Wave on String

A wave on a string is a mechanical transverse wave, with vibrations moving sideways as energy travels along the string. Its speed depends on the string’s tension and thickness. When the wave reflects at the end, overlapping waves form standing patterns with fixed points (nodes) and moving points (antinodes). These reveal the string’s natural frequencies, key in music and physics. Waves also carry energy, explaining how instruments and machines work.

1.0Transverse Waves on a String

• A transverse wave on a string happens when the particles of the string move up and down, while the wave itself travels horizontally along the string. When you give one end of a stretched string a quick disturbance, the wave moves along the string, creating small vertical movements in the string as it passes.                                         

2.0Wave Equation on a String

Assumptions

1. The string is perfectly flexible and offers no resistance to bending.
2. Points on the string move only in the vertical direction; there is no motion in the horizontal direction.
3. Gravitational Forces on the string are neglected.

$\begin{array}{rl}& -T_1\cos \alpha +T_2\cos \beta =0\\ & \Rightarrow T_1\cos \alpha =T_2\cos \beta =T_{\dots ..}\end{array}$

From Newton's Second Law

Resultant Force=Mass ✕ Acceleration

$\begin{array}{rl}& \begin{array}{rl}& \frac{-T_1\sin \alpha }{T}+\frac{T_2\sin \beta }{T}=\frac{\rho \delta x}{T}\frac{{\partial }^{2}u}{\partial t^2}\\ & \frac{-T_1\sin \alpha }{T_1\cos \alpha }+\frac{T_2\sin \beta }{T_2\cos \beta }=\frac{\rho \delta x}{T}\frac{{\partial }^{2}u}{\partial t^2}\\ & -\mathrm{Tan}\alpha +\mathrm{Tan}\beta =\frac{\rho \delta x}{T}\frac{{\partial }^{2}u}{\partial t^2}\\ & \mathrm{Tan}\alpha ={\left(\frac{\partial u}{\partial x}\right)}_x,\mathrm{Tan}\beta ={\left(\frac{\partial u}{\partial x}\right)}_{x+\delta x}\\ & \frac{\partial u(x+\delta x,t)}{\partial x}-\frac{\partial u(x,t)}{\partial x}=\frac{\rho \delta x}{T}\frac{{\partial }^{2}u}{\partial t^2}\end{array}\\ & \begin{array}{rl}& \frac{{\partial }^{2}u}{\partial t^2}=\frac{T}{\rho }\frac{\left(\frac{\partial u(x+\delta x,t)}{\partial x}-\frac{\partial u(x,t)}{\partial x}\right)}{\delta x}\\ & \delta x\to 0\Rightarrow \frac{{\partial }^{2}u}{\partial t^2}=\frac{T}{\rho }\frac{{\partial }^{2}u}{\partial x^2}\\ & \frac{{\partial }^{2}u}{\partial t^2}=c\frac{{\partial }^{2}u}{\partial x^2}\left(c^2=\frac{T}{\rho }\right)\end{array}\end{array}$

3.0Wave Speed on a String

Velocity (v) of a wave on a string depends on:

(a) Tension in the string (elasticity) — F

(b) Mass per unit length (inertia) —  μ

String Element Dynamics

• Consider a small arc segment of the string of length Δλ , forming part of a circle with radius R.
• The tension forces act tangentially at both ends.
• Horizontal components of tension cancel out.
• Vertical components combine to provide a restoring force.
• This restoring force is what allows the wave to propagate.

$\begin{array}{rl}& F_r=F\sin \theta +F\sin \theta \\ & F_r=2F\sin \theta \approx 2F\cdot \frac{\frac{\Delta l}{2}}{R}=\frac{F\Delta l}{R}\end{array}$

If $\mu$ is mass per unit length of string and $\Delta m$ is the mass of element then downward acceleration is given by

$a=\frac{F_r}{\Delta m}=\frac{\frac{F\Delta l}{R}}{\Delta m}=\frac{F}{\mu R}\text{ where }\mu =\frac{\Delta m}{\Delta l}$

But element is moving in a circle of radius R. So Centripetal acceleration $a=\frac{v^2}{R}$

$\frac{v^2}{R}=\frac{F}{\mu R}\Rightarrow v=\sqrt{\frac{F}{\mu }}$

Note:

If A is area of cross-section and   is density

$=A\mu =\rho A$

So,velocity $v=\sqrt{\frac{F}{\rho A}}$

4.0Energy Density in Travelling Waves

• Snapshot of a travelling wave on a string at time t = 0 t = 0.Consider a string element of mass dm.

$\begin{array}{rl}& \text{ Kinetic Energy: }dK=\frac{1}{2}dm{u}^2\\ & y(x,t)=y_m\sin (kx-\omega t)\\ & u=\frac{\partial y}{\partial t}=-\omega y_m\cos (kx-\omega t)\\ & \text{ since }dm=\mu dx\\ & dK=\frac{1}{2}(\mu dx){\left(-\omega y_m\right)}^2{\cos }^2(kx-\omega t)\end{array}$

Potential energy:

• Potential energy is carried in the string when it is stretched. Stretching is largest when the displacement has the largest gradient.
• Hence, the potential energy is also maximum at the y = 0 y = 0 position. This is different from the harmonic oscillator, in which case energy is conserved.

Consider the extension $\Delta s$ of a string element.

