Driven systems and resonance

A driven oscillator receives energy from an external time-dependent force. Its response depends on drive frequency, natural frequency, damping, and what observable is measured. Resonance is a frequency-response phenomenon, not simply the largest displacement seen during startup.

The driven damped equation

For a sinusoidal force,

\[m\ddot x+c\dot x+kx=F_0\cos(\omega t)\]

The mass is \(m\), stiffness is \(k\), damping coefficient is \(c\), and drive amplitude is \(F_0\). Drive angular frequency \(\omega\) is in radians per second. Ordinary frequency is

\[f=\frac{\omega}{2\pi}\]

Confusing hertz with radians per second shifts a sweep by a factor of \(2\pi\).

The state equations are

\[\dot x=v\qquad\dot v=\frac{F_0\cos(\omega t)-cv-kx}{m}\]

import numpy as np

mass, stiffness, damping, amplitude, frequency = 1.0, 4.0, 0.4, 1.0, 2.0
time, position, velocity = 0.5, 0.2, 0.1
drive_force = amplitude * np.cos(frequency * time)
acceleration = (drive_force - damping * velocity - stiffness * position) / mass
print("Acceleration (m/s^2):", acceleration)

Drive phase matters during transients and for timestamped comparisons. In the settled linear response, changing only the arbitrary time origin does not change amplitude.

Transient plus steady response

The complete solution is the sum of a homogeneous transient and a forced steady response:

\[x(t)=x_{\mathrm{transient}}(t)+x_{\mathrm{steady}}(t)\]

With positive damping, the transient decays. The steady displacement is sinusoidal at the drive frequency, not the natural frequency:

\[x_{\mathrm{steady}}(t)=A(\omega)\cos(\omega t-\delta)\]

A startup maximum can include constructive addition of transient and steady components. It should not be used as the settled response without a documented reason.

Steady displacement amplitude

Substitution into the equation gives

\[A(\omega)=\frac{F_0}{\sqrt{(k-m\omega^2)^2+(c\omega)^2}}\]

At very low frequency, the response approaches the static displacement

\[A(0)=\frac{F_0}{k}\]

At very high frequency, inertia dominates and amplitude falls approximately as \(F_0/(m\omega^2)\). Damping keeps the response finite near the natural frequency.

def response(omega):
    return amplitude / np.hypot(stiffness - mass * omega**2, damping * omega)


print(f"Steady amplitude: {response(frequency):.3f} m")

This analytic formula is an excellent pointwise test of a linear numerical solver after transients are removed.

Phase lag

The phase lag obeys a quadrant-aware relation

\[\delta=\operatorname{atan2}(c\omega,\,k-m\omega^2)\]

At low frequency displacement is nearly in phase with force. Near the undamped natural frequency it lags by about one quarter cycle. At high frequency it approaches opposition to the force. Use a two-argument arctangent so the phase stays in the correct quadrant.

Velocity leads displacement by one quarter cycle, so force-velocity phase controls average input power.

Which frequency is resonant?

The undamped natural frequency is

\[\omega_0=\sqrt{\frac{k}{m}}\]

For sufficiently light viscous damping, the displacement-amplitude maximum occurs at

\[\omega_{\mathrm r}=\sqrt{\omega_0^2-\frac{c^2}{2m^2}}=\omega_0\sqrt{1-2\zeta^2}\]

where damping ratio is \(\zeta=c/(2\sqrt{mk})\). This real nonzero maximum exists only when damping is small enough. It lies below \(\omega_0\).

Other observables have different peaks. Velocity amplitude and average absorbed power peak at \(\omega_0\) for this ideal force-driven linear model. Always say whether “resonance frequency” refers to displacement, velocity, acceleration, or power.

