SL & HL Wave Behaviour BETA C.4

Standing Waves and Resonance

Standing waves are formed by the superposition of two counter-propagating waves of equal frequency and amplitude, creating localized energy states. This topic explores how boundary conditions in strings and pipes determine harmonic frequencies, and how damping affects resonant systems.

Part of IB Physics (2025-2030 syllabus) — Standard and Higher Level.

Key points

  • Standing waves do not transfer energy across the medium; they store energy in localized patterns of oscillation.
  • Nodes are positions of permanent zero displacement caused by continuous destructive interference, whereas antinodes are positions of maximum displacement caused by continuous constructive interference.
  • The phase difference between any two oscillating points on a standing wave is either 00 (if they are within the same loop or separated by an even number of nodes) or π rad\pi\ \text{rad} (if they are separated by an odd number of nodes).
  • Boundary conditions dictate the wave patterns: fixed boundaries (or closed pipe ends) force displacement nodes, while free boundaries (or open pipe ends) force displacement antinodes.
  • Adjacent nodes (or adjacent antinodes) are separated by exactly half a wavelength, λ2\frac{\lambda}{2}, so the wavelength of any standing-wave pattern can be read directly from the node spacing.
  • Resonance occurs when the frequency of an external periodic driving force matches a system's natural frequency, maximizing the transfer of energy and resulting in large-amplitude oscillations.
  • Damping forces remove mechanical energy from an oscillating system, reducing the amplitude of standing waves and broadening the resonance peak.

Subtopic by subtopic

Formation of standing waves

A standing (stationary) wave forms when two waves of equal frequency and amplitude travel through each other in opposite directions and superpose. In practice the second wave is almost always the reflection of the first: a wave sent along a string reflects from the far end and overlaps with the waves still arriving. At certain frequencies the superposition settles into a stable pattern that appears to stand still rather than travel.

Standing waves rely entirely on the phase change that occurs when a wave reflects at a boundary. When a transverse wave on a string meets a fixed boundary, it undergoes a phase change of π rad\pi\ \text{rad} (180180^\circ); the incoming and reflected waves interfere destructively at that point, establishing a permanent node.

Conversely, when a wave meets a free boundary, or a sound wave reaches the open end of a pipe, it reflects with zero phase change, so the interference there is constructive and an antinode forms.

Unlike a travelling wave, a standing wave transfers no net energy along the medium: the energy is trapped within each loop, cycling between kinetic and potential forms.

You must be able to explain nodes and antinodes in terms of superposition and state the formation conditions:

  • same frequency
  • similar amplitude
  • opposite directions

Nodes, antinodes and harmonics

Nodes are points of permanently zero displacement, produced by continuous destructive interference; antinodes are points of maximum amplitude, produced by continuous constructive interference.

The pattern has a strict geometry: adjacent nodes are λ2\frac{\lambda}{2} apart, and a node and its neighbouring antinode are λ4\frac{\lambda}{4} apart. This is how you extract a wavelength from a sketch or photograph of a standing wave.

Two features separate standing waves from travelling waves. First, amplitude depends on position: a point at an antinode swings widely while a point near a node barely moves, whereas on a travelling wave every point eventually oscillates with the same amplitude.

Second, phase comes in only two values: all points within one loop oscillate in phase (00 phase difference), while points in adjacent loops, separated by an odd number of nodes, are π rad\pi\ \text{rad} out of phase.

Because the boundaries fix where nodes and antinodes must sit, only certain wavelengths fit on the system, so only a discrete set of frequencies, the harmonics, can be sustained. For a guitar string fixed at both ends, the second harmonic has a node at the midpoint and twice the fundamental frequency.

Be ready to compare the amplitude and phase of two labelled points on a diagram.

Standing waves in strings and pipes

Boundary conditions determine which harmonics exist. A fixed end of a string, or the closed end of a pipe, forces a displacement node; a free end, or the open end of a pipe, forces a displacement antinode.

For a string fixed at both ends, whole numbers of half-wavelengths must fit in the length LL, so λn=2Ln\lambda_n = \frac{2L}{n} and the harmonic frequencies are given by:

fn=nv2L,n=1,2,3...f_n = \frac{nv}{2L}, \quad n = 1, 2, 3...

This gives every harmonic. A pipe open at both ends has antinodes at both ends and produces exactly the same harmonic series.

