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Solving coupled equations of motion

Equations of motion are coupled when the acceleration of one coordinate or body depends on another coordinate or body. A spring between two carts, masses connected by a rope, and planets attracting one another are coupled systems. They must be solved as one state: advancing one body before evaluating the others mixes different physical times.

One state for the whole system

For two particles in one dimension, a useful state is

\[\mathbf y=(x_1,v_1,x_2,v_2)^T\]

Its derivative is

\[\dot{\mathbf y}=(v_1,a_1,v_2,a_2)^T\]

The accelerations may depend on both positions and both velocities. State order is a data contract: a solver that interprets the second component as another position rather than velocity will return numerically smooth nonsense. Named records can improve readability, but the mathematical coupling remains the same.

import numpy as np

masses = np.array([1.0, 2.0])


def derivative(state):
    x1, x2, v1, v2 = state
    force = 10.0 * (x2 - x1 - 1.0)
    return np.array([v1, v2, force / masses[0], -force / masses[1]])


state = np.array([0.0, 1.2, 0.0, 0.0])
print("State derivative:", derivative(state))

The function returns rates; it does not update state. Time stepping is a separate operation.

Example: two carts joined by a spring

Let cart positions be \(x_1\) and \(x_2\). A spring of rest length \(L_0\) has extension

\[\Delta L=x_2-x_1-L_0\]

assuming cart two remains to the right. Hooke's law gives equal and opposite forces:

\[m_1\ddot x_1=k\Delta L,\qquad m_2\ddot x_2=-k\Delta L\]

If the spring is stretched, it pulls the first cart right and the second cart left. The signs follow from geometry and Newton's third law; they should not be assigned by two unrelated functions.

The center of mass

\[X=\frac{m_1x_1+m_2x_2}{m_1+m_2}\]

moves at constant velocity when no external horizontal force acts. The relative coordinate \(q=x_2-x_1-L_0\) obeys

\[\ddot q=-k\left(\frac{1}{m_1}+\frac{1}{m_2}\right)q\]

This reduction separates uniform center-of-mass motion from internal oscillation. It also supplies two independent checks: center-of-mass momentum remains constant, and the relative motion has angular frequency

\[\omega=\sqrt{k\left(\frac{1}{m_1}+\frac{1}{m_2}\right)}\]

Evaluate pair forces once

A robust interaction loop evaluates each unordered pair once and applies opposite vectors to both bodies.

pair_force = 10.0 * (state[1] - state[0] - 1.0)
net_forces = np.array([pair_force, -pair_force])
accelerations = net_forces / masses
assert np.isclose(net_forces.sum(), 0.0)
print("Accelerations (m/s^2):", accelerations)

If the second force is calculated independently, roundoff or inconsistent branches can violate the third-law pair. If body one is moved before the pair force on body two is calculated, the interaction uses two separations in one nominal instant.

Coupling through a rope constraint

Consider two masses connected by an ideal massless inextensible rope over an ideal pulley. With each coordinate chosen positive in its direction of motion, fixed rope length implies equal acceleration magnitudes. For a simple Atwood machine with \(m_2>m_1\),

\[T-m_1g=m_1a,\qquad m_2g-T=m_2a\]

Adding eliminates tension:

\[a=\frac{(m_2-m_1)g}{m_1+m_2}\]

and substitution gives

\[T=\frac{2m_1m_2}{m_1+m_2}g\]

These results depend on the pulley and rope assumptions. Pulley rotational inertia makes the two tensions differ; rope stretch adds another dynamical degree of freedom. The kinematic rope constraint must be written alongside the force equations.

first_mass, second_mass, gravity = 2.0, 3.0, 9.81
acceleration = (second_mass - first_mass) * gravity / (first_mass + second_mass)
tension = first_mass * (gravity + acceleration)
print(f"Acceleration: {acceleration:.3f} m/s^2; tension: {tension:.3f} N")

The expression for tension uses the first mass equation and the chosen sign convention. Substituting it into the second equation is an independent algebra check.

Matrix form for linear coupled motion

Many small-oscillation systems can be written

\[\mathbf M\ddot{\mathbf q}+\mathbf C\dot{\mathbf q}+\mathbf K\mathbf q=\mathbf f(t)\]

The mass matrix \(\mathbf M\) couples accelerations, \(\mathbf C\) couples velocities, and \(\mathbf K\) couples displacements. Units differ between these matrices, so treating them as interchangeable arrays is incorrect.

To calculate acceleration,

\[\ddot{\mathbf q}=\mathbf M^{-1}(\mathbf f-\mathbf C\dot{\mathbf q}-\mathbf K\mathbf q)\]

but software should normally solve the linear system rather than form an explicit matrix inverse. A singular mass matrix can indicate redundant coordinates, algebraic constraints, or a malformed model.

