Mass-spring systems, simple harmonic motion, damping, forced oscillations, resonance, and energy — the physical language of second-order differential equations
Chapter 7 applies the second-order linear differential equations developed in Chapter 6 to one of the most important families of physical systems: vibrating and oscillating systems. Students who have learned to solve homogeneous second-order equations with constant coefficients will now see exactly why those solutions take the shapes they do, and what each shape means in terms of physical motion.
This chapter studies mass-spring systems, simple harmonic motion, damping, forced oscillations, and a first look at resonance. It connects the algebraic structure of characteristic roots to the physical behavior of a system: whether it oscillates forever, oscillates while decaying, decays without oscillating, or responds dramatically to an external force. By the end of this chapter, students should be able to set up a differential equation from a physical description, classify its solutions, interpret the behavior of the system, and understand why differential equations are the natural language for studying dynamic mechanical systems.
A spring is hanging from a fixed support. A mass is attached to the lower end, pulled downward from its rest position, and released. The mass begins to bounce. Will it bounce forever? Will the bouncing slow down and stop? Does it matter how far it was pulled? Does it matter whether there is friction or air resistance? What happens if someone pushes the mass periodically while it is bouncing? These are not casual questions. They are precise mathematical questions, and differential equations answer them completely.
Before writing any differential equation, it is worth pausing to ask what an oscillation actually is and why it is worth studying carefully. An oscillation is a back-and-forth motion around some central position. The position of the oscillating object varies periodically over time, returning to the same values at regular intervals. A pendulum swings left and right. A spring compresses and extends. A string vibrates up and down. In all of these cases, there is a restoring force: a force that pushes or pulls the object back toward its equilibrium position whenever it is displaced.
The key feature that produces oscillation is the combination of a restoring force and inertia. The restoring force drives the object back toward equilibrium. Inertia causes the object to overshoot. The overshoot leads to displacement in the opposite direction, which generates a restoring force in the opposite direction, which drives the object back again. This cycle repeats.
Real oscillations are also affected by dissipation, the loss of energy to friction, air resistance, or other damping effects. Dissipation causes oscillations to die out over time unless energy is supplied by an external force. All of these phenomena emerge from a single family of second-order linear differential equations. That is the subject of this chapter.
The mass-spring system is not just about springs and masses. The mathematical structure of restoring force, inertia, damping, and external forcing appears in electrical circuits, biological rhythms, structural mechanics, and control systems. Mastering this chapter means mastering the governing equation of oscillatory behavior across all of science and engineering.
Let x(t) denote the position of a mass at time t. The choice of coordinate system matters. For a vertical spring system, it is conventional to define the positive direction as downward from the equilibrium position, so that a positive value of x means the mass is below equilibrium. What matters is consistency: once a positive direction is chosen, it must be maintained throughout the analysis.
The equilibrium position is the position at which the net force on the mass is zero. The displacement x(t) measures how far the mass is from equilibrium, not from a fixed reference point on a wall or ceiling.
Velocity: v(t) = x′(t) = dx/dt
Acceleration: a(t) = x″(t) = d²x/dt²
Newton's Second Law: F_net = m · x″(t)
The task in each problem is to identify all forces, express them in terms of x and x′, substitute into Newton's law, and obtain a second-order differential equation.
Because the governing equation is second-order, every initial value problem requires two conditions, typically x(0) = x₀ (initial displacement) and x′(0) = v₀ (initial velocity). If the mass starts from rest at a displaced position: x₀ ≠ 0, v₀ = 0. If the mass starts at equilibrium with a push: x₀ = 0, v₀ ≠ 0.
It is essential to measure displacement from equilibrium, not from the natural length of the spring. When a mass is attached to a vertical spring, gravity stretches the spring to a new equilibrium position. Measuring from that equilibrium conveniently removes the gravitational force from the equation, as Section 7.4 will show.
The key physical ingredient in the mass-spring model is the restoring force provided by the spring. Hooke's Law states that for a spring within its elastic limit, the force exerted by the spring is proportional to the displacement from the spring's natural (unstretched) length and directed opposite to the displacement.
