What is a Wave?

Is it particles moving from one place to another?

When a person is speaking to another person, sound travels but the air in better dos not.

A wave does not represent movement of particles.

A wave is a disturbance that travels from one place to another!

Dose a wave mean that the distarbance has to travel from one place to another?

wave is a distarbance created at one place and travelling to onther place (travelling waves)

An extanded distarbance (stationary waves)

$y(x,t) = A \sin 2\pi (\frac{x}{\lambda}-ft)$

or $y(x,t) =A \cos 2\pi (\frac{x}{\lambda}-ft)$

$v= f\lambda$

where:

$\lambda$ = wavelength

f = frequency

$y(x)= a^2-x^2 \quad |x|\geq a$

$\quad$ = 0 otherwise

$y(x,y)= a^2 - (x-x_0)^2 \quad$ for $|x-x_0|\geq a$

$\quad$ =0 otherwise

$y(x) = a^2- x^2 \quad |x|\geq a$

$\quad$ = 0 otherwise

$y(x, t)=a^2-(x-x_0)^2 \quad|x-x_0| \leqslant a$

$\quad$ = 0

$a^2-(x-v t)^2 \mid x-vt\mid \leqslant$

$\quad$ = 0 otherwise

$y(x, t) = a^2-(x+v t)^2 \quad |x+v t|\leqslant a$

$\quad$ = 0 otherwise

$y(x, t)$ is a function of either $(x-u t) or(x+vt)$

$f(x, t)=f(x+v t, t=0)$

A disturbance created as $f(x)$ at time $t=0$ will be changing / will be give a a function of time as

$f(x, t)=f(x-v t, t=0)$

$y$ it travels $t$ the right $f(x, t)$

$=f(x+v t, t=0)$ if it travels $t$ the left.

$f(x-v t, t=0)=f(x, t)$

$f(x+v t, 0)=f(x t)$

Condition:

The disturbance travels undistorted with constant speed $v$.

Undistorted $\Longleftrightarrow$ Dispersionless

Another way of looking at wave travel.

$\Delta t=\frac{x}{v}$

$f(x, t)=f\left(x=0, t-\frac{x}{v}\right)$ If travel is to the right.

$f(x, 1)=f(x=0, t + \frac{x}{v}) \quad x<0$

wave as a function

$f(t+\frac{x}{v})$, wave disturbance travelling to the left.

$f(t-\frac{x}{v})$, wave disturbance travelling to the right.

$f(t+\frac{x}{v})=f(\frac{v t+x}{v})=f(x+v t)$

$f(t-\frac{x}{v})=f(\frac{v t-x}{v})=f(v t-x)$

A wave can be described as $f(x+v t), f(x-v t)$

$f(t+x / v)$, or $(t-\frac{x}{v})$.

$y(x, t) = A \sin (2 \pi(\frac{x}{\lambda}-f t)$

$f(x-v t)$ are functions of $x$, and $t$ both.

• A travelling is $a$ function of $x$ ane $t$

Look at $f(x-v t)$ as a function of time by looking at $a$ fixed position $x_0$

See how wave disturbance to changing with time.

Fix 't', $t=t_0$ and observe the disturbance look as a function of $x$.

$A \text { sin } 2 \pi\left(\frac{x}{\lambda}-f t\right)$

$f(x=0, t)= A \sin (\frac{0}{\lambda}-f t)$

$=-A \sin 2 \pi f t$

$=-A \sin \omega t$

As a function of position: $t=0$.

$f(x, t=0) =A \sin \frac{2 \pi}{\lambda} x$

$x \rightarrow (x+\lambda) \quad A \sin \frac{2 \pi}{\lambda}(x+\lambda)$

$=A \sin \left(\frac{2 \pi}{\lambda} x+2 \pi\right)$

$=A \sin \frac{2 \pi}{\lambda} x$

How are $\lambda, v$ and frequency $f$ of the ware related

$x=x_0$

$A \sin \omega t$

After time $t=T=\frac{1}{f}$

The displacement is the same.

In time $T$, the ware has moves by distance $\lambda$

$v=\frac{\lambda}{T}=f \lambda$

How to generate $a$ wave like $A \sin 2 \pi\left(\frac{x}{\lambda}-f t\right)$.

This is called a sine wave.

$y(x=0, t) =A \sin 2 \pi f t$

$y(x, t) =y(x=0, t-\frac{x}{v})$

$=A \sin 2 \pi f(t-\frac{x}{v})$

$=A \sin 2 \pi(f t-\frac{x f}{v})$

$=-A \sin 2 \pi(\frac{x}{\lambda}-f t)$

$y(x, t)= y(x=0, t+\frac{x}{v})$

$={A} \sin 2 \pi+(t+\frac{x}{v})$

$=A \sin 2 \pi(f t+\frac{\pi}{\lambda})$.

(1) We have looked at a disturbance travelling with speed $v$ are undisturbed.

$f(x-v t) or f(t-\frac{x}{v})$

If it travels to the right (+x axis)

$f(x+v t) \text { or } f(t+\frac{x}{v})$

y travelling to the left.

