Engineering Physics - Waves

Introduction to Waves

  • Major Aim: Understanding the classification of waves based on physical properties (mechanical, water, EM, sound, and matter waves).
  • Discussion Criteria: Longitudinal, transverse, and mixed waves.
  • Dimensionality: 1-D, 2-D, and 3-D waves.
  • Topics: Waves on a string, wave equations, harmonic waves, reflection and transmission at boundaries, standing waves and eigenfrequencies.
  • Reference: Chapter 5, Section 5.2 from 'The Physics of Vibrations and Waves’, 1st Edition, H J Pain and P Rankin, John Wiley and Sons Ltd, 2015.

Waves: The Concept

  • Definition: A wave is a disturbance that propagates in a medium and transfers energy without a net transfer of medium particles.
  • Forms: Elastic deformation, pressure variation, electric or magnetic intensity, electric potential, or temperature.
  • Examples: Sound waves, water waves, seismic waves, radio waves, light, and microwaves.
  • Crowd Wave: An example of a wave.

Characteristics of Waves

  • Crest: The highest point of a wave.
  • Trough: The lowest point of a wave.
  • Time Period (T): Time taken for the wave to complete one oscillation.
  • Velocity (v):
    v=λ/Tv = λ / T

Classification of Waves

  • Based on Medium:
    • Mechanical Waves
    • Non-mechanical Waves (Electromagnetic Waves)
    • Matter Waves
  • Based on Vibrations:
    • Transverse Waves
    • Longitudinal Waves
    • Mixed Waves
  • Based on Propagation Dimension:
    • 1D Waves
    • 2D Waves
    • 3D Waves

Classification by Medium

  • Mechanical Waves: Require a medium to travel; cannot travel in a vacuum.

  • Electromagnetic Waves: Do not require a medium; can travel in a vacuum. Examples include microwaves, X-rays, radio waves, ultraviolet waves, and light.

  • Matter Waves: Associated with each particle; described by the De-Broglie wavelength.

    \λ = h / p

    • Cannot be understood by classical theories; require Quantum Physics.

Classification by Vibration

  • Transverse Waves: Vibrations are perpendicular to the direction of propagation. Examples include EM waves and water waves.
  • Longitudinal Waves: Vibrations are along the direction of propagation. Example: Sound waves.
    *Note: Electromagnetic waves can only be transverse waves.

Sound Waves: Longitudinal Waves

  • Description: A wave of compression and rarefaction that propagates in an elastic medium such as air.

EM Waves: Transverse Waves

  • Propagation: Propagate by varying electric and magnetic fields perpendicular to each other.

Surface Waves

  • Characteristics: Exhibit properties of both longitudinal and transverse waves.
  • Motion: Particles move up and down as well as back and forth, resulting in an overall circular motion.

1D, 2D & 3D Waves

  • Classification: Based on dimensionality.

Mathematical Interpretation of Waves

  • Description: Representing a disturbance moving with speed vv in the forward direction.
  • Equation:
    • At t=0t = 0, y(x,0)=f(x)y(x, 0) = f(x)
    • At time t=t<em>1t = t<em>1, y(x,t</em>1)=y(xvt,0)=f(xvt)y(x, t</em>1) = y(x - vt, 0) = f(x - vt)
    • General form: y(x,t)=f(x±vt)y(x, t) = f(x ± vt)

Vector Resolution

  • Definition: Splitting a single vector into two or more vectors in different directions which together produce a similar effect as is produced by a single vector itself.
  • Component Vectors: The vectors formed after splitting.

Concept of Partial Derivative

  • For any function of two independent variables, g(x,y)g(x,y)
  • First order partial derivative
    gx=limΔx0g(x+Δx,y)g(x,y)Δx\frac{{\partial g}}{{\partial x}} = \lim_{\Delta x \to 0} \frac{{g(x+\Delta x,y) - g(x,y)}}{{\Delta x}}
  • Second order partial derivative
    2gx2=limΔx0gx(x+Δx,y)gx(x,y)Δx\frac{{\partial^2 g}}{{\partial x^2}} = \lim_{\Delta x \to 0} \frac{{\frac{{\partial g}}{{\partial x}}(x+\Delta x, y) - \frac{{\partial g}}{{\partial x}}(x, y)}}{{\Delta x}}

