How to solve a standing wave problem

A standing wave problem requires matching the wave's spatial variation to the physical boundary conditions of the medium.

This method applies to strings, acoustic tubes, and electromagnetic cavities where displacement or pressure nodes and antinodes are constrained at specific physical locations.

The setup

Identify the medium length LL, the wave speed vv, and the boundary conditions at both ends. Boundaries are either fixed (nodes) or free (antinodes). Symmetric systems have identical boundaries at both ends (e.g., fixed-fixed), while asymmetric systems have different boundaries (fixed-free).

The steps

  1. Assign boundary conditions. Fixed ends enforce a node; free ends enforce an antinode.
  2. Express the wavelength λn\lambda_n in terms of LL. For symmetric boundaries, λn=2Ln\lambda_n = \frac{2L}{n} for n=1,2,3,n = 1, 2, 3, \dots. For asymmetric boundaries, λn=4Ln\lambda_n = \frac{4L}{n} for n=1,3,5,n = 1, 3, 5, \dots.
  3. Relate wavelength to frequency using the wave equation v=fλv = f \lambda.
  4. Substitute λn\lambda_n to find the harmonic frequencies: fn=nv2Lf_n = \frac{n v}{2L} (symmetric) or fn=nv4Lf_n = \frac{n v}{4L} (asymmetric).
  5. Evaluate the expression using the given numerical values.

Checking the result

Verify that asymmetric systems (like a tube closed at one end) only produce odd harmonics. Check that higher harmonics in a symmetric system are exact integer multiples of the fundamental frequency (fn=nf1f_n = n f_1).

Common errors

Using even values of nn for asymmetric fixed-free boundaries is physically impossible and mathematically invalid. Another frequent error is confusing the harmonic number nn with the number of nodes; a fixed-fixed string in the nn-th harmonic has n+1n+1 nodes.

Worked example

A string of length L=0.8extmL = 0.8 ext{ m} is fixed at both ends. The wave speed on the string is v=120extm/sv = 120 ext{ m/s}. Calculate the frequency of the third harmonic.

Boundary conditions: Fixed-fixed (symmetric). Harmonic number: n=3n = 3. Wavelength equation: λ3=2L3\lambda_3 = \frac{2L}{3} Substitute length: λ3=2(0.8extm)3=0.5333extm\lambda_3 = \frac{2(0.8 ext{ m})}{3} = 0.5333 ext{ m} Frequency equation: f3=vλ3f_3 = \frac{v}{\lambda_3} Calculate frequency: f3=120extm/s0.5333extm=225extHzf_3 = \frac{120 ext{ m/s}}{0.5333 ext{ m}} = 225 ext{ Hz}

FAQ

Run your own problem

References: University Physics Volume 1 (OpenStax) · Fundamentals of Physics (Halliday, Resnick, Walker)

See also