F_spring = −kx
where k > 0 is the spring constant (stiffness constant). Units: N/m in SI. The negative sign reflects the restoring nature of the force: if x > 0, the spring force is negative (directed toward equilibrium); if x < 0, the spring force is positive.
| Quantity | Description | Example Value |
|---|---|---|
| Spring constant k | Stiffness; force needed per unit displacement | 400 N/m (car spring) |
| Displacement x | Distance from equilibrium; positive or negative | 0.05 m |
| Spring force | −kx; always directed back toward equilibrium | −20 N |
| Static deflection δ | Equilibrium stretch under gravity: kδ = mg | δ = mg/k |
Hang a known mass m from the spring and measure the equilibrium stretch δ. At equilibrium, kδ = mg, so k = mg/δ. This is the simplest way to determine a spring constant in a laboratory.
With Newton's second law and Hooke's law in hand, we can derive the fundamental differential equation for a mass-spring system. Consider a mass m attached to a vertical spring. When the mass is at equilibrium, the static deflection δ satisfies kδ = mg. Let x(t) be the displacement from this equilibrium position, measured positive downward. Two forces act on the displaced mass: gravity mg downward, and spring force k(δ + x) upward.
Applying Newton's second law with positive direction downward:
mx″ = mg − k(δ + x) = mg − kδ − kx
Since kδ = mg, the gravitational term and static deflection term cancel exactly, leaving:
mx″ + kx = 0
Dividing by m and defining the natural angular frequency:
ω₀ = √(k/m)
the equation becomes x″ + ω₀² x = 0. The natural frequency depends only on k and m, not on the initial conditions or amplitude.
By measuring displacement from the equilibrium position rather than from the spring's natural length, the equilibrium absorbs the effect of gravity. The equation mx″ + kx = 0 is identical for a vertical or a horizontal spring system, because the gravitational and static spring forces cancel at the equilibrium point.
The equation x″ + ω₀² x = 0 is a second-order linear homogeneous equation with constant coefficients. From Chapter 6, we know how to solve it using the characteristic equation.
Characteristic equation:
r² + ω₀² = 0 → r = ±ω₀ i
Purely imaginary roots give the general solution:
x(t) = C₁ cos(ω₀ t) + C₂ sin(ω₀ t)
Amplitude-phase form (equivalent):
x(t) = A cos(ω₀ t − φ) where A = √(C₁² + C₂²)
This is simple harmonic motion (SHM) — a pure sinusoidal oscillation at the natural frequency ω₀.
k = 4 N/m has a mass of 1 kg attached. The mass is pulled 0.3 m below equilibrium and released from rest. Find x(t) and describe the motion.
ω₀ = √(k/m) = √(4/1) = 2 rad/s. Governing equation: x″ + 4x = 0. Initial conditions: x(0) = 0.3, x′(0) = 0.
x(t) = C₁ cos 2t + C₂ sin 2t
x(0) = C₁ = 0.3
x′(t) = −2C₁ sin 2t + 2C₂ cos 2t
x′(0) = 2C₂ = 0 → C₂ = 0
Therefore: x(t) = 0.3 cos 2t
x(0) = 0.3 cos 0 = 0.3 ✓. x′(0) = −0.6 sin 0 = 0 ✓. Second derivative: x″ = −1.2 cos 2t; then x″ + 4x = −1.2 cos 2t + 1.2 cos 2t = 0 ✓.
0.3 m at angular frequency 2 rad/s. The period is T = 2π/2 = π ≈ 3.14 s. Since there is no damping, the oscillation continues forever with unchanged amplitude.
| Parameter | Formula | Physical Meaning |
|---|---|---|
| Amplitude A | A = √(C₁² + C₂²) | Maximum displacement from equilibrium; set by initial conditions |
| Angular frequency ω₀ | ω₀ = √(k/m) | Rate of oscillation in rad/s; set by spring and mass |
| Period T | T = 2π/ω₀ = 2π√(m/k) | Time for one complete oscillation |
| Frequency f | f = ω₀/(2π) | Oscillations per second (Hz) |
| Phase angle φ | tan φ = C₂/C₁ | Time offset of the oscillation; set by initial conditions |
Increasing mass m increases the period (heavier mass oscillates more slowly). Increasing spring constant k decreases the period (stiffer spring oscillates more quickly). Amplitude has no effect on period — this is a special property of linear restoring forces.