Wave Equation: 1D Wave on a String

  • Assumptions:

    1. The string is perfectly flexible and offers no resistance to bending; tension is tangential.
    2. Points move only vertically; no horizontal motion.
    3. Gravitational forces are negligible.
  • Parameters:

    • Linear mass density: ρ\rho
    • Tension: TT
  • Derivation:

    • Net force along x direction = 0

      F<em>1cos(θ</em>1)=F<em>2cos(θ</em>2)=TF<em>1 \cos(\theta</em>1) = F<em>2 \cos(\theta</em>2) = T

    • Net force along y direction:

      ΔF=F<em>1sin(θ</em>1)F<em>2sin(θ</em>2)=ρΔxay\Delta F = F<em>1 \sin(\theta</em>1) - F<em>2 \sin(\theta</em>2) = \rho \Delta x a_y

    • Approximation:

      ΔF=T[tan(θ<em>1)tan(θ</em>2)]=T[yx<em>x+Δxyx</em>x]\Delta F = T[\tan(\theta<em>1) - \tan(\theta</em>2)] = T [\frac{\partial y}{\partial x}<em>{x+ \Delta x} - \frac{\partial y}{\partial x}</em>{x}]
      ΔF=T2yx2Δx=ρΔx2yt2\Delta F = T \frac{\partial^2 y}{\partial x^2} \Delta x = \rho \Delta x \frac{\partial^2 y}{\partial t^2}

    • Wave Equation:

      2yx2=ρT2yt2\frac{\partial^2 y}{\partial x^2} = \frac{\rho}{T} \frac{\partial^2 y}{\partial t^2}

    • Wave Velocity:

      v=Tρv = \sqrt{\frac{T}{\rho}}

    • Final Wave Equation:

      2yx2=1v22yt2\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}

  • Dimensional Analysis:

    • [v]=[LT1][v] = [LT^{-1}]

Wave Equation: General

  • Description: A second-order linear partial differential equation.
  • Origin: Newton’s second law of motion, F=maF = ma
  • Application: Vibrations of a String.
  • Derivation Steps: Using a 1-D infinite string.

Examples

  • Problem: A copper wire is pulled by a tension T=0.98T = 0.98 N. The density is ρ=9.86g/cm\rho = 9.86 g/cm. Compute the speed of the wave.

  • Solution:

    v=Tρ=1.01m/sv = \sqrt{\frac{T}{\rho}} = 1.01 m/s

Verifying Wave Equation Solutions

  • Problem: Show that f(x,t)=Asin[B(xvt)]f(x, t) = A \sin[B(x - vt)] satisfies the wave equation.
  • Problem: Show that the function y(x,t)=Aeb(xvt)2y(x,t) = Ae^{-b (x-vt)^2} satisfies the wave equation.
  • Problem: Show that the function y(x,t)=A1+B(x+vt)y(x,t) = \frac{A}{1+B(x+vt)} satisfies the wave equation. What is the direction of the wave in this case?

Simple Harmonic Motions

  • Definition: A motion where the restoring force is directly proportional to the displacement from the mean position.

    F=kyF = -ky

  • Harmonic waves are sinusoidal.

Harmonic Wave - Displacement

  • Model: Amplitude modeled by a sine function.
  • Propagation: Propagates one wavelength (λ\lambda) along the x-axis during one time period (T).
  • Speed: Constant, v=λTv = \frac{\lambda}{T}
  • Amplitude Range: -A to +A.
  • Mathematical Representation:
    • ω=2πT    T=2πω\omega = \frac{2\pi}{T} \implies T = \frac{2\pi}{\omega}
    • k=2πλk = \frac{2\pi}{\lambda}
    • y=Asin(2πλx+ϕ)=Asin(kx+ϕ)y = A \sin(\frac{2\pi}{\lambda} x + \phi) = A \sin(kx + \phi)

Harmonic Wave - General Solution and Direction

  • General Wave Equation:

    y(x,t)=Asin(kx±ωt+ϕ)y(x,t) = A \sin(k x \pm \omega t + \phi)