$y(x, t)=A \sin 2 \pi\left(\frac{x}{\lambda}-f t\right)$

$V=f \lambda=(2 \pi f) \frac{\lambda}{2 \pi}$

$\frac{2 \pi}{\lambda}= k$ = wave vector

wave number = $\frac{\omega}{k}$

$\omega=v k$

$y(x, t)=A \sin(k x-\omega t)$

Transverse waves:

$y(x,t)$ is perpendicular to the travel direction wave on a string are example of transverse waves.

Longitudinal waves:

Distarbance is in the same direction as the direction of motion of wave.

Sound waves are longitudinal.

Question : Do waves Carry particles as they travel?

Question : Do waves carry energy from one place to another?

Speed of waves on a string.

$y(x, t) =A sin(k x-\omega t)$

$v=\frac{\omega}{k} \text { or } \omega=v k$

(Assume $y \ll \lambda)$ $\theta \ll 1$

$\Rightarrow \sin \theta=\tan \theta =(\frac{d y}{d x})$

$\cos \theta =1$

Net vertical force:

${[T(\frac{d y}{d x})_2-T(\frac{d y}{d x})_2] \hat{y}}$

= 0 in the x direction

Taylor's theorm $(\frac{d y}{d x})_2 = (\frac{d y}{d x})_1 + (\frac{d^2 y}{d x}) \Delta x$

Vertical force on the section of string

$=T(\frac{d^2 y}{d x^2}) \Delta x$

Acceleation of the section = $\frac{d^2y}{dt^2}$

Force $=T(\frac{d^2 y}{d x^2})_t=T(\frac{\partial^2 y}{\partial x^2})_t {\Delta x}$

Acceleration : $(\frac{d^2 y}{d t^2})_x =(\frac{\partial^2 y}{\partial t^2})_x$

Force $=\mu \Delta x$ (accelaration)

$\mu=$ mass per unit length of string

$T(\frac{d^2 y}{d x^2})_t {\Delta x}=\mu \Delta x(\frac{d^2 y}{d t^2})_x$

$y(x, t)=A \sin (k x-\omega t)$

$T(\frac{d^2 y}{d x})_t \Delta x=\mu \Delta x^2(\frac{d^2 y}{d t^2})_x$

$y=A \sin(k x-\omega t)$

$(\frac{d^2 y}{d x^2})_t=-k^2 A \sin (k x-\omega t)$

$(\frac{d^2 y}{d t^2})_x=-\omega^2 A \sin (k x-\omega t)$

$T k^2= \mu \omega^2-(\frac{\omega}{k})=\sqrt{\frac{T}{\mu}}$

$\Rightarrow \quad v=\frac{\omega}{k}=\sqrt{\frac{T}{\mu}}$

$\Rightarrow \lambda=\frac{v}{f}=\frac{1}{f}\sqrt{\frac{T}{\mu}}$

$p+\Delta p$

A= Cross sectional area

$(\Delta p) A$

$m(\frac{\partial^2 z}{\partial t^2})_x=-\Delta p A$

Changes the volume

m = $\rho A \Delta x$, acceleration ( $\frac{\partial^2 f}{\partial t^2})$

$(\Delta P A)=$ Force

$A\rho \Delta x\frac{\partial^2 f}{\partial t^2}=-\Delta Pf A$

$\rho \frac{\partial^2 f}{\partial \omega^2} =-(\frac{\Delta P}{\Delta x})$

Pressure ' $P$ ' change the volume

$B=-V(\frac{\partial P}{\partial V}) =-V(\frac{\Delta P}{\Delta V})$

$\overline{\Delta p} =p$

$\Delta V =A(z+\Delta z)-A z$

$=(A \cdot \Delta z)$

$= A(\frac{d z}{d x}) \Delta x$

$-V(\frac{P}{\Delta V}) = B=-A \Delta x \cdot \frac{p}{A(\frac{d z}{d x}) \Delta x}$

$\Rightarrow P = - B \frac{dz}{dx}$

$\rho \frac{\partial^2 f}{\partial t^2}=-(\frac{\Delta p}{\Delta x})=-(\frac{d p}{d x})$

$P=-B(\frac{d z}{d x})$

$\rho \frac{\partial^2 z}{\partial t^2}=-(\frac{d p}{d x})_t=+B(\frac{d^2 z}{d x^2})$

$(\frac{d^2 z}{d x^2})=(\frac{\rho}{B})(\frac{d^2 z}{d t^2})$

$\frac{d^2 z}{d t^2}=(\frac{B}{\rho})(\frac{d^2 z}{d x^2}) \quad p \propto z$

Taking Sinusoidal wave

$Z(x, t)=A \sin (k x-\omega t)$

$\frac{d^2 z}{d x^2}|_{-2}=-k^2 A \sin (k x-\omega t)$

$\frac{d^2 z}{d t^2}|_x=-\omega^2 A \sin (k x-\omega t)$

$\frac{d^2 f}{d t^2}=\frac{d^2 z}{d t^2}=\frac{B}{\rho} \frac{d^2 z}{d x^2}$

$-\omega^2=\frac{\beta}{\rho} \cdot k^2$

$v^2=(\frac{\omega^2}{k^2})$$=B / \rho-\sqrt{v}=\sqrt{\frac{B}{\rho}}$

B = bulk modulus

(Adiabatic bulk modulus)

$v=\sqrt{\frac{B}{\rho}}$

We introduced the sinusoidal wave.

Calculated speed of waves by relating acceleration of a portion of medium (string and bulk medium for sound waves) of the force generated due to wave disturbance.