0.5 kg is attached to a spring with k = 8 N/m. At t = 0, the mass is at equilibrium (x = 0) with velocity v₀ = 0.4 m/s. Find the amplitude, period, and position function.
ω₀ = √(8/0.5) = √16 = 4 rad/s. General solution: x(t) = C₁ cos 4t + C₂ sin 4t.
x(0) = C₁ = 0 → C₁ = 0
x(t) = C₂ sin 4t
x′(t) = 4C₂ cos 4t
x′(0) = 4C₂ = 0.4 → C₂ = 0.1
Amplitude: A = √(0² + 0.1²) = 0.1 m. Period: T = 2π/4 = π/2 ≈ 1.57 s.
x(0) = 0 ✓. x′(0) = 4(0.1)(1) = 0.4 ✓. Substituting: x″ + 16x = −1.6 sin 4t + 1.6 sin 4t = 0 ✓.
0.1 m because it was given an initial push. The mass passes through equilibrium with speed 0.4 m/s at t = 0 and again every half-period thereafter.
In any real physical system, some energy is dissipated over time through air resistance, internal friction in the spring, friction with a surface, or viscous damping from a fluid. The simplest model assumes a damping force proportional to velocity, directed opposite to the direction of motion:
Damping force (viscous): F_damping = −bx′
where b > 0 is the damping coefficient (N·s/m in SI units). Applying Newton's second law:
mx″ + bx′ + kx = 0
Characteristic equation:
mr² + br + k = 0 → r = (−b ± √(b² − 4mk)) / (2m)
The discriminant Δ = b² − 4mk determines whether the system oscillates.
Three cases of damped motion correspond to three physically distinct behaviors. Each produces a qualitatively different type of motion determined entirely by the sign of the discriminant.
| Case | Discriminant | Roots | General Solution | Behavior |
|---|---|---|---|---|
| Underdamped | Negative | Complex: −λ ± ω_d i | e^{−λt}(C₁ cos ω_d t + C₂ sin ω_d t) |
Decaying oscillation |
| Critical | Zero | Repeated: r = −λ | (C₁ + C₂ t)e^{−λt} |
Fastest non-oscillatory return |
| Overdamped | Positive | Two real: r₁, r₂ < 0 | C₁ e^{r₁ t} + C₂ e^{r₂ t} |
Slow, non-oscillatory return |
Sketch the graph of x(t) = e^{−0.5t} cos(2t) for t ≥ 0. The function oscillates back and forth, with each successive maximum and minimum smaller in magnitude. Draw dashed curves ±e^{−0.5t} as the decaying envelope. Label these envelope curves. Note that the oscillation crosses zero at regular intervals even as the amplitude shrinks toward zero over time.
On the same axes, sketch the critically damped solution x(t) = (1 + t)e^{−t} and an overdamped solution such as x(t) = 1.5e^{−t} − 0.5e^{−3t}, both with x(0) = 1 and x′(0) = 0. The critically damped curve reaches zero sooner. The overdamped curve descends more gradually. Neither curve oscillates or crosses zero. Label each curve and mark approximately where each reaches x = 0.05.
1 kg is attached to a spring with k = 5 N/m and a damper with b = 2 N·s/m. The mass starts at x(0) = 1 m with x′(0) = 0. Classify the motion and find x(t).
x″ + 2x′ + 5x = 0. Check discriminant: b² − 4mk = 4 − 20 = −16 < 0. Underdamped.
r² + 2r + 5 = 0
r = (−2 ± √(4 − 20))/2 = (−2 ± 4i)/2 = −1 ± 2i
So λ = 1 and ω_d = 2. General solution:
x(t) = e^{−t}(C₁ cos 2t + C₂ sin 2t)
Applying x(0) = 1: C₁ = 1. Differentiating with the product rule:
x′(t) = −e^{−t}(C₁ cos 2t + C₂ sin 2t)
+ e^{−t}(−2C₁ sin 2t + 2C₂ cos 2t)
x′(0) = −C₁ + 2C₂ = 0 → C₂ = 1/2
Therefore:
x(t) = e^{−t}(cos 2t + (1/2) sin 2t)
x(0) = e⁰(1 + 0) = 1 ✓. x′(0) = −1 + 2(1/2) = 0 ✓. The discriminant is negative, confirming underdamped behavior ✓.