Normal modes

For an undamped unforced linear system, try \(\mathbf q=\mathbf u e^{i\omega t}\). This leads to

\[(\mathbf K-\omega^2\mathbf M)\mathbf u=\mathbf 0\]

The eigenvectors give collective motion patterns, and the eigenvalues give squared angular frequencies when the physical matrices satisfy the required positivity conditions. Normal modes do not mean the original coordinates were uncoupled; they are a transformed basis that decouples this linear idealization.

Numerical integration of a coupled state

Each stage of a Runge-Kutta method must evaluate the complete derivative from one staged whole-system state. It is wrong to complete all stages for one body while keeping the others at the old state.

dt = 0.01
for _ in range(100):
    first_slope = derivative(state)
    midpoint = state + 0.5 * dt * first_slope
    state = state + dt * derivative(midpoint)
print("State after 1 s:", state)
assert np.isclose(np.dot(masses, state[2:]), 0.0, atol=1e-10)

The same principle applies to implicit methods: solve the coupled unknown next state to the requested tolerance rather than performing one undocumented sequential pass.

Coupling can create fast and slow modes. The step must resolve the fastest relevant mode even if the plotted coordinate mainly shows slow motion. A stable-looking coarse result may suppress or distort an unresolved fast oscillation.

Scaling and solver diagnostics

Coupled states can contain quantities with very different numerical scales. A position measured in kilometres, a small angle in radians, and a velocity in metres per second should not be compared with one raw absolute tolerance. Choose characteristic scales for each kind of variable or use unit-aware absolute tolerances together with a relative tolerance. Rescaling changes numerical conditioning but must not change the physical solution after units are converted back.

An implicit or algebraic solve should report more than a returned state. Record the residual of the equations, iteration count, and whether convergence was achieved. A solver that reaches its iteration limit has not proved the equations are satisfied. If the residual is a vector, scale its components before reducing them to one norm so that the largest numerical unit does not hide failure elsewhere.

Parameter changes can also reveal structural mistakes. Make one mass extremely large and verify that it approaches a fixed support in the appropriate limit. Make the spring constant small and recover nearly independent motion. Exchange labels on two identical bodies and confirm that the solution exchanges in the same way. These tests examine the coupling itself, while time-step refinement examines the numerical approximation.

For long calculations, store derived diagnostics separately from the evolving state. Rounding a displayed center-of-mass position and feeding it back into the next update silently changes the system. The integrator should retain full internal precision, while output routines decide how many digits are justified for reporting.

Initial conditions must satisfy constraints

Positions and velocities should satisfy any rope, joint, or manifold constraint at the initial time. Starting with an inextensible rope at the wrong length or with incompatible endpoint velocities can create an artificial impulse or persistent solver failure.

For differential-algebraic systems, not every arbitrary state vector is admissible. Initialization may require solving constraint equations for dependent coordinates, velocities, and multipliers.

Checks for coupled systems

Coupled motion is not a collection of unrelated one-body updates. It is one larger initial-value problem whose forces, constraints, and numerical stages must all refer to the same system state.

A worked Python notebook

Keep one Atwood-machine mass at 2 kilograms and change the other mass. The question for this calculation is precise: how does acceleration change when we change second hanging mass, 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 rope and pulley are ideal, and positive acceleration follows the second mass downward. 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 0.5 to 5 kg. 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 second hanging mass, expressed in kg. The output is acceleration, expressed in m/s^2. 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.

\[f(q) = \frac{9.81 q - 19.62}{q + 2}\]

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 3 kg, the expression gives 1.962 m/s^2. 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("9.81*(q-2)/(q+2)", locals={"q": q})
reference_input = 3.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "kg")
print("Predicted acceleration:", reference_output, "m/s^2")

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 -5.886 m/s^2. The midpoint, at 2.75 kg, gives 1.54895 m/s^2. The upper endpoint gives 4.20429 m/s^2. Read these together with the assumptions: they describe different choices of second hanging mass 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([0.5, 2.75, 5.0])
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} kg -> {output_value:.6g} m/s^2")
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 second hanging mass; the vertical axis shows acceleration. 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(0.5, 5.0, 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("Second hanging mass (kg)")
ax.set_ylabel("Acceleration (m/s^2)")
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 acceleration change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 3 kg is 1.5696 in those ratio units. Locally, the output increases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = \frac{9.81}{q + 2} - \frac{9.81 q - 19.62}{\left(q + 2\right)^{2}}\]

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 = 3.0225
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, "m/s^2")
print("Full expression:", full_prediction, "m/s^2")
print("Approximation difference:", full_prediction - linear_prediction, "m/s^2")

Here the changed input is 3.0225 kg. The tangent estimate is 1.99732 m/s^2, while the full expression gives 1.99716 m/s^2. 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 = 4.5
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 second hanging mass inside the stated interval, predict whether acceleration 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 rope and pulley are ideal, and positive acceleration follows the second mass downward. 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 m/s^2: 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(5.0) - evaluate(0.5))
for points in [101, 201, 401]:
    grid = np.linspace(0.5, 5.0, 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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