Dispersionless waves

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

$f(t-v / v)\quad$ waves travelling in the positive x direction

For the wave travelling in the $-x$ direction

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

$f(t+x / v)$

Question:

a) What happens when two or more waves arrive at a place at the same time?

b) what is the displacement or pressure when two different waves arrive at place at the same time?

If there are two or more waves arriving at a point at the same time

$f_1(x, t), \quad f_2(x, t)+\cdots$

Then the net displacement / pressure at that point is given by the sum of individual displacement/pressure.

$f(x, t)=f_1(x, t)+f_2(x, t)+\cdots$

Waves satisfying linear differential equation

$f(x, t)=f_1(x, t)+f_2(x, t)+\ldots$

Specialize in sinusoidal waves

$f(x,t)=$

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

A $G_s(k x-\omega t)\rightarrow$, A $\sin (k x+\omega t)\leftarrow$

$k=\frac{\omega}{v}, \quad k=\frac{2 \pi}{\lambda}, \quad v=2 \lambda$

$f(x, t)= A_1 \sin (k_1 x-\omega_1 t)$

$+A_2 \sin (k_2 x-\omega_2 t)$, $+A_3 \sin (k_2 x+\omega_3 t)$, and $+B_1 \cos (k_4 x+\omega_4 t)$

x = 0 by a hard wall

displacement at the point when the string is tied = 0

At the boundary at x = 0

Net displacement = 0

Net displacement = Sum of individual displacement

= displacement due to incoming wave + displacement due to the reflected wave

0 = displacement due to in coming wave + displacement due to the reflected wave

$\Rightarrow$ Displacement of the reflaction wave = -( displacement due to the incoming wave)

Incoming wave

$y_{incoming}(x, t)=A \sin (k x-\omega t)$

$y_{\text {reflecter }}(x, t)=B \sin (k x+\omega t)$

We are looking at $x=0$

$y_{\text {net }}(x=0)=-A \sin \omega t+B \sin \omega t=0 \Longrightarrow B=A$

In coming wave $=A \sin (k x-\omega t)$

Reflector wave = A Sin $(k x+\omega t)$

Reflected wave at $x=0$

Incoming wave

$A \sin \omega t =-(-A \sin \omega t) =-y_{incoming}$

$y_{reflected } =A \sin \omega t =A \sin (-\omega t+\pi)$

$A \sin (-\omega t+\pi) =-A \sin (-\omega t) -A \sin \omega t$

$y_{refected } =A \sin (-\omega t+\pi)$

To get the reflected wave, we add a phase of $\pi$ to the phase of incoming wave

$\Rightarrow$ There is a phase difference of $\pi$ between phase of incoming and reflected wave

The represents not a travelling wave but it represents a Standing Wave

What happens to the displacement at the points?

Superposition of incoming and reflected waves gives rise to standing waves

Net displacement at the boundary $\cancel=$ 0

The $\Delta P$ of the boundary $\cancel=$0

A wave moving in the + x direction

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

Superpose this with a wave travelling in the - x direction

$A \sin (k x+\omega t)$

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

$y(x, t)= 2 A \sin k x \cos \omega t$

This is not of the from $f(x-v)$

$f(x+v t)$ - The represents not a travelling wave but it represents a STANDING WAVE

y(x,t) = (2A) $\sin k x \cos \omega t B\sin k x \cos \omega t$

$y(x, t) =B \sin k x \cos \omega t=C \cos k x \text { Cost }=D \sin k x \sin \omega t$

Different example of standing wave

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

(1) Standing wave on a string

(1) String tied at Both Ends

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

Length of the string is $-L$

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

$\text { At } x=0 \quad y=0$

(1) String tied at Both Ends

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

Length of the string is $-L$

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

$\text { At } x=0 \quad y=0$

$\omega =2\pi\nu$

$\nu_n = n(\frac{v}{2 L}\nu_n =n(\frac{v}{2 L})=\frac{n}{2 L} \sqrt{\frac{T}{\mu}}\nu_n =\frac{n}{2 L} \sqrt{\frac{T}{\mu}}$

$\text { SinkL }=0$

$k =\frac{2 \pi}{\lambda} L=n \pi\lambda=(\frac{2 L}{n})\frac{\lambda}{2} \times n=L \Rightarrow \lambda=\left(\frac{2 L}{n}\right)$

Distance between nods $=\frac{\lambda}{2}$

Distance detween two antinods $=\frac{\lambda}{2}$

$\nu_n=\frac{n}{2 L} \sqrt{\frac{T}{\mu}}$

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

$y(0, t)=0$

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

maximum displacement at $x=L$

$k L=(2 n+1) \frac{\pi}{2}$

$k=\frac{2 \pi}{\lambda}=\frac{(2 n+1) \pi}{2 L}$

$\lambda=\frac{4 L}{(2 n+1)}$

$\nu_n=\frac{v}{\lambda} =\frac{v}{4 L / 2 n+1}=\frac{v}{4 L}(2 n+1) =\left(n+\frac{1}{2}\right) \cdot \frac{v}{2 L}\frac{\lambda}{4}(2 n+1)=L\Rightarrow \lambda=\left(\frac{4 L}{2 n+1}\right)$

Conclude

(i) Principle of superposition

$f(x, y)=\sum_i f_i(x, t)$

(ii) Reflection of a wave from a boundary phase difference of $\pi$ between the phase

of incoming and reflected waves when the boundary is hard $\Rightarrow$ No displacement at the boundary

(III) Standing waves: Standing's wars on a string.