 2A$\rho$gy

$\frac{d^2y}{dt^2}= - \frac{2g}{l}y$

$\omega^2 = \frac{2g}{l}$

Time = $2 \pi \sqrt{\frac{l}{2 g}}$

Simple pendulum m:

Time period for oscilation of a simple pendulum.

Motion of the pendulum is periodic.

Question: Is the motion simple harmonic motion?

Force in the x direction:

$F_\perp$ to string = mg $sin\theta$

for $\theta \ll 1 \quad, mg sin\theta \cong m g \theta$

$F_\perp = -\frac{m g x}{l}$

$m \frac{d^2x}{dt^2} = -\frac{mg}{l}x$

$\frac{d^2x}{dt^2} = - \frac{g}{l}x$

$\omega^2=\frac{g}{l} \quad \omega=\sqrt{\frac{g}{l}}$

Time period $T=2 \pi \sqrt{\frac{l}{g}}$

Find the length of a pendulum that has time period of 1 second.

$1=2 \pi \sqrt{\frac{l}{9}}$

$l=\frac{g}{4 \pi^2} \simeq \frac{1}{4} m$ = 25 cm

So for we have looked at Point masses performing simple harmonic motion.

What happens if we deal with extended bodies or rigid bodies.

Is it simple harmonic motion?

When dealing with extended bodies or rigid bodies, we use to torque equation.

A uniform rod of mass m and length l is pivoted at one end. It is hanging vertically in equalibrium. If displaced by an angle $\theta$ from the vertical, write is equation of motion when it is released.

Find if for $\theta$<< 1, it performs SHM and find its time period.

Torque = $\frac{mgl}{2} sin \theta$

I $\alpha$ = -$\frac{mgl}{2} sin \theta$, $\alpha = \frac{d^2\theta}{dt^2}$

Equation of motion

I $\frac{d^2\theta}{dt^2} = - \frac{mgl}{2} sin\theta$

If $\theta$<< 1, $sin \theta \simeq \theta$

$\frac{d^2\theta}{dt^2} = - \frac{mgl}{2I} \theta$

$\omega^2 = \frac{mgl}{2I}$, $I= \frac{ml^2}{3}$

$\omega^2=\frac{m g l}{2 \times \frac{ml}{3}}=\left(\frac{3 g}{2 l}\right)$

$T=\frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{2 l}{3 g}}$

Torque = mg r $sin \theta \simeq$ mgr$\theta$

Equation of motion: I$\alpha$ = - mgr$\theta$

I $\frac{d^2\theta}{dt^2} = - mgr \theta$

I = $\frac{mr^2}{2} + mr^2 = \frac{3mr^2}{2}$

$\frac{3mr^2}{2} \frac{d^2\theta}{dt^2} = -mgr\theta$

$\frac{d^2\theta}{dt^2} = - \frac{2g}{3r} \theta$

$\omega^2 = \frac{2g}{3r}, T= 2\pi \sqrt\frac{3r}{2g}$

Plasma: Collection of positive and negative charges.

Plasma Oscillation $\omega_p$ = plasma frequancy

Shown below is a collection of positive and negative changes ( positive ions and electrions) is slab geometry. The positive charges are fixed while negative charge are mobile. If the slab of negative charge is displaced as shown, its starts oscillation. find the frequency of oscillations.

These is a restoring force due to the electric field.

Let, n = number density of the charges.

Volume = Ax

Number of charges = nAx

Charge = - neAx, e: electronic charge.

Charge / area =$\sigma = \frac{neAx}{A} = nex$

$E = \frac{\sigma}{\epsilon} = \frac{nex}{\epsilon}$

$F=+(n A l) e \times \frac{nex}{\epsilon} = \frac{n^2Ale^2}{\epsilon}$

$m_e \ddot{x}=-\frac{n e^2 x}{\epsilon}$.

$\ddot{x}=-\frac{n e^2 }{m_e \epsilon} x$

Plasma frequency: $\omega_p^2 = \frac{n e^2 }{m_e \epsilon}$

$I \ddot{\theta}=I \alpha=\tau$

$\tau=-m g \frac{a}{\sqrt{2}} \sin \alpha \simeq-\frac{m g a}{\sqrt{2} } \theta \quad (\theta - small )$

$I \ddot{\theta}=\frac{-m g^a}{\sqrt{2}} \theta$

$\omega^2=\left(\frac{m g a}{\sqrt{2} I}\right)$

$I =I_{CM}+I_{\text{about CM}}$

$I =\frac{m a^2}{2}+\frac{m a^2}{6} =\frac{2 m a^2}{3}$

$\omega^2=\frac{m g^2}{\sqrt{2} \cdot \frac{2 p^2}{3}}=\left(\frac{3 g}{2 \sqrt{2} a}\right)$

$\omega=\left[\frac{3g}{2\sqrt{2} x}\right]^{1 / 2}$