1 kg is on a spring with k = 4 N/m. What value of b produces critical damping? If the mass starts at x(0) = 2 m with x′(0) = −1 m/s, find x(t).
b² = 4mk = 4(1)(4) = 16, so b = 4 N·s/m. The equation is x″ + 4x′ + 4x = 0.
r² + 4r + 4 = (r + 2)² = 0 → r = −2 (repeated)
General solution:
x(t) = (C₁ + C₂ t)e^{−2t}
Applying x(0) = 2: C₁ = 2. Differentiating:
x′(t) = C₂ e^{−2t} − 2(C₁ + C₂ t)e^{−2t}
x′(0) = C₂ − 2C₁ = C₂ − 4 = −1 → C₂ = 3
Therefore:
x(t) = (2 + 3t)e^{−2t}
x(0) = 2 ✓. x′(0) = 3 − 4 = −1 ✓.
(2 + 3t) and a decaying exponential e^{−2t}. Since e^{−2t} decays faster than any polynomial grows, x(t) → 0 as t → ∞. The mass returns to equilibrium without oscillating. This is the fastest non-oscillatory return possible for these spring and mass values.
Engineers often express damping in terms of the damping ratio ζ = b / (2√(mk)). Underdamped: ζ < 1. Critically damped: ζ = 1. Overdamped: ζ > 1. Most structural systems have ζ between 0.01 and 0.05, making underdamped analysis especially important in practice.
In many applications, a mass-spring system is not simply released from an initial position but is subjected to an ongoing external force. A machine might vibrate a platform. An earthquake might shake a building's foundation. An electromagnetic driver might push a speaker cone. In all of these cases, the governing equation has a nonzero right-hand side.
mx″ + bx′ + kx = F(t)
For sinusoidal forcing F(t) = F₀ cos(ωt), the general solution splits into two parts:
x(t) = x_h(t) + x_p(t)
Transient response x_h(t): the complementary solution; decays to zero when b > 0.
Steady-state response x_p(t): the particular solution; persists for all time; oscillates at the forcing frequency ω.
For sinusoidal forcing of a damped system, the steady-state amplitude is:
X = (F₀/m) / √((ω₀² − ω²)² + (2λω)²)
where ω₀ = √(k/m) is the natural frequency and λ = b/(2m) is the decay rate. The methods for deriving this formula using undetermined coefficients will be developed in Chapter 8. For now, the key observation is how X depends on the forcing frequency ω.
In any damped forced system, the transient response (natural oscillation) dies away after a few time constants 1/λ. What remains is the steady-state response oscillating at the forcing frequency ω, not the natural frequency ω₀. The system "forgets" its initial conditions and "remembers" only the driving force.
One of the most important and dramatic phenomena in vibration analysis is resonance: the large-amplitude response that occurs when a system is forced near its natural frequency.
Imagine pushing a child on a swing. If you push at the right moment in each cycle, matching the natural frequency of the swing, the amplitude grows. If you push at the wrong frequency, your pushes sometimes help and sometimes hinder, and the amplitude stays small. Matching the forcing frequency to the natural frequency is the key to large-amplitude response.
When ω = ω₀, the standard particular solution form is a solution to the homogeneous equation (not a new particular solution). Multiplying by t gives the correct resonant particular solution:
x_p(t) = (F₀ / (2mω₀)) · t · sin(ω₀ t)
The factor t causes the amplitude to grow linearly without bound. This is pure resonance.
Sketch the graph of x(t) = t sin(2t) for t ≥ 0. The function oscillates, but each successive maximum and minimum is larger in magnitude because of the growing factor t. The oscillation is contained within the envelope ±t, which grows linearly. Label the envelope y = ±t and mark several oscillation crests showing their increasing height.
