Standing Waves on a String
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You should be comfortable with the wave equation \(\partial_t^2 u = c^2 \partial_x^2 u\) and the concept of boundary conditions. Familiarity with simple harmonic motion and trigonometric identities is assumed.
Why this matters
Standing waves are the fundamental building blocks of all wave phenomena, from musical instruments to quantum mechanics.
While travelling waves describe energy moving through space, standing waves describe how energy is stored in specific spatial patterns. Understanding these patterns is crucial because they explain why strings and air columns produce discrete notes rather than a continuous range of sounds. This lecture bridges the gap between the general wave equation and the specific, quantized frequencies observed in physical systems.
The Idea of a Standing Wave
Imagine plucking a guitar string. The disturbance travels to the end, reflects, and travels back. When the outgoing and returning waves overlap, they interfere. If the timing is right, the interference creates a pattern where certain points on the string never move (nodes) and others oscillate with maximum amplitude (antinodes). This is a standing wave. Unlike a travelling wave, which carries energy from one end to the other, a standing wave stores energy locally, sloshing between kinetic and potential forms within fixed spatial regions. The key insight is that the boundary conditions—the fixed ends of the string—force the wave to fit into the space in specific, discrete ways.
Boundary Conditions and Quantization
For a string of length \( L \) fixed at both ends, the displacement \( u(x,t) \) must be zero at \( x=0 \) and \( x=L \) for all time \( t \). These are the boundary conditions: \( u(0,t) = 0 \) and \( u(L,t) = 0 \). When we seek solutions to the wave equation that satisfy these conditions, we find that not every frequency is allowed. The spatial part of the solution must be a sine function that starts and ends at zero. This geometric constraint leads to the quantization of wavenumbers. Only wavelengths that fit an integer number of half-waves into the length \( L \) are permitted. This is the origin of the discrete spectrum of frequencies.
Here, \( n \) is the mode number. The fundamental mode (\( n=1 \)) has the longest wavelength and lowest frequency. Higher modes (\( n=2, 3, \dots \)) have shorter wavelengths and higher frequencies. The corresponding angular frequencies are given by \( \omega_n = c k_n \), where \( c \) is the wave speed on the string.
Normal Modes and Superposition
Each allowed wavenumber \( k_n \) corresponds to a specific spatial shape called a normal mode. The displacement for the \( n \)-th mode is given by a sine function in space and a harmonic oscillation in time. The general solution for a single mode is:
The constants \( a_n \) and \( b_n \) determine the amplitude and phase of the oscillation for that specific mode. A crucial property of linear wave equations is superposition. The total displacement of the string is the sum of all possible normal modes:
This means that any arbitrary initial shape of the string (that satisfies the boundary conditions) can be decomposed into a sum of these sine waves. This is the foundation of Fourier analysis, which will be explored in the next lecture.
Nodes and Antinodes
In a standing wave, the points that do not move are called nodes. For the \( n \)-th mode, there are \( n+1 \) nodes, including the two fixed ends. The points of maximum displacement are called antinodes. The distance between adjacent nodes is half a wavelength, \( \lambda_n / 2 \). Since the string length \( L \) contains \( n \) half-wavelengths, we have \( L = n \lambda_n / 2 \), which leads to \( \lambda_n = 2L/n \). The frequency of the \( n \)-th mode is \( f_n = c / \lambda_n = nc / 2L \). Notice that the frequencies are integer multiples of the fundamental frequency \( f_1 = c / 2L \). This harmonic series is responsible for the timbre of musical instruments.
The derivation
Standing Waves and Boundary Quantization
For the transverse displacement \( u(x,t) \) of a uniform string of length \( L \) held fixed at both ends, the wave equation \( \partial_t^2 u = c^2\,\partial_x^2 u \) together with the boundary conditions \( u(0,t)=u(L,t)=0 \) admits nontrivial time-harmonic solutions only for a discrete set of wavenumbers \( k_n = n\pi/L \) with \( n = 1,2,3,\dots \); the corresponding angular frequencies form the spectrum \( \omega_n = n\pi c/L \), and the associated spatial profiles (normal modes) are \( \sin(k_n x) \).
A common misconception is that a standing wave carries no energy at its nodes. The kinetic energy density is zero at a node, but the slope of the string is largest there, so the potential (stretching) energy density peaks at the nodes; at the antinodes it is the other way round. Averaged over a cycle, the energy density of a single mode is uniform along the string, and energy shuttles between the node and antinode regions every quarter period with no net flow along the string. C
Worked example
Problem. A guitar string of length \( L = 0.65 \ \mathrm{m} \) is fixed at both ends. The wave speed on the string is \( c = 400 \ \mathrm{m/s} \). Calculate the fundamental frequency and the frequency of the third harmonic.
- Identify the formula for the fundamental frequency: \( f_1 = \frac{c}{2L} \).
- Substitute the given values: \( f_1 = \frac{400 \ \mathrm{m/s}}{2 \times 0.65 \ \mathrm{m}} \).
- Calculate \( f_1 \): \( f_1 = \frac{400}{1.3} \approx 307.7 \ \mathrm{Hz} \).
- The third harmonic corresponds to \( n=3 \), so \( f_3 = 3 f_1 \).
- Calculate \( f_3 \): \( f_3 = 3 \times 307.7 \approx 923.1 \ \mathrm{Hz} \).
Units check. The wave speed is in \( \mathrm{m/s} \) and length is in \( \mathrm{m} \). The ratio \( \mathrm{m/s} / \mathrm{m} \) yields \( 1/\mathrm{s} \), which is Hertz (\( \mathrm{Hz} \)). The units are consistent.
Three quick tests
- What is the relationship between the wavelength of the \( n \)-th mode and the length of the string?
Answer
The wavelength is \( \lambda_n = 2L/n \). - How many nodes does the second harmonic (\( n=2 \)) have?
Answer
It has 3 nodes (including the two ends). - If the tension in the string is increased, what happens to the fundamental frequency?
Answer
The wave speed \( c \) increases, so the fundamental frequency \( f_1 = c/2L \) increases.
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.