$\begin{array}{rl}& \Delta s=\sqrt{(dx{)}^2+[y(x+dx,t)-y(x,t){]}^2}-dx\\ & \Delta s=\sqrt{(dx{)}^2+{\left(\frac{\partial y}{\partial x}dx\right)}^2}-dx=\left[\sqrt{1+{\left(\frac{\partial y}{\partial x}\right)}^2}-1\right]dx\end{array}$

Using approximation

$\Delta s\approx \left[1+\frac{1}{2}{\left(\frac{\partial y}{\partial x}\right)}^2-1\right]dx=\frac{1}{2}{\left(\frac{\partial y}{\partial x}\right)}^{2}dx$

The potential energy of the string element is given by the work done in extending the string element,

$dU=dW=T\Delta s\approx \frac{T}{2}{\left(\frac{\partial y}{\partial x}\right)}^{2}dx=\frac{T}{2}{k}^2{y}_m^2{\cos }^2(kx-\omega t)dx$

Note:

dU = dKE  dU = dKE only valid for travelling waves.

$\begin{array}{ll}dU=\frac{1}{2}T{\left(\frac{v_p}{v_w}\right)}^{2}dx=\frac{1}{2}\left(\mu v_w^2\right){\left(\frac{v_p}{v_w}\right)}^{2}dx=\frac{1}{2}\mu v_P^{2}dx& \\ dKE=\frac{1}{2}\mu v_P^{2}dx& \\ \frac{dE}{dx}=\frac{dU}{dx}+\frac{dKE}{dx}=\mu v_P^2& \\ \frac{dE}{dx}=\mu {\left(\frac{dy}{dt}\right)}^2=T{\left(\frac{dy}{dx}\right)}^2& (\because \frac{dE}{dx}=2\frac{dKE}{dx}=\frac{\mu dx{v}_P^2}{dx}\end{array}$

Key Points for travelling wave in string only

• KE and PE of an element are always equal to each other. Mechanical energy (KE + PE) of an element is not constant.
•  At an instant when an element is passing through the mean position its KE and PE are maximum.
• At an instant when an element is at an extreme position its KE and PE are minimum.

5.0Power Transmission by  String

• Suppose a simple harmonic progressive wave is propagating in x direction

$y=Asint-kxy=A\sin (\omega t-kx)$

• The string on the left of the point x exerts a force on to the right of position x.This force acts along the tangent to the string at x.x.
• Rate of change of energy of the particle $=Fvycos90+=F{v}_y\cos (90+\theta )$
• {Here we are taking velocity component in the direction of force

$=-Fvysin=-F{v}_y\sin \theta$

• This is the equation of instantaneous power transmitted to the neighbouring particles. i.e. 

$P=-F{v}_y\sin \theta$

$v_y=\frac{\partial y}{\partial t}$

$\text{And }\sin \theta \approx \theta \approx \tan \theta =\frac{\partial y}{\partial x}$

$\text{Hence, }P=-T\frac{\partial y}{\partial x}\frac{\partial y}{\partial t}$

$P=F[Ak\cos (\omega t-kx)]\cdot \frac{\partial }{\partial t}[A\sin (\omega t-kx)]$

$P=F[Ak\cos (\omega t-kx)]\cdot [A\omega \cos (\omega t-kx)]$

$P=\frac{{\omega }^2{A}^{2}F}{\nu }{\cos }^2(\omega t-kx)$

$⟨P⟩=P_{avg}=\frac{{\omega }^2{A}^{2}F}{\nu }⟨{\cos }^2(\omega t-kx)⟩$

$⟨P⟩=\frac{1}{2}\frac{{\omega }^2{A}^{2}F}{\nu }=\frac{1}{2}{\omega }^2{A}^2\mu \nu$

$⟨P⟩=2{\pi }^2\mu \nu A^2{f}^2$

$\text{Hence, }{P}_{avg}=2{\pi }^2\mu \nu A^2{f}^2$

$\text{i. e. }{P}_{avg}\propto A^2{f}^2$

$P=-F_y{v}_y\sin \theta$

$\frac{\partial z}{\partial x}$

And $\sin \theta \approx \theta \approx \tan \theta =\frac{\partial y}{\partial x}$

$\text{Hence, }P\approx -F\frac{\partial y}{\partial t}\frac{\partial y}{\partial x}$

$P=F[Ak\cos (\omega t-kx)][A\sin (\omega t-kx)]$

$P=F[Ak\cos (\omega t-kx)]\cdot [A\omega \cos (\omega t-kx)]$

$P=\frac{{\omega }^2{A}^2}{\nu }{\cos }^2(\omega t-kx)$

${⟨P⟩}_{\text{avg}}=\frac{1}{2}\frac{{\omega }^2{A}^2}{\nu }F{\cos }^2(\omega t-kx)$

$⟨P⟩=\frac{1}{2}{\omega }^2{A}^2\frac{F}{\nu }=\frac{1}{2}{\omega }^2{A}^2\nu \mu$

$⟨P⟩=2{\pi }^2{\nu }^2{A}^2\sqrt{F\mu }$

$\text{Hence, }{P}_{avg}=2{\pi }^2\mu \nu A^2{f}^2$

$\text{i. e. }{P}_{avg}\propto A^2{f}^2$

6.0Wave Intensity on String

Intensity of transverse wave on string

$I=\frac{\text{Power}}{\text{Area}}$

Since power is variable for a point and we have taken its average value

So, Intensity at any point is also taken average.

$<I>=<P>Area=12vA22Area⟨I⟩=\frac{⟨P⟩}{\text{Area}}=\frac{1}{2}\frac{\mu \nu A^2{\omega }^2}{\text{Area}}$

$<I>=12vA22⟨I⟩=\frac{1}{2}\rho \nu A^2{\omega }^2$

$\text{Where }\rho =\text{Density of the material of string.}$