Waves • Topic 2 of 3

Superposition & Standing Waves

What happens when two waves arrive at the same point at the same time? The principle of superposition answers this: when two or more waves overlap, the resultant displacement at any point is the vector sum of the individual displacements. The waves pass through each other and emerge unchanged — they do not collide or scatter. This single principle underlies interference, beats and standing waves.

Interference is the result of superposing two waves of the same frequency travelling in the same direction:

  • Constructive interference occurs when the waves arrive in phase (crest meets crest). The amplitudes add, giving a louder sound or brighter light. This needs a path difference of a whole number of wavelengths: $\Delta x=n\lambda$.
  • Destructive interference occurs when the waves arrive out of phase by half a cycle (crest meets trough). The amplitudes cancel, giving silence or darkness. This needs a path difference of an odd number of half-wavelengths: $\Delta x=(2n+1)\frac{\lambda}{2}$.

Reflection of waves. When a wave hits a boundary it is reflected. The nature of the reflection depends on the boundary:

  • At a rigid (fixed) boundary, the reflected wave suffers a phase change of $\pi$ (a crest returns as a trough). A node always forms there.
  • At a free (open) boundary, the reflected wave has no phase change. An antinode forms there.

Standing (stationary) waves form when two identical waves travel in opposite directions and superpose — typically an incident wave and its reflection. The result is a pattern that appears to stand still: $y=2A\sin(kx)\cos(\omega t)$. Energy is not transported along the medium; it stays trapped. The pattern has:

  • Nodes: points of zero displacement (always at rest). Spacing between consecutive nodes is $\frac{\lambda}{2}$.
  • Antinodes: points of maximum displacement (vibrating most). They lie midway between nodes.

Standing waves on a string fixed at both ends (length $L$): both ends must be nodes, so only certain wavelengths fit. The allowed frequencies are $f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu}}$ for $n=1,2,3,\dots$ The lowest is the fundamental (first harmonic), $f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$, and the higher ones ($2f_1, 3f_1, \dots$) are harmonics. An overtone is any frequency above the fundamental; the first overtone is the second harmonic.

Air columns. Wind instruments use standing waves of sound:

  • An open pipe (open at both ends) has antinodes at both ends. It produces all harmonics: $f_n=\frac{nv}{2L}$ ($n=1,2,3,\dots$).
  • A closed pipe (one end closed) has a node at the closed end and an antinode at the open end. It produces only odd harmonics: $f_n=\frac{nv}{4L}$ ($n=1,3,5,\dots$). Its fundamental is $\frac{v}{4L}$ — half the open pipe's, which is why a closed pipe of the same length sounds an octave lower.
Standing wave on a string with nodes and antinodesNNNNAAAλ/2N = node, A = antinode
1
Worked Example
A string fixed at both ends is 0.5 m long and a transverse wave travels on it at 100 m/s. Find the fundamental frequency.
Solution
  1. Step 1: For a string fixed at both ends, $f_1=\frac{v}{2L}$.
  2. Step 2: Substitute: $f_1=\frac{100}{2\times0.5}=\frac{100}{1}$.
  3. Step 3: Compute: $f_1=100\ \text{Hz}$.

Answer: Fundamental frequency $f_1=100\ \text{Hz}$.

2
Worked Example
An open pipe of length 0.34 m resonates with air. Find its fundamental frequency. (Speed of sound $=340\ \text{m/s}$.)
Solution
  1. Step 1: For an open pipe, $f_1=\frac{v}{2L}$.
  2. Step 2: Substitute: $f_1=\frac{340}{2\times0.34}=\frac{340}{0.68}$.
  3. Step 3: Compute: $f_1=500\ \text{Hz}$.

Answer: $f_1=500\ \text{Hz}$.

3
Worked Example
A closed pipe of length 0.34 m is sounded. Find its fundamental frequency. (Speed of sound $=340\ \text{m/s}$.)
Solution
  1. Step 1: For a pipe closed at one end, $f_1=\frac{v}{4L}$.
  2. Step 2: Substitute: $f_1=\frac{340}{4\times0.34}=\frac{340}{1.36}$.
  3. Step 3: Compute: $f_1=250\ \text{Hz}$ — exactly half that of an open pipe of the same length.

Answer: $f_1=250\ \text{Hz}$.

4
Worked Example
Two waves of equal amplitude interfere. Find the resultant amplitude when they are (a) in phase and (b) exactly out of phase, with each amplitude $A$.
Solution
  1. Step 1: For in-phase (constructive) superposition, amplitudes add: $A_{res}=A+A=2A$.
  2. Step 2: For out-of-phase (destructive) superposition, amplitudes subtract: $A_{res}=A-A=0$.
  3. Step 3: So intensity (proportional to amplitude squared) is $4$ times one wave when constructive, and zero when destructive.

Answer: (a) $2A$ (constructive); (b) $0$ (destructive).

5
Worked Example
The first overtone of a string fixed at both ends is 240 Hz. What is its fundamental frequency?
Solution
  1. Step 1: For a string fixed at both ends, all harmonics are present: $f_n=nf_1$.
  2. Step 2: The first overtone is the second harmonic, so $f_2=2f_1=240\ \text{Hz}$.
  3. Step 3: Therefore $f_1=\frac{240}{2}=120\ \text{Hz}$.

Answer: Fundamental frequency $=120\ \text{Hz}$.

6
Worked Example
A closed pipe has a fundamental frequency of 200 Hz. Find the frequency of its first overtone.
Solution
  1. Step 1: A closed pipe produces only odd harmonics: $f_1, 3f_1, 5f_1,\dots$
  2. Step 2: The first overtone is the next allowed mode, the third harmonic: $f=3f_1$.
  3. Step 3: Compute: $f=3\times200=600\ \text{Hz}$.

Answer: First overtone $=600\ \text{Hz}$ (the third harmonic).

Key Points

  • Principle of superposition: the resultant displacement is the vector sum of the individual wave displacements.
  • Constructive interference (in phase, $\Delta x=n\lambda$) adds amplitudes; destructive ($\Delta x=(2n+1)\frac{\lambda}{2}$) cancels them.
  • Standing waves form from two opposite-travelling waves: $y=2A\sin(kx)\cos(\omega t)$; nodes are at rest, antinodes vibrate most, node spacing is $\frac{\lambda}{2}$.
  • String fixed at both ends and an open pipe give all harmonics; a closed pipe gives only odd harmonics ($f_1=\frac{v}{4L}$).
  • The first overtone is the second harmonic for a string/open pipe, but the third harmonic for a closed pipe.
Tap an option to check your answer0 / 4
Q1.When two waves arrive at a point exactly out of phase, the interference is:
Explanation: Out-of-phase waves (crest meets trough) cancel, giving destructive interference.
Q2.In a standing wave, the distance between two consecutive nodes is:
Explanation: Nodes occur every half wavelength, so consecutive nodes are $\frac{\lambda}{2}$ apart.
Q3.A pipe closed at one end produces only:
Explanation: A closed pipe has a node at the closed end and antinode at the open end, allowing only odd harmonics ($f_n=\frac{nv}{4L}$, $n$ odd).
Q4.The fundamental frequency of a string fixed at both ends of length $L$ is:
Explanation: Both ends are nodes, so $L=\frac{\lambda}{2}$ for the fundamental, giving $f_1=\frac{v}{2L}$.