The Tacoma Narrows Bridge collapsed in 1940 after wind-induced oscillations drove the bridge at a frequency close to a natural frequency of the structure. Even with damping, forcing near the natural frequency produces very large amplitudes. Engineering design must ensure that natural frequencies do not coincide with likely forcing frequencies from wind, machinery, earthquakes, or traffic.
| Case | Forcing Frequency | Amplitude | Result |
|---|---|---|---|
| Undamped, ω ≠ ω₀ | Different from natural | Finite | Oscillation at both frequencies |
| Undamped, ω = ω₀ | Equals natural | Grows without bound (∝ t) | Pure resonance — system failure risk |
| Damped, ω near ω₀ | Near natural | Large but finite | Near-resonance peak; still hazardous |
| Damped, ω far from ω₀ | Far from natural | Small | Normal forced oscillation |
The solution to a differential equation describes how position evolves over time. Energy provides a complementary perspective that reveals why the system behaves as it does.
Kinetic energy of the mass:
KE = (1/2) m [x′(t)]²
Potential energy stored in the spring:
PE = (1/2) k [x(t)]²
Total mechanical energy:
E(t) = KE + PE = (1/2) m(x′)² + (1/2) kx²
For simple harmonic motion, substituting x(t) = A cos(ω₀ t − φ) and using ω₀² = k/m gives E = (1/2) kA² = constant. The undamped system is conservative: energy transfers between kinetic and potential forms but the total never changes.
In a damped system, the damping force does negative work on the mass, removing energy at a rate equal to b[x′(t)]²:
dE/dt = −b[x′(t)]² ≤ 0
This is always nonpositive, confirming that energy is only dissipated, never created, by damping. Energy is lost most rapidly when the mass is moving quickly (near equilibrium) and least rapidly when the mass is momentarily stationary (at maximum displacement). For an underdamped system with amplitude envelope Ae^{−λt}, the energy decays as E(t) ≈ (1/2)kA²e^{−2λt} — at twice the rate of the amplitude.
Because energy is proportional to amplitude squared, if the amplitude decreases by half, the energy decreases by a factor of four. This quadratic relationship between amplitude and energy is a key fact in acoustics, optics, and signal processing as well as mechanics.
The mass-spring model is a mathematical archetype that appears throughout physics and engineering. Despite their apparent differences, a wide range of systems share the same second-order governing equation.
For small oscillations, the simple pendulum (a mass on a string of length L) obeys θ″ + (g/L)θ = 0, where θ is the angle from vertical. This is identical in structure to the undamped mass-spring equation, with ω₀ = √(g/L) and period T = 2π√(L/g). This linearized approximation is valid for small angles (typically θ ≲ 15°).
A series RLC circuit with resistance R, inductance L, and capacitance C obeys: Lq″ + Rq′ + (1/C)q = V(t), where q(t) is the charge on the capacitor and V(t) is the applied voltage. This is mathematically identical to the forced, damped mass-spring equation.
| Mechanical | Electrical |
|---|---|
| Mass m | Inductance L |
| Damping coefficient b | Resistance R |
| Spring constant k | 1/C (reciprocal capacitance) |
| External force F(t) | Applied voltage V(t) |
| Displacement x(t) | Charge q(t) |
| Velocity x′(t) | Current I(t) = q′(t) |
Oscillations appear throughout biology: the heartbeat is driven by oscillating electrical signals in cardiac muscle, neurons fire in rhythmic bursts, circadian rhythms in cells have periods of approximately 24 hours, and hearing depends on resonance of the basilar membrane in the cochlea at different frequencies. Predator-prey population cycles involve coupled oscillations, which are studied in Chapter 9.
Buildings and bridges have natural frequencies. Earthquake engineering must ensure that building natural frequencies do not match the dominant frequencies of local ground motion. In control engineering, the step response of a second-order system is characterized by its natural frequency and damping ratio. Designers must choose ζ carefully: too little damping leads to large oscillations before settling, while too much damping leads to a slow response.
If displacement is measured from the natural length, the gravitational force does not cancel and the equation has an extra constant term. Always define x = 0 at the equilibrium position. The gravitational and static spring forces cancel there automatically.
ω₀ and ω_d are angular frequencies in rad/s. The frequency in Hz is f = ω/(2π). For underdamped systems, the actual oscillation frequency is ω_d = √(ω₀² − λ²) < ω₀. Use ω_d in the solution, not ω₀.
This guess is a solution to the homogeneous equation, not a new particular solution. At resonance in the undamped case, multiply by t: x_p = t(A cos ω₀ t + B sin ω₀ t). This produces the linearly growing amplitude that characterizes pure resonance.