Average power and energy balance

Instantaneous power supplied by the drive is

\[P_{\mathrm{in}}(t)=F(t)\dot x(t)\]

and damping removes mechanical energy at rate

\[P_{\mathrm d}(t)=-c\dot x^2(t)\]

In settled periodic motion, average mechanical energy no longer grows, so average input power equals average damping loss in magnitude:

\[\langle P_{\mathrm{in}}\rangle=\frac12c\omega^2A^2\]

period = 2 * np.pi / frequency
times = np.linspace(0, period, 2001)
phase = np.arctan2(damping * frequency, stiffness - mass * frequency**2)
velocity = -response(frequency) * frequency * np.sin(frequency * times - phase)
drive_work = np.trapezoid(amplitude * np.cos(frequency * times) * velocity, times)
damping_transfer = np.trapezoid(damping * velocity**2, times)
assert np.isclose(drive_work, damping_transfer)
print("Drive work and dissipated energy per cycle (J):", drive_work, damping_transfer)

Over an integer number of settled cycles, drive work and damping transfer should agree within integration error.

Quality factor and bandwidth

For light damping, quality factor is

\[Q=\frac{m\omega_0}{c}=\frac{1}{2\zeta}\]

A large \(Q\) gives a tall, narrow resonance and a long transient decay. The approximate half-power bandwidth is

\[\Delta\omega\approx\frac{\omega_0}{Q}=\frac{c}{m}\]

These bandwidth relations are approximations for light damping. A narrow peak requires a fine frequency grid and long settling time, so high quality factor makes numerical measurement more demanding.

Build a reproducible frequency sweep

At each drive frequency, define whether the oscillator restarts from the same initial state or continues from the previous frequency. Restarted sweeps isolate the linear frequency response when they run long enough. Continued sweeps can reveal hysteresis in nonlinear systems but make history part of the input.

frequencies = np.linspace(0.1, 4.0, 100)
amplitudes = response(frequencies)
peak = np.argmax(amplitudes)
print("Largest sampled response:", frequencies[peak], amplitudes[peak])

Use the same estimator everywhere. Half peak-to-peak amplitude works for a clean settled sinusoid. A least-squares sine-cosine fit at the known drive frequency gives amplitude and phase and can tolerate sampling that misses exact extrema. RMS must be converted carefully if comparing with peak amplitude.

Decide when the transient is gone

A fixed number of periods may be inadequate near a narrow resonance because the free transient decays on a damping timescale. Compare amplitude and phase in successive windows. Declare settled only when both change less than tolerances for enough cycles.

# Compare estimates over complete cycles of the analytic steady response.
for cycles in [1, 2, 4]:
    times = np.linspace(0, cycles * period, cycles * 1000 + 1)
    signal = response(frequency) * np.cos(frequency * times - phase)
    design = np.column_stack([np.cos(frequency * times), np.sin(frequency * times)])
    coefficients = np.linalg.lstsq(design, signal, rcond=None)[0]
    print(cycles, "fitted amplitude:", np.linalg.norm(coefficients))

If the duration limit is reached, report the point as unsettled instead of silently using a startup peak.

Numerical and physical checks

Refine both time step and frequency spacing. A small time step does not prevent a coarse sweep from missing a narrow maximum, and a dense sweep does not repair an inaccurate integrator. Compare selected frequencies with the analytic amplitude and phase.

Zero drive must recover free damped motion. Zero damping at exact resonance has no finite settled amplitude in the linear ideal model; amplitude grows with time, so the steady formula's denominator vanishes. Any finite plateau there comes from finite duration, nonlinearity, numerical damping, or another omitted effect.

Beyond the linear model

Nonlinear stiffness can shift resonance with amplitude, produce harmonics, multiple steady responses, jumps, and sweep-direction hysteresis. Coupled systems have several normal modes. Real actuators may supply displacement rather than force or have their own frequency response.

Large motion can change damping, geometry, and material behavior. A frequency response is therefore tied to drive amplitude, boundary conditions, measurement definition, and sweep history. A reliable resonance claim includes all of them plus convergence and energy evidence.

Reading a response curve

A response curve should show more than a connected line through a few samples. Mark the actual drive frequencies, distinguish settled from unsettled points, and include the amplitude definition. If a peak lies between sampled frequencies, the largest sampled point is only a lower bound on the true maximum under the continuous model. Refine around the candidate peak or fit the justified analytic response.

At frequencies well below resonance, stiffness dominates and displacement follows the force almost quasistatically. Through the resonance region, response grows and phase changes rapidly. Well above resonance, inertia dominates, displacement becomes small, and it is nearly opposite the applied force. Seeing all three regions is a useful qualitative check on sweep range and units.

