Driven oscillations and resonance

Start with a forced spring

A 2 kg mass moves horizontally against a spring fixed to a support. Let x be its signed displacement from the unforced spring equilibrium, positive to the right. Supply a spring stiffness k = 8 N/m and a viscous resistance coefficient b = 4 kg/s. The spring force is −kx and the resistance force is −bv, where v = dx/dt.

An external drive begins at t = 0 s: F_drive(t) = (8 N) cos(Ωt). Its force amplitude is F₀ = 8 N and its chosen angular frequency Ω is nonnegative, expressed in rad/s. The cosine argument is the numerical phase in radians. Ω = 0 gives a constant force, not an oscillating drive. Initially x(0) = 0 m and v(0) = 0 m/s.

For this exact ideal model the support is fixed, vertical forces balance, and these are the only horizontal forces on the chosen mass. There is no dry friction, nonlinear spring or changing coefficient. Newton’s law gives m*x_ddot + b*x_dot + k*x = F0*cos(Omega*t). Dots denote derivatives with respect to actual time t; each term has units of newtons.

Units, standards, and physical constants and Algebra, functions, graphs, and logarithms prepare the units and graph reading used here. Newton’s laws and free-body diagrams and Friction, tension, drag, springs, and contact forces prepare the signs of the forces. Geometry, trigonometry, and coordinate systems is a published return path for sine and cosine. Locally, cos θ and sin θ mean the horizontal and vertical coordinates of a unit-circle point at angle θ; the response coefficients and amplitudes are defined below.

On a first reading, follow the single-frequency calculation, the two panels and their exact table. Derivatives, Taylor series, and local approximation and Ordinary differential equations are published optional return paths for verifying extrema and startup motion. No later oscillation lesson is needed to interpret this example.

Solve one forcing frequency

First choose Ω = 1 rad/s. Write τ = t/(1 s), a dimensionless numerical time, and try an eventual response x_p(t) = A cos τ + B sin τ. The subscript p denotes a particular solution of the driven equation; A and B have units of metres.

The spring and acceleration terms together multiply displacement by 6 N/m, while the resistance contributes 4 N/m to the cross terms. Matching cosine coefficients gives (6 N/m)A + (4 N/m)B = 8 N; matching sine coefficients gives (6 N/m)B − (4 N/m)A = 0 N.

The second relation gives B = 2A/3. Substitution into the first gives A = 12/13 m and B = 8/13 m. Thus the particular displacement has amplitude D = √(A²+B²) = 4/√13 m, approximately 1.1094 m. Amplitude is nonnegative even when an individual sine or cosine coefficient is negative.

Differentiating this particular solution gives v_p(t) = −(12/13 m/s) sin τ + (8/13 m/s) cos τ. Its velocity amplitude is V = 4/√13 m/s. These two amplitudes have equal numerical values at this chosen frequency but different units. Neither is the initial velocity of the resting mass: v_p(0) = 8/13 m/s.

Define the two response amplitudes

For any Ω > 0, use x_p(t) = A cos(Ωt) + B sin(Ωt). Define a = k−mΩ² and d = bΩ; both have units of N/m. Coefficient matching gives aA+dB = F₀ and aB−dA = 0 N. Therefore A = F₀a/(a²+d²) and B = F₀d/(a²+d²). The denominator is positive for this positive-mass, positive-spring, positive-damping model.

The eventual displacement amplitude is D = F₀/√(a²+d²), and the eventual velocity amplitude is V = ΩD. In this amplitude relation Ω is used as its inverse-time value; radians are dimensionless. For the periodic particular solution, V is also its greatest speed. An amplitude does not carry the sign of the instantaneous displacement or velocity.

Let ρ = Ω/(1 rad/s), the dimensionless numerical frequency. For the supplied coefficients define g(ρ) = ρ⁴−4ρ²+16 = (ρ²−2)²+12. The amplitudes become D = (4/sqrt(g(rho))) m; V = (4*rho/sqrt(g(rho))) m/s. The strict lower bound g ≥ 12 makes both finite at every ρ ≥ 0.

A lag δ can write the same particular motion as x_p = D cos(Ωt−δ). Here δ = atan2(4ρ,8−2ρ²), using both numerical components in the same N/m units. Choose 0 ≤ δ ≤ π, so the sine coefficient is nonnegative and the cosine sign is retained. At ρ = 2 the lag is π/2, although the largest displacement occurs elsewhere.

At Ω = 0 the eventual state is the static deflection x_p = F₀/k = 1 m, with zero velocity. Use D(0) = 1 m and V(0) = 0 m/s as the endpoints of the comparison. They do not describe a cycle or an instantaneous jump of the initially resting mass.