Compute Δ = b² − 4mk carefully with the actual values of m and k. If Δ < 0: underdamped; if Δ = 0: critical; if Δ > 0: overdamped. Using the wrong case leads to an entirely wrong solution form.
The underdamped solution is a product of two functions. The derivative requires the product rule: [uv]′ = u′v + uv′ where u = e^{−λt} and v = C₁ cos ω_d t + C₂ sin ω_d t. Skipping this leads to wrong values of C₁ and C₂ from the initial conditions.
The actual oscillation frequency in a damped system is ω_d = √(ω₀² − λ²), which is always less than ω₀ when damping is present. Compute ω_d from the imaginary part of the characteristic roots; do not substitute ω₀.
Differential equations are sensitive to unit consistency. Work in one system throughout: SI (kg, N, m, s) or US customary (slug, lb, ft, s). Convert all given quantities before writing the equation.
Chapter 7 has developed the complete theory of undamped and damped free vibrations, and has introduced forced oscillations and resonance. The next step is to study nonhomogeneous equations in full generality.
Forced oscillations, where the right-hand side F(t) is not zero, require the methods of Chapter 8: Nonhomogeneous Second-Order Equations. Chapter 8 will develop the method of undetermined coefficients for polynomial, exponential, and trigonometric forcing; the modification rule for the resonance case; and variation of parameters as a more general alternative.
Students who understand Chapter 7 physically will find that Chapter 8 makes sense as a set of tools for computing what Chapter 7 already describes: the transient and steady-state responses of a driven oscillatory system. Every worked example in Chapter 8 will connect back to the physical picture established here.
Read physical meaning from given solutions. No calculation required for problems 1–4.
x(t) = 0.5 cos(3t) m. Identify the amplitude, angular frequency, period, and frequency in Hz. At what times does the mass reach maximum positive displacement?x(t) = 2 sin(4t) − cos(4t) m. Find the amplitude and write the solution in the form A cos(ωt − φ).x(t) = 3e^{−t} cos(2t) m. What is the amplitude at t = 0? At t = 1 s? At t = 3 s? Describe the long-term behavior.x(t) = (1 + 2t)e^{−3t} m. What type of damping is this? What is the position as t → ∞? Does the mass ever pass through equilibrium after t = 0?ω₁ = 2 rad/s and ω₂ = 6 rad/s, with amplitudes A₁ = 0.4 m and A₂ = 0.1 m respectively. Which system has greater total mechanical energy? (Use E = (1/2)kA² and k = mω₀².)x(t) = e^{−0.5t}(3 cos 2t − sin 2t) m. Find the amplitude-form Ae^{−0.5t} cos(2t − φ). At approximately what time has the envelope amplitude decreased to half its initial value?Write the differential equation and identify key parameters from the physical description.
2 kg is attached to a spring with spring constant 18 N/m. The system is undamped. Write the differential equation and identify ω₀, T, and f.0.05 m when a 1 kg mass is attached. Find the spring constant k. Write the undamped mass-spring equation and find the natural period.0.5 kg mass is on a spring with k = 12.5 N/m and a damper with b = 1 N·s/m. Write the differential equation. Compute the discriminant and classify the motion.m = 4 kg, k = 100 N/m, and b = 8 N·s/m. Write and classify the equation. Find ω₀, λ, ω_d, and the damped period.b produces critical damping for a 2 kg mass on a spring with k = 50 N/m?1 kg is connected to two springs in parallel with k₁ = 3 N/m and k₂ = 5 N/m. Parallel springs obey k_eff = k₁ + k₂. Write the equation and find ω₀.1 kg is connected to two springs in series with k₁ = 4 N/m and k₂ = 4 N/m. Series springs obey 1/k_eff = 1/k₁ + 1/k₂. Write the equation and compare ω₀ to the parallel case above.Classify each system and find the characteristic roots. Write the general solution.
x″ + 2x′ + 5x = 0x″ + 6x′ + 9x = 0x″ + 10x′ + 9x = 02x″ + 4x′ + 8x = 0x″ + 4x′ + 4x = 03x″ + 6x′ + 3x = 0x″ + 4x′ + 13x = 0, x(0) = 1, x′(0) = 0. Identify the damped frequency and write the amplitude-form solution.x″ + 6x′ + 9x = 0, x(0) = