Waves: Lecture 1 - Transverse Waves

Traveling Waves

• A traveling wave is an organized disturbance that moves with a well-defined wave speed.
• The medium of a mechanical wave is the substance through which the wave propagates.
• The displacement description uses a function D(x, t) (or y(x, t)) to denote the displacement at position x and time t for a given particle.
• In the context of waves on a string, the transverse displacement is typically denoted by y(x, t).

Sinusoidal Waves

• The wavelength λ is the distance over which the wave’s shape repeats (distance spanned by one cycle).
• Wavelength λ is measured in meters.
• A sinusoidal wave has a well-defined period T, during which each crest advances forward by one wavelength.
• Wave speed v is distance per unit time. For periodic motion, v = distance/time = λ/T.
• Since frequency f = 1/T, the wave speed can also be expressed as
  $v = f\,\lambda.$
• The wave frequency f is in Hz (s^{-1}).

Key quantities and relations

• Angular frequency ω and wave number k:
  $\omega = 2\pi f, \quad k = \frac{2\pi}{\lambda}.$
• For a traveling sinusoidal wave, the displacement can be written (for a wave traveling in the +x direction) as
  $y(x,t) = A \sin(kx - \omega t + \phi),$
  where A is the amplitude and φ is the phase constant.
• The phase velocity relation:
  $v = \frac{\omega}{k}.$
• The wave on a physical system that obeys the wave equation supports sinusoidal traveling waves with speed v.

Wave parameters on a string

• A transverse wave on a string has speed
  $v = \sqrt{\frac{T<em>s}{\mu}},$ where Ts is the string tension and μ is the linear (mass-per-length) density.
• The linear density is
  $\mu = \frac{m}{L},$
  with m being the mass of the string segment and L its length.

Generating a sinusoidal wave (example setup)

• Scenario: A very long string with linear density μ = 0.002 kg/m (2.0 g/m) is stretched along the x-axis with a tension T_s = 5.0 N.
• A simple harmonic oscillator at x = 0 m vibrates perpendicular to the string with frequency f = 100 Hz and amplitude A = 2.0 mm, and at t = 0 s the oscillator is at its maximum positive displacement.
• Questions:
  a) Write the displacement equation for the traveling wave on the string.
  b) At t = 5.0 ms, what is the string’s displacement at a point x = 2.7 m from the oscillator?

Step-by-step solution to the example

• Given data:

  • μ = 0.0020 kg/m
  • T_s = 5.0 N
  • f = 100 Hz
  • A = 2.0 mm = 0.0020 m
  • x = 2.7 m
  • t = 5.0 ms = 0.0050 s

• Wave speed on the string:
  $v = \sqrt{\frac{T_s}{\mu}} = \sqrt{\frac{5.0}{0.0020}} = \sqrt{2500} = 50\ \text{m/s}.$

• Wavelength from v and f:
  $\lambda = \frac{v}{f} = \frac{50}{100} = 0.50\ \text{m}.$

• Wave number:
  $k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.50} = 4\pi\ \text{rad/m}.$

• Angular frequency:
  $\omega = 2\pi f = 2\pi\cdot 100 = 200\pi\ \text{rad/s}.$

• Phase constant φ for a wave that has maximum displacement at t = 0 and x = 0:

  • The oscillator displacement at t = 0 is A with y(0,0) = A, which requires sin(φ) = 1, so φ = \frac{\pi}{2}.

• The displacement equation becomes
  $y(x,t) = A \sin(kx - \omega t + \tfrac{\pi}{2}) = (0.0020)\ \sin(4\pi x - 200\pi t + \tfrac{\pi}{2}).$

• Alternatively, using the identity sin(θ + π/2) = cos(θ), we can write
  $y(x,t) = A \cos(kx - \omega t).$

• Part (a) result (displacement equation):
  $y(x,t) = (0.0020)\ \sin(4\pi x - 200\pi t + \tfrac{\pi}{2}).$

• Part (b) displacement at t = 0.0050 s and x = 2.7 m:

  • Compute the argument: kx - ωt + φ = 4\pi(2.7) - (200\pi)(0.0050) + \tfrac{\pi}{2} = 10.8\pi - \pi + \tfrac{\pi}{2} = 10.8\pi - \tfrac{\pi}{2}.
  • Since 10.8π modulo 2π equals 0.8π, the angle becomes 0.8π - 0.5π = 0.3π.
  • Therefore
    $y(2.7, 0.005) = (0.0020) \sin(0.3\pi) \approx (0.0020) \cdot 0.809 = 1.618\times 10^{-3}\ \text{m} \approx 1.62\ \text{mm}.$

• Summary of the final displacement for the example:
  $y(x,t) = (0.0020)\sin(4\pi x - 200\pi t + \tfrac{\pi}{2}) = (0.0020)\cos(4\pi x - 200\pi t).$
  $y(2.7\ \text{m}, 0.0050\ \text{s}) \approx 1.62\ \text{mm}.$

Important cross-checks and notes

• The wave speed v is determined by string tension and linear density, not by the amplitude:
  $v = \sqrt{T_s/\mu}.$
• The relationship v = f λ confirms consistency between f, λ, and v.
• In the given example, there was a transcription inconsistency in the angular frequency value; the correct ω is $\omega = 2\pi f = 200\pi\ \text{rad/s}.$
• The phase constant φ is chosen to ensure the boundary condition at the source (x = 0) matches the source motion at t = 0 (maximum displacement corresponds to sin(φ) = 1).
• These results illustrate how amplitude, phase, and wavenumber combine to describe a traveling sinusoidal wave on a string and how to compute the displacement at any point and time.

Connections to broader concepts

• This analysis exemplifies the general solution to the one-dimensional wave equation for a string and shows how physical parameters (T_s, μ) set the wave speed.
• The sinusoidal wave form is a fundamental building block due to linear superposition; any complex wave on a string can be decomposed into sinusoidal components (Fourier analysis).
• Understanding the relationships among f, λ, v, ω, and k is essential for analyzing wave phenomena in acoustics, optics, and vibrating systems.