  • Relating to General Solution:

    y(x,t)=Asin(2π(xλ±tT))y(x,t) = A \sin(2\pi(\frac{x}{\lambda} \pm \frac{t}{T}))
    y(x,t)=Asin(kx±ωt)y(x,t) = A \sin(k x \pm \omega t)
    Here, v = \frac{\omega}{k} => \omega = kv

  • Direction of Propagation:

    • ++ sign: negative x-direction.
    • - sign: positive x-direction.
    • Note: k=ωv=ω1vk = \frac{\omega}{v} = \omega*\frac{1}{v}, [ωv]=[ωx][\frac{\omega}{v}] = [\frac{\omega}{x}] and ω\omega = [rad/sec]

Concept of Phase

  • General Form:

    y(x,t)=Asin(kx±ωt+ϕ0)y(x, t) = A \sin(kx \pm \omega t + \phi_0)

    • AA = Amplitude
    • kk = Wavenumber
    • ω\omega = Angular frequency
    • ϕ0\phi_0 = Initial phase
  • Temporal Term ωt\omega t: Phase of the wave with respect to the reference oscillator.

  • Spatial Term kxkx: Displacement of the oscillator.

Properties of SH Waves

  1. Amplitude (A): Maximum displacement from the undisturbed position.
  2. Wavelength (λ\lambda) / Wavenumber (k):
    • Wavelength is the distance between successive crests/troughs.
    • Wavenumber: k=2πλk = \frac{2\pi}{\lambda}
  3. Period (T) and Frequency (f):
    • Period is the time between arrivals of two successive crests/troughs.
    • Frequency is the number of crests/troughs passing a point per unit time.
    • Relationships:
      • f=1Tf = \frac{1}{T}
      • ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f
  4. Velocity (v):
    • Relationship: v=λT=λf=ωkv = \frac{\lambda}{T} = \lambda f = \frac{\omega}{k}

Simple Harmonic Wave Representation

  • Problem: For a given wave y(x,t)=0.02sin(x0.01+t0.05)y(x, t) = 0.02 \sin(\frac{x}{0.01} + \frac{t}{0.05}), find amplitude, wavelength, frequency, and velocity.

Simple Harmonic Wave: Mathematical Representation

  • Equivalent Expressions:
    • y(x,t)=f(x±vt)y(x, t) = f(x ± vt)
    • y(x,t)=Asin(kx±ωt+ϕ)y(x, t) = A \sin(kx ± \omega t + \phi)
    • y(x,t)=Asin(kx±ωt)y(x, t) = A \sin(kx ± \omega t)
    • y(x,t)=Asin(ω(xv±t))y(x, t) = A \sin(\omega(\frac{x}{v} ± t))
    • y(x,t)=Asin(2π(xλ±νt))y(x, t) = A \sin(2\pi(\frac{x}{\lambda} ± \nu t))
    • y(x,t)=Asin(2πλ(x±vt))y(x, t) = A \sin(\frac{2\pi}{\lambda} (x ± vt))

Simple Harmonic Wave: Wave Velocity & Particle Velocity

  • Right-Moving Wave:

    yt=vyx\frac{\partial y}{\partial t} = -v \frac{\partial y}{\partial x}

  • Left-Moving Wave:

    yt=vyx\frac{\partial y}{\partial t} = v \frac{\partial y}{\partial x}

  • Particle Velocity: direction × wave velocity × slope

Simple Harmonic Wave: Particle Motion

  • Determining which particles move up/down based on the wave's shape.

Characteristic Impedance of a String

  • Impedance: Opposition to wave motion offered by a medium.

  • Characteristic Impedance (Z): Impedance offered by the string to transverse waves.

    Z=TransverseForceTransverseVelocity=F<em>yv</em>y=FyytZ = \frac{Transverse Force}{Transverse Velocity} = \frac{F<em>y}{v</em>y} = \frac{F_y}{\frac{\partial y}{\partial t}}

Characteristic Impedance Details

  • Transverse Force: F<em>y=Tsin(θ</em>1)F<em>y = T \sin(\theta</em>1)

  • For small angles: FyTtan(θ)=T(yx)F_y \approx -T \tan(\theta) = -T(\frac{\partial y}{\partial x})

  • Impedance:

    Z=F<em>yv</em>y= Tyxvyx=Tv=TρZ = \frac{F<em>y}{v</em>y} = \ - \frac{-T \frac{\partial y}{\partial x}}{-v \frac{\partial y}{\partial x}} = \frac{T}{v} = \sqrt{T\rho}

Wave Travel in Different Medium

  • Understanding wave properties when traveling between two different mediums in terms of impedance.