A pipe closed at one end must have a node at the closed end and an antinode at the open end, so only odd numbers of quarter-wavelengths fit:

fn=nv4L,n=1,3,5...f_n = \frac{nv}{4L}, \quad n = 1, 3, 5...

Only odd harmonics exist. This is why a clarinet, which behaves like a closed--open pipe, sounds roughly an octave below a flute of similar length, which behaves like an open--open pipe.

For sound, remember that a displacement node is a pressure antinode and vice versa. You must be able to:

  • sketch the first few harmonics for each system
  • label nodes (NN) and antinodes (AA)
  • calculate harmonic frequencies and wavelengths from LL and vv

Resonance and natural frequency

Every oscillating system has a natural frequency: the frequency at which it vibrates freely after a single disturbance, set by its physical properties such as mass, stiffness or length. If instead a periodic external force drives the system, it is forced to oscillate at the driving frequency, usually with small amplitude.

Resonance occurs when the driving frequency matches the natural frequency. Energy then transfers from the driver to the system most efficiently, and the amplitude builds up to a maximum, limited only by damping.

A child's swing illustrates this: small pushes timed once per swing cycle produce a large amplitude, while mistimed pushes achieve little. Other examples include:

  • a wine glass driven loudly at its natural frequency
  • parts of a vehicle that buzz at particular engine speeds
  • structures that must be designed so wind or footsteps cannot drive them near a natural frequency

Standing waves are resonance phenomena: a string or air column responds strongly only when driven at one of its harmonic frequencies, which are its natural frequencies.

You should be able to sketch a resonance curve of amplitude against driving frequency, showing a sharp peak at the natural frequency, and describe what happens to the amplitude on either side of the peak.

Damping

Damping is the removal of energy from an oscillating system by resistive forces such as friction, air resistance or viscous drag. In a freely oscillating damped system the amplitude decays over time while the period stays almost constant.

Qualitatively:

  • light damping gives many oscillations of gradually shrinking amplitude
  • critical damping returns the system to equilibrium in the shortest time without oscillating
  • heavy damping also prevents oscillation but the return is slow

Car shock absorbers are designed close to critical damping so the car settles quickly after a bump.

Damping does not merely reduce the amplitude of free oscillations over time; it fundamentally alters the system's resonance response curve. Under forced oscillation, increasing the level of damping:

  • lowers the maximum peak amplitude of the resonance curve
  • broadens the frequency range over which a significant response occurs
  • shifts the peak resonant frequency slightly below the undamped natural frequency

This is why engineers add dampers to bridges and tall buildings: even if wind drives the structure near a natural frequency, the resonant amplitude stays small.

You must be able to sketch resonance curves for different amounts of damping on the same axes and describe these effects qualitatively; no damping calculations are required.

Formulae

fn=nv2Lf_n = \frac{n v}{2L}

To calculate the frequency of the nn-th harmonic (n=1,2,3...n = 1, 2, 3...) for a string fixed at both ends, or for a pipe open at both ends, where vv is the wave speed and LL is the length.

fn=nv4Lf_n = \frac{n v}{4L}

To calculate the frequency of the nn-th harmonic (n=1,3,5...n = 1, 3, 5...) for a pipe closed at one end and open at the other. Note that only odd harmonics exist for this boundary condition.

dnode–node=λ2d_{\text{node--node}} = \frac{\lambda}{2}

To find the wavelength from a sketch or a measurement of a standing wave: adjacent nodes (or adjacent antinodes) are half a wavelength apart, and a node-to-antinode distance is λ4\frac{\lambda}{4}.

Definitions

Standing Wave
A stationary wave pattern formed by the superposition of two coherent waves traveling in opposite directions, characterized by fixed nodes and antinodes.
Node
A point along a standing wave where the displacement of the medium is always zero.
Antinode
A point along a standing wave where the medium oscillates with maximum amplitude.
Natural Frequency
The frequency at which a system oscillates freely after being disturbed, with no external driving force acting on it.
Harmonic
One of the discrete standing-wave modes allowed by a system's boundary conditions; the nn-th harmonic has a frequency that is nn times the fundamental (first harmonic) frequency.
Resonance
The condition under which an oscillating system undergoes maximum amplitude oscillations when driven by an external periodic force at its natural frequency.
Damping
The dissipative process by which energy is removed from an oscillating system by resistive forces (such as friction or drag), causing the amplitude to decay over time.

