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Standing waves

A standing wave is an oscillating pattern whose nodes and antinodes stay at fixed positions. The medium still moves: points between nodes oscillate back and forth. What stands still is the spatial pattern, not the material. Standing waves appear when waves traveling in opposite directions have the same frequency, wavelength, and amplitude and maintain a stable phase relationship. Reflection at a boundary is a common way to create the returning wave.

From traveling waves to a standing pattern

Take two sinusoidal waves with amplitude \(A\), wave number \(k\), and angular frequency \(\omega\). Let one travel right and the other left:

\[y_1=A\sin(kx-\omega t),\qquad y_2=A\sin(kx+\omega t)\]

The superposition identity gives

\[y(x,t)=y_1+y_2=2A\sin(kx)\cos(\omega t)\]

This form separates into a spatial factor and a time factor. At positions where \(\sin(kx)=0\), displacement is always zero. Those positions are nodes. Halfway between neighboring nodes, \(|\sin(kx)|=1\), so the oscillation amplitude is largest; these positions are antinodes. Neighboring regions oscillate in opposite phase because the spatial sine changes sign across a node.

The coefficient \(2A\) is the maximum displacement amplitude at an antinode. It does not say that every point moves with that amplitude. Nor does it mean that energy was created: two traveling waves, each with amplitude \(A\), supplied the standing pattern.

Fixed ends select discrete modes

Consider an ideal string of length \(L\) fixed at \(x=0\) and \(x=L\). The boundary conditions are

\[y(0,t)=0,\qquad y(L,t)=0\]

The first condition is already satisfied by \(\sin(kx)\). The second requires \(\sin(kL)=0\), so \(kL=n\pi\) for positive integers \(n\). Therefore

\[k_n=\frac{n\pi}{L},\qquad \lambda_n=\frac{2L}{n},\qquad n=1,2,3,\ldots\]

For a nondispersive string, \(v=f\lambda\). The corresponding mode frequencies are

\[f_n=\frac{nv}{2L}=n f_1\]

Mode \(n=1\) is the fundamental. It has nodes at both ends and one antinode. Mode \(n=2\) also has a node at the center. In general, mode \(n\) contains \(n\) half-wavelengths along the string, \(n-1\) interior nodes, and \(n\) antinodes.

For example, let \(L=2\,\mathrm{m}\) and \(v=12\,\mathrm{m/s}\). Then \(f_1=3\,\mathrm{Hz}\) and \(f_3=9\,\mathrm{Hz}\). Doubling the length while preserving wave speed halves every frequency.

One fixed end and one free end

A free end is not constrained to zero displacement. For an ideal transverse string, the transverse force vanishes there, which is represented by zero spatial slope:

\[y(0,t)=0,\qquad \left.\frac{\partial y}{\partial x}\right|_{x=L}=0\]

The fixed end is a node and the free end is an antinode. Only an odd number of quarter wavelengths fits:

\[L=\frac{(2n-1)\lambda_n}{4},\qquad f_n=\frac{(2n-1)v}{4L},\qquad n=1,2,3,\ldots\]

The allowed frequency ratios are \(1:3:5:\cdots\), not every positive integer. This is why boundary conditions must be stated before quoting a harmonic series. Pipes follow related patterns, although whether an end is a pressure node or displacement node requires care; pressure and air-displacement descriptions swap node and antinode roles.

Normal modes and resonance

Each allowed spatial pattern is a normal mode. In an ideal linear system, an arbitrary motion can be written as a sum of modes, and each modal amplitude oscillates at its own frequency. A single mode of a fixed-fixed string has the form

\[y_n(x,t)=\sin\left(\frac{n\pi x}{L}\right)\left[C_n\cos(\omega_n t)+D_n\sin(\omega_n t)\right]\]

The constants \(C_n\) and \(D_n\) are set by the initial displacement and velocity. Calling the mode frequencies “natural” does not mean other frequencies cannot be applied. It means free motion decomposes into these frequencies and a periodic driver transfers energy especially efficiently when its frequency lies near one of them. Damping limits the resonant amplitude and broadens the frequency response of a real system.

import numpy as np

length, speed, time = 1.0, 2.0, 0.1
x = np.linspace(0, length, 101)
displacement = np.zeros_like(x)
for mode, amplitude in [(1, 0.01), (3, 0.003)]:
    frequency = mode * speed / (2 * length)
    displacement += (
        amplitude
        * np.cos(2 * np.pi * frequency * time)
        * np.sin(mode * np.pi * x / length)
    )
print("Endpoint displacements (m):", displacement[[0, -1]])

This Python is a modal calculation, not a finite-difference string update. It works when the chosen modes satisfy the same boundary conditions and the system is linear.