Plotting phase with amplitude helps diagnose mistakes. A supposed linear settled response that shows a discontinuous arbitrary phase jump may have used a one-argument inverse tangent or inconsistent wrapping. A response containing strong harmonics may indicate nonlinearity, clipping, or incomplete transient removal. Inspect the time trace and its frequency content instead of forcing every result into a single sinusoidal amplitude.

Sensitivity and uncertainty

Uncertainty in mass and stiffness changes natural frequency, while uncertainty in damping strongly affects peak height and width. Near a sharp resonance, a small frequency error can create a large amplitude error. Report input uncertainty and sweep resolution rather than presenting the peak frequency with more digits than the experiment or grid supports.

Measurement duration also controls frequency resolution. A short window cannot cleanly distinguish nearby frequencies and may capture a beat between transient and drive. Use an integer or sufficiently long number of drive cycles where possible, retain timestamps, and disclose windowing or fitting choices.

If the driving device is coupled to the oscillator, its force may fall as response grows. Treating measured actuator command as a constant applied force can then exaggerate the ideal response. Measure force at the system boundary or include actuator dynamics. The same caution applies to sensors whose calibration or bandwidth changes near the peak.

Finally, compare energy per cycle. A high amplitude without corresponding drive work is an accounting warning. In settled motion the oscillator returns to the same mechanical energy after each cycle, so all net drive work over that cycle must be transferred by damping or another explicitly modeled loss.

A worked Python notebook

Drive a one kilogram oscillator with stiffness 4 and damping coefficient 0.4 using a one newton sinusoid. The question for this calculation is precise: how does steady displacement amplitude change when we change drive angular frequency, 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 expression describes the steady response after transients decay. 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 0.1 to 4 rad/s. 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 drive angular frequency, expressed in rad/s. The output is steady displacement amplitude, expressed in m. 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.

\[f(q) = \frac{1}{\sqrt{0.16 q^{2} + \left(4 - q^{2}\right)^{2}}}\]

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 2 rad/s, the expression gives 1.25 m. 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("1/sqrt((4-q**2)**2+(0.4*q)**2)", locals={"q": q})
reference_input = 2.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "rad/s")
print("Predicted steady displacement amplitude:", reference_output, "m")

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 0.250614 m. The midpoint, at 2.05 rad/s, gives 1.18395 m. The upper endpoint gives 0.0826023 m. Read these together with the assumptions: they describe different choices of drive angular frequency 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([0.1, 2.05, 4.0])
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} rad/s -> {output_value:.6g} m")
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 drive angular frequency; the vertical axis shows steady displacement amplitude. 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(0.1, 4.0, 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("Drive angular frequency (rad/s)")
ax.set_ylabel("Steady displacement amplitude (m)")
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 steady displacement amplitude change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 2 rad/s is -0.625 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.

\[f'(q) = \frac{2 q \left(4 - q^{2}\right) - 0.16 q}{\left(0.16 q^{2} + \left(4 - q^{2}\right)^{2}\right)^{\frac{3}{2}}}\]

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 = 2.0195
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")
print("Full expression:", full_prediction, "m")
print("Approximation difference:", full_prediction - linear_prediction, "m")

Here the changed input is 2.0195 rad/s. The tangent estimate is 1.23781 m, while the full expression gives 1.23214 m. 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 = 3.9
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 drive angular frequency inside the stated interval, predict whether steady displacement amplitude 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 expression describes the steady response after transients decay. 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: 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(4.0) - evaluate(0.1))
for points in [101, 201, 401]:
    grid = np.linspace(0.1, 4.0, 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

• OpenStax, University Physics Volume 1, Section 15.6, "Forced Oscillations".
• MIT OpenCourseWare, 8.01SC Classical Mechanics, Chapter 23, Section 23.6, "Forced Damped Oscillator".
• Benjamin Crowell, Light and Matter, resonance and driven-oscillation discussions.
• The Physics Hypertextbook, "Resonance".
• Physics LibreTexts, "Forced Oscillations".