Oscillations & WavesTopic #14 of 35

Wave Properties

Wavelength, frequency, amplitude, wave speed, and the wave equation.

Overview

A wave is a disturbance that transfers energy through a medium (or through space) without transferring matter. Understanding wave properties is fundamental to physics.

Types of Waves

Mechanical Waves

  • Require a medium to travel
  • Examples: sound, water waves, waves on a string

Electromagnetic Waves

  • Do not require a medium
  • Examples: light, radio waves, X-rays

By Direction of Oscillation

Transverse waves: Oscillation perpendicular to wave direction

  • Example: light, waves on a string

Longitudinal waves: Oscillation parallel to wave direction

  • Example: sound waves

Wave Parameters

Wavelength (λ\lambda)

Distance between two consecutive points in phase (e.g., crest to crest)

  • Unit: meters (m)

Frequency (ff)

Number of complete waves passing a point per second

  • Unit: Hertz (Hz) = 1/s

Period (TT)

Time for one complete wave to pass

  • Unit: seconds (s)
T=1fT = \frac{1}{f}

Amplitude (AA)

Maximum displacement from equilibrium

  • Unit: meters (m)

Wave Speed (vv)

Speed at which the wave propagates:

v=fλ=λTv = f\lambda = \frac{\lambda}{T}

Wave Equation

For a wave traveling in the +x direction:

y(x,t)=Asin(kxωt+ϕ)y(x,t) = A \sin(kx - \omega t + \phi)

For a wave traveling in the -x direction:

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

Wave Number

k=2πλk = \frac{2\pi}{\lambda}

Angular Frequency

ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}

Relationship

v=ωk=fλv = \frac{\omega}{k} = f\lambda

Wave Speed in Different Media

Wave on a String

v=Tμv = \sqrt{\frac{T}{\mu}}

Where TT = tension, μ\mu = linear mass density (kg/m)

Sound in Air

v=331 m/s×T273 Kv = 331 \text{ m/s} \times \sqrt{\frac{T}{273 \text{ K}}}

At 20°C: v343v \approx 343 m/s

Sound in General

v=Bρv = \sqrt{\frac{B}{\rho}}

Where BB = bulk modulus, ρ\rho = density

Energy in Waves

Intensity

Power per unit area:

I=PAI = \frac{P}{A}

Unit: W/m²

Intensity and Amplitude

IA2I \propto A^2

Intensity and Distance (Point Source)

I=P4πr2I = \frac{P}{4\pi r^2}

Intensity decreases as 1/r21/r^2 from a point source

Superposition Principle

When two waves overlap, the net displacement is the sum of individual displacements:

ytotal=y1+y2y_{\text{total}} = y_1 + y_2

Constructive Interference

Waves in phase add up:

  • Path difference = nλn\lambda (n=0,1,2,...n = 0, 1, 2, ...)
  • Maximum amplitude: A1+A2A_1 + A_2

Destructive Interference

Waves out of phase cancel:

  • Path difference = (n+12)λ(n + \frac{1}{2})\lambda
  • Minimum amplitude: A1A2\lvert A_1 - A_2 \rvert

Standing Waves

Formed by superposition of two waves traveling in opposite directions.

Nodes

Points of zero displacement

  • Occur at x=nλ2x = \frac{n\lambda}{2}

Antinodes

Points of maximum displacement

  • Occur at x=(n+12)λ2x = \frac{(n + \frac{1}{2})\lambda}{2}

Standing Wave Equation

y(x,t)=2Asin(kx)cos(ωt)y(x,t) = 2A \sin(kx) \cos(\omega t)

Standing Waves on a String

For a string fixed at both ends, length LL:

Wavelengths:

λn=2Ln(n=1,2,3,...)\lambda_n = \frac{2L}{n} \quad (n = 1, 2, 3, ...)

Frequencies:

fn=nv2L=nf1f_n = \frac{nv}{2L} = nf_1

Where f1f_1 = fundamental frequency (first harmonic)

Examples

Example 1: Wave Parameters

A wave has λ=2\lambda = 2 m and f=5f = 5 Hz. Find wave speed and period.

v=fλ=5×2=10 m/sv = f\lambda = 5 \times 2 = 10 \text{ m/s} T=1f=0.2 sT = \frac{1}{f} = 0.2 \text{ s}

Example 2: Wave on String

A string (μ=0.01\mu = 0.01 kg/m) under 100 N tension. Find wave speed.

v=Tμ=1000.01=100 m/sv = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.01}} = 100 \text{ m/s}

Example 3: Standing Waves

A 1.2 m guitar string produces a fundamental frequency of 220 Hz. Find wave speed.

λ1=2L=2.4 m\lambda_1 = 2L = 2.4 \text{ m} v=f1λ1=220×2.4=528 m/sv = f_1\lambda_1 = 220 \times 2.4 = 528 \text{ m/s}

Example 4: Harmonics

Find the first three harmonic frequencies for Example 3.

f1=220 Hzf_1 = 220 \text{ Hz} f2=2f1=440 Hzf_2 = 2f_1 = 440 \text{ Hz} f3=3f1=660 Hzf_3 = 3f_1 = 660 \text{ Hz}

Example 5: Intensity

Sound intensity at 5 m from a source is 0.4 W/m². Find intensity at 10 m.

I1r12=I2r22I_1 r_1^2 = I_2 r_2^2 I2=I1(r1r2)2=0.4×(510)2=0.1 W/m2I_2 = I_1 \left(\frac{r_1}{r_2}\right)^2 = 0.4 \times \left(\frac{5}{10}\right)^2 = 0.1 \text{ W/m}^2