Energy moves even when the pattern does not

At an antinode, displacement is momentarily greatest when every point is instantaneously at rest. The string then has mostly elastic potential energy. A quarter period later it passes through equilibrium fastest and has mostly kinetic energy. The two forms exchange throughout the cycle.

An ideal standing wave has zero time-averaged net energy transport along the string because its equal traveling components carry equal average power in opposite directions. That does not mean its local energy density is zero. It also does not mean instantaneous energy flow is zero everywhere. A source that builds a standing wave must supply energy, and damping can continuously remove it.

Computing a standing wave on a grid

A finite-difference solver can update the one-dimensional wave equation. For grid spacing \(\Delta x\) and time step \(\Delta t\), define the Courant number

\[r=\frac{v\Delta t}{\Delta x}\]

The standard centered explicit update is stable only under the appropriate Courant condition; for the uniform one-dimensional wave equation, require \(r\le 1\).

dx, dt = x[1] - x[0], 0.002
courant = speed * dt / dx
assert courant <= 1
current = 0.01 * np.sin(np.pi * x / length)
previous = current * np.cos(np.pi * speed * dt / length)
for _ in range(100):
    following = np.zeros_like(current)
    following[1:-1] = (
        2 * current[1:-1]
        - previous[1:-1]
        + courant**2 * (current[2:] - 2 * current[1:-1] + current[:-2])
    )
    previous, current = current, following
assert current[0] == current[-1] == 0
print("Midpoint displacement:", current[50])

The initial second time level needs its own formula that includes initial velocity; simply copying the initial displacement silently imposes an unintended velocity approximation. The numerical grid also has dispersion, so high-frequency modes may propagate at the wrong speed even when the update is stable.

Checks that expose mistakes

Boundary residuals should remain near zero after every update. For a pure analytic mode, sample a point away from a node and measure the period; the estimate should approach \(1/f_n\) under refinement. Check node positions against \(x=mL/n\) for integers \(m=0,1,\ldots,n\).

Also repeat a run with half the grid spacing and a correspondingly safe time step. A stable answer that fails to converge is not yet reliable. Monitor energy: modest bounded oscillation may reflect discrete energy definitions, but systematic growth suggests an unstable update or bad boundary enforcement, while systematic decay in an undamped model suggests numerical dissipation.

Real strings depart from this ideal through stiffness, damping, changing tension, nonuniform linear density, finite support motion, air loading, and large-amplitude nonlinearity. These effects can shift frequencies away from exact integer ratios. The ideal modes remain useful because they provide a clear baseline and strong tests for a more detailed model.

A worked Python notebook

Inspect the third standing-wave mode of a string fixed at both ends. The question for this calculation is precise: how does third-mode shape change when we change position, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

The shape is normalised; a separate amplitude and time factor set the physical displacement. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0 to 1 m. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for position, expressed in m. The output is third-mode shape, expressed in dimensionless. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = \sin{\left(3 \pi q \right)}\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 0.25 m, the expression gives 0.707107 dimensionless. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("sin(3*pi*q)", locals={"q": q})
reference_input = 0.25
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "m")
print("Predicted third-mode shape:", reference_output, "dimensionless")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 0 dimensionless. The midpoint, at 0.5 m, gives -1 dimensionless. The upper endpoint gives 0 dimensionless. Read these together with the assumptions: they describe different choices of position within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.0, 0.5, 1.0])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} m -> {output_value:.6g} dimensionless")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows position; the vertical axis shows third-mode shape. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.0, 1.0, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Position (m)")
ax.set_ylabel("Third-mode shape (dimensionless)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does third-mode shape change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 0.25 m is -6.66432 in those ratio units. Locally, the output decreases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = 3 \pi \cos{\left(3 \pi q \right)}\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 0.255
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "dimensionless")
print("Full expression:", full_prediction, "dimensionless")
print("Approximation difference:", full_prediction - linear_prediction, "dimensionless")

Here the changed input is 0.255 m. The tangent estimate is 0.673785 dimensionless, while the full expression gives 0.673013 dimensionless. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 1.0
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of position inside the stated interval, predict whether third-mode shape will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: The shape is normalised; a separate amplitude and time factor set the physical displacement. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of dimensionless: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(1.0) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 1.0, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources


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