Compare frequencies and quantities

The upper panel compares D against forcing frequency; the lower compares V against that same frequency. Blue circles and green squares distinguish the selected values. The diamond marks the global displacement maximum and the triangle the global velocity maximum; dotted guides only locate their frequencies.

For this model the displacement maximum is D = 2/√3 m at Ω = √2 rad/s. The velocity maximum is V = 2 m/s at Ω = 2 rad/s. At that second frequency the displacement amplitude is 1 m, below its own maximum. Name the response quantity when calling a frequency resonant.

These are separate constant-frequency cases. The horizontal coordinate is not time, and a line joining frequency samples is not a path of the mass or a record of sweeping the drive. Each ordinate has its own units and scale; equal drawn heights in different panels do not compare physical quantities.

The mathematical comparison covers every Ω ≥ 0, but the drawing shows only 0 ≤ Ω ≤ 4 rad/s. Its short line segments approximate the analytic curves. The hollow markers and table are selected model values, not experimental observations.

Check when a displacement peak exists

The two peaks can be checked without a plot. Differentiating the numerical amplitude functions with respect to ρ gives D′ = −8ρ(ρ²−2)/g^(3/2), carrying metres, and V′ = 4(16−ρ⁴)/g^(3/2), carrying m/s. Each changes from positive to negative at its stated positive peak. Both amplitudes approach zero at arbitrarily high frequency.

For general positive m, k and b with F₀ > 0, a positive-frequency displacement maximum exists only when k/m−b²/(2m²) > 0. It then occurs at Ω_D = √(k/m−b²/(2m²)). If that expression is zero or negative, the greatest displacement amplitude is at the static endpoint Ω = 0; there is no positive interior displacement peak.

Here the strict threshold is b < 4√2 kg/s. This is not the unforced critical-damping threshold b = 2√(mk) = 8 kg/s. A decaying free oscillation and a positive-frequency displacement resonance are different questions.

For a concrete comparison, change only b to 8 kg/s. The eventual amplitudes simplify to D = 4/(ρ²+4) m and V = 4ρ/(ρ²+4) m/s. D decreases for every ρ > 0, while V still has a maximum: 1 m/s at ρ = 2. More damping has removed the interior displacement peak without removing that velocity peak.

The undamped natural angular frequency √(k/m) = 2 rad/s is an equation parameter. The free damped carrier here is √3 rad/s, and the displacement resonance is √2 rad/s. They answer three different questions. Replacing the exact displacement peak by the natural frequency would require an appropriate weak-damping approximation, which this comparison does not use.

Restore the initial state

An eventual particular response does not usually satisfy the initial data. Choose Ω = 2 rad/s. The coefficient equations give x_p = (1 m) sin(2τ) and v_p = (2 m/s) cos(2τ). At t = 0 s its displacement is zero but its velocity is 2 m/s, contrary to the supplied resting state.

Add a solution of the unforced equation, so the sum still solves the driven equation. For the main coefficients that transient has the form e^(−τ)[C cos(√3τ)+S sin(√3τ)], with C and S in metres. The initial displacement requires C = 0 m; the initial velocity requires S = −2/√3 m.

The selected initially resting displacement is x(t) = [sin(2*tau) - (2/sqrt(3))*exp(-tau)*sin(sqrt(3)*tau)] m. Its initial acceleration is +4 m/s² because the initial spring and resistance forces are zero while the drive is +8 N.

The difference x−x_p has magnitude at most (2/√3)e^(−τ) m. It tends to zero, but no finite startup duration was supplied for discarding it. The plotted D and V are eventual amplitudes, not bounds on every displacement or speed during startup. Different initial states change the transient, while this positive-damping model has the same eventual particular response.

Follow the drive and dissipation

Use mechanical energy E = mv²/2+kx²/2, with zero spring energy at x = 0. Differentiating and using the force equation gives dE/dt = F_drive v−bv². The drive can add or remove energy at an instant; the viscous term always removes mechanical energy or is zero. With a drive present, E need not decrease monotonically.

In the eventual periodic response at Ω > 0, average over one full forcing period 2π/Ω. E returns to the same value, so mean supplied power equals mean viscous dissipation: ⟨P_drive⟩ = bV²/2. This identity is about the eventual cycle, not an arbitrary startup interval.

At the velocity peak, V = 2 m/s gives mean supplied power (4 kg/s)(2 m/s)²/2 = 8 W. At the displacement peak V = 2√(2/3) m/s gives 16/3 W. These quantities reinforce why the chosen response measure matters.

