Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse waves in stretched strings.

Stationary wave : When two progressive waves of same amplitude and frequency travelling along the same straight line in opposite direction superimposed, a stationary wave is formed.
Formation of Stationary wave : A string is fixed between two rigid supports. It is excited perpendicular to its length to generate transverse waves which travel in opposite directions from point of excitation. These transverse waves get reflected at rigid supports and overlap on each other and produce stationary waves.
Theory : Let ${\mathrm{Y}}_1=\mathrm{Asin}(\mathrm{kx}-\mathrm{\omega t})$ and ${\mathrm{Y}}_2=\mathrm{Asin}(\mathrm{kx}+\mathrm{\omega t})$ are the displacements of incident and reflected waves from rigid supports. According to principle of superposition resultant displacement:

  y = y₁ + y₂
$\begin{array}{r}\mathrm{Y}=\mathrm{Asin}(\mathrm{kx}-\mathrm{\omega t})+\mathrm{Asin}(\mathrm{kx}+\mathrm{\omega t})\\ \mathrm{Y}=\mathrm{A}\left[\right(\sin (\mathrm{kx}-\mathrm{\omega t})+\sin (\mathrm{kx}+\mathrm{\omega t})]\end{array}$
From, $\sin (\mathrm{A}+\mathrm{B})+\sin (\mathrm{A}-\mathrm{B})=2\sin \mathrm{Acos}\mathrm{B}$,
$\mathrm{Y}=2\mathrm{Asin}\mathrm{kxcos}\mathrm{\omega t}$

The resultant Amplitude = $2\mathrm{Asin}\mathrm{kx}$ 

at positions,
$\mathrm{x}=0,\frac{\mathrm{\lambda }}{2},\frac{2\mathrm{\lambda }}{2}\dots \dots$nodes are formed
and at positions, $\mathrm{x}=\frac{\mathrm{\lambda }}{4},\frac{3\mathrm{\lambda }}{4},\frac{5\mathrm{\lambda }}{4}$.....antinodes are formed
Consider a string of length l stretched between two rigid supports P and Q. It may vibrate in form of one (or) more number of loops with velocity,
$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}$
The distance between adjacent nodes is $\frac{\mathrm{\lambda }}{2}$. So that in string fixed at both ends, there must be integral number ‘p’ of half wavelengths.
$\text{ i.e., }\mathrm{p}\left(\frac{\mathrm{\lambda }}{2}\right)=\mathrm{l};\mathrm{\lambda }=\frac{2\mathrm{l}}{\mathrm{p}}\dots$(1)
where p = 1, 2, .....
But $\mathrm{v}=\mathrm{n\lambda }\text{ and }\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}\dots$ (2)
$\mathrm{n}=\frac{\mathrm{v}}{\mathrm{\lambda }}$

[from equation (1) & (2)]:
$\mathrm{n}=\frac{\sqrt{\mathrm{T}/\mathrm{\mu }}}{2\mathrm{l}/\mathrm{p}}=\frac{\mathrm{p}}{2\mathrm{l}}\sqrt{\mathrm{T}/\mathrm{\mu }}$
From the equation, $\mathrm{n}=\frac{\mathrm{p}}{2\mathrm{I}}\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}$ laws of Transverse wave obtained.
I Law : The fundamental frequency of a vibrating string is inversely proportional to the length of the string.
$\mathrm{n}\propto \frac{1}{\mathrm{l}}(\mathrm{T},\mathrm{\mu }\text{ are constant });\mathrm{nl}=\mathrm{constant}$
II Law : The fundamental frequency of a vibrating string is directly proportional to the square root of tension.
$\mathrm{n}\propto \sqrt{\mathrm{T}}(\mathrm{l},\mathrm{\mu }\text{ are constant })$
III Law : The fundamental frequency of a vibrating string is inversely proportional to square root of linear mass density of the string.
$\mathrm{n}\propto \frac{1}{\sqrt{\mathrm{\mu }}}(\mathrm{l},\mathrm{T}\text{ are constant });\mathrm{n}\sqrt{\mathrm{\mu }}=\mathrm{Constant}$