Worked examples

1

An organ pipe is closed at one end and open at the other. The length of the pipe is 0.85 m0.85\ \text{m}. Taking the speed of sound in air to be 340 m s1340\ \text{m s}^{-1}, calculate the frequency of the third harmonic of this pipe.

  1. 1
    Identify the correct boundary condition: the pipe is closed at one end and open at the other, so we must use the formula fn=nv4Lf_n = \frac{n v}{4L}.
  2. 2
    Recall that for a closed-open pipe, only odd harmonics exist. The third harmonic corresponds to n=3n = 3.
  3. 3
    Substitute the given values into the equation: L=0.85 mL = 0.85\ \text{m}, v=340 m s1v = 340\ \text{m s}^{-1}, and n=3n = 3.
  4. 4
    Calculate the value: f3=3×3404×0.85f_3 = \frac{3 \times 340}{4 \times 0.85}.
  5. 5
    Simplify the denominator: 4×0.85=3.4 m4 \times 0.85 = 3.4\ \text{m}.
  6. 6
    Complete the calculation: f3=10203.4=300 Hzf_3 = \frac{1020}{3.4} = 300\ \text{Hz}.

Answer: 300 Hz300\ \text{Hz}

2

A string of length 0.65 m0.65\ \text{m} is stretched between two fixed supports. When plucked, it vibrates at its fundamental frequency of 110 Hz110\ \text{Hz}. (a) Calculate the speed of transverse waves on the string. (b) State the frequency of the second harmonic.

  1. 1
    For a string fixed at both ends the fundamental (n=1n = 1) obeys f1=v2Lf_1 = \frac{v}{2L}, which rearranges to v=2Lf1v = 2 L f_1.
  2. 2
    Substitute the values: v=2×0.65×110v = 2 \times 0.65 \times 110.
  3. 3
    Evaluate: v=1.3×110=143 m s1v = 1.3 \times 110 = 143\ \text{m s}^{-1}.
  4. 4
    A string fixed at both ends supports all integer harmonics, so fn=nf1f_n = n f_1.
  5. 5
    For the second harmonic: f2=2×110=220 Hzf_2 = 2 \times 110 = 220\ \text{Hz}.

Answer: v=143 m s1v = 143\ \text{m s}^{-1}; f2=220 Hzf_2 = 220\ \text{Hz}

3

A microwave source faces a flat metal reflector, setting up a standing wave between them. A small detector moved along the line between the source and the reflector registers adjacent minima (nodes) separated by 6.0 cm6.0\ \text{cm}. Calculate the frequency of the microwaves.

  1. 1
    Adjacent nodes on a standing wave are half a wavelength apart, so λ=2×0.060=0.12 m\lambda = 2 \times 0.060 = 0.12\ \text{m}.
  2. 2
    Microwaves are electromagnetic waves, so their speed is c=3.00×108 m s1c = 3.00 \times 10^8\ \text{m s}^{-1}.
  3. 3
    Rearrange the wave equation v=fλv = f\lambda to give f=cλf = \frac{c}{\lambda}.
  4. 4
    Substitute and evaluate: f=3.00×1080.12=2.5×109 Hzf = \frac{3.00 \times 10^8}{0.12} = 2.5 \times 10^9\ \text{Hz}.

Answer: 2.5×109 Hz2.5 \times 10^9\ \text{Hz} (i.e. 2.5 GHz2.5\ \text{GHz})

Common mistakes

  • ×Using the incorrect harmonic number (nn) for pipes closed at one end. Remember that the next harmonic above the fundamental (n=1n=1) is the third harmonic (n=3n=3); there is no second harmonic (n=2n=2) for a closed-open pipe.
  • ×Confusing displacement standing wave patterns with pressure standing wave patterns in pipes. A displacement node always corresponds to a pressure antinode, and vice versa.
  • ×Assuming that energy propagates through a standing wave. Unlike traveling waves, the net energy transmission in a standing wave is zero because energy is trapped within the loops between nodes.

Exam tips

  • When asked to **sketch** standing waves in pipes, always clearly label your nodes (NN) and antinodes (AA) to ensure full marks.
  • Be ready to **distinguish** between traveling and standing waves by comparing their energy transfers, phase differences, and amplitude variations across the wave profile.
  • When you **calculate** harmonic wavelengths from a diagram, first count the number of loops. Remember that each loop (from node to node) represents exactly half a wavelength (λ2\frac{\lambda}{2}).

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