Now halve only the drive amplitude to 4 N, keeping coefficients, frequency and the zero initial data. Linearity halves the displacement and velocity, including the transient. Peak frequencies are unchanged, while the mean power is quartered: its velocity-peak value is 2 W. This scaling is conditional on the unchanged linear model, not a rule for an arbitrary real oscillator.

Keep boundaries and return paths explicit

Constant forcing is a separate boundary case. At Ω = 0 the eventual deflection is 1 m, but the initially resting solution is x(t) = [1−e^(−τ)(cos(√3τ)+(1/√3)sin(√3τ))] m. At t = π/√3 s it reaches 1+e^(−π/√3) m, above 1 m. The static endpoint of the frequency plot does not rule out startup overshoot.

Removing damping changes the resonant problem. For b = 0, Ω = 2 rad/s and the same resting data, direct substitution gives x(t) = [τ sin(2τ)] m. Its oscillation envelope grows without bound. There is no bounded periodic steady response at that undamped resonant frequency; the positive-damping amplitude formula cannot be used there with a zero denominator.

For a formal return, coefficient matching explains the two amplitude curves, their derivatives explain the maxima, and the added transient checks both initial data. An optional complex-number method replaces the cosine drive by a complex exponential and finally takes the real displacement; linearity permits that calculation, but the imaginary part is not an additional physical direction.

The spring and resistance laws were supplied exactly. Their useful ranges for a material device, dry-friction transitions, nonlinear resonance, changing parameters, coupled modes and safety limits need additional evidence and models. A finite amplitude graph alone does not establish any of them.

Worked examples

Try it yourself

1. For the main model at Ω = 2 rad/s, find A, B, D, V and the lag. Does that particular response satisfy the resting initial data?

   Show the answer and reasoning

   A = 0 m, B = 1 m, D = 1 m, V = 2 m/s and δ = π/2 rad. It does not satisfy zero initial velocity.

   The dynamic spring-minus-inertia coefficient is zero and d = 8 N/m. Thus x_p = (1 m) sin(2τ), with x_p(0) = 0 m but v_p(0) = 2 m/s.

2. At Ω = 3 rad/s, calculate D and V. What is the sign of the cosine coefficient A, and can D be negative?

   Show the answer and reasoning

   D = 4/√61 m, V = 12/√61 m/s and A = −20/61 m. D is positive.

   g(3) = 81−36+16 = 61 and 4(4−9)/61 = −20/61. A coefficient records a phase sign; √(A²+B²) is an amplitude and is nonnegative.

3. Give both global positive-frequency peaks of the main model and the velocity amplitude at the displacement peak.

   Show the answer and reasoning

   D peaks at Ω = √2 rad/s with D = 2/√3 m. V peaks at Ω = 2 rad/s with V = 2 m/s. At the displacement peak V = 2√(2/3) m/s.

   The derivative factors ρ²−2 and 16−ρ⁴ locate different extrema. Multiplying D at ρ = √2 by √2 s⁻¹ gives the stated velocity amplitude.

4. Change only F₀ from 8 N to 4 N. At Ω = 2 rad/s, find the eventual D, V and mean supplied power. Does the velocity-peak frequency change?

   Show the answer and reasoning

   D = 1/2 m, V = 1 m/s and mean supplied power is 2 W. The velocity peak stays at Ω = 2 rad/s.

   The linear response amplitudes scale with F₀, while their frequency extrema do not. Power bV²/2 scales with the square of the amplitude, so 8 W becomes 2 W.

5. At Ω = 0, compare the eventual displacement with the initially resting solution at t = π/√3 s. What is the velocity at that time?

   Show the answer and reasoning

   The eventual displacement is 1 m. At that time x = [1+e^(−π/√3)] m > 1 m and v = 0 m/s.

   In the static-forcing startup solution the sine is zero and the cosine is −1 at √3τ = π. Differentiation gives v = (4/√3 m/s)e^(−τ)sin(√3τ), which is zero there. This startup turn is not an oscillating-drive resonance.

6. For m = 2 kg and k = 8 N/m with positive drive amplitude, what strict condition on b gives a positive-frequency displacement peak? Is equality sufficient, and is this the unforced critical threshold?

   Show the answer and reasoning

   The strict condition is 0 < b < 4√2 kg/s. Equality gives no positive interior displacement peak. Unforced critical damping is instead b = 8 kg/s.

   The peak frequency squared is k/m−b²/(2m²). It must be positive. The separate unforced discriminant vanishes when b² = 4mk, rather than the displacement-peak boundary b² = 2mk.

What to carry forward

• State which response amplitude defines resonance; displacement and velocity can peak at different forcing frequencies.
• An eventual particular solution still needs a transient to meet the chosen initial state.
• Static endpoints, damping thresholds and undamped resonance change the question, so keep their conditions explicit.