Boundary Conditions

  • At the boundary, x=0x = 0: Displacement and Tension must be continuous.

  • Displacement Continuous:

    y<em>i(x,t)+y</em>r(x,t)=yt(x,t)    A+B=Cy<em>i(x, t) + y</em>r(x, t) = y_t(x, t) \implies A + B = C

  • Tension Continuous:

    T(yx)<em>i+T(yx)</em>r=T(yx)<em>t    AB=k</em>2k1CT(\frac{\partial y}{\partial x})<em>i + T(\frac{\partial y}{\partial x})</em>r = T(\frac{\partial y}{\partial x})<em>t \implies A - B = \frac{k</em>2}{k_1} C
    Where:

  • Medium 1: Z<em>1,ρ</em>1,v<em>1,k</em>1,λ1Z<em>1, \rho</em>1, v<em>1, k</em>1, \lambda_1

  • Medium 2: Z<em>2,ρ</em>2,v<em>2,k</em>2,λ2Z<em>2, \rho</em>2, v<em>2, k</em>2, \lambda_2

Reflection and Transmission Coefficients

  • Using Boundary Conditions:

    • AB=k<em>2k</em>1CA - B = \frac{k<em>2}{k</em>1} C
    • A+B=CA + B = C
  • Reflection Coefficient:

    BA=k<em>1k</em>2k<em>1+k</em>2=Z<em>1Z</em>2Z<em>1+Z</em>2=v<em>2v</em>1v<em>2+v</em>1\frac{B}{A} = \frac{k<em>1 - k</em>2}{k<em>1 + k</em>2} = \frac{Z<em>1 - Z</em>2}{Z<em>1 + Z</em>2} = \frac{v<em>2 - v</em>1}{v<em>2 + v</em>1}

    • Independent of ω\omega (k<em>1=ωv</em>1k<em>1 = \frac{\omega}{v</em>1} and k<em>2=ωv</em>2k<em>2 = \frac{\omega}{v</em>2})
  • Transmission Coefficient:

    CA=2k<em>1k</em>1+k<em>2=2Z</em>1Z<em>1+Z</em>2=2v<em>2v</em>1+v2\frac{C}{A} = \frac{2k<em>1}{k</em>1 + k<em>2} = \frac{2Z</em>1}{Z<em>1 + Z</em>2} = \frac{2v<em>2}{v</em>1 + v_2}

Wave Travel in Different Medium: Rigid End

  • Infinite Impedance: Z2Z_2 \rightarrow \infty

  • Transmission Coefficient:

    CA=0\frac{C}{A} = 0

  • Reflection Coefficient:

    BA=1\frac{B}{A} = -1

    • Phase shift of π\pi (180 degrees).
  • Qualitatively transmitted amplitude approaches zero and wave reflects with a 180-degree phase shift.

Wave Travel in Different Medium: Free End

  • Low Impedance: Z20Z_2 \rightarrow 0

  • Transmission Coefficient:

    CA=2\frac{C}{A} = 2

    • The coefficient of the transmitted wave will be as large as it can possibly be: TWICE the size of the incident wave
  • Reflection Coefficient:

    BA=1\frac{B}{A} = 1

Boundary with Same Medium

  • Equal Impedance: Z<em>1=Z</em>2Z<em>1 = Z</em>2

  • Transmission Coefficient:

    CA=1\frac{C}{A} = 1

  • Reflection Coefficient:

    BA=0\frac{B}{A} = 0

  • No boundary exists; the incident wave keeps traveling.

Reflection and Transmission of Wave in String: Free End

PropertiesReflectionTransmission
Velocity (v)SameChange
Frequency (f)SameSame
Wavelength ($\lambda$)SameChange
Phase ($\phi$)0 or π\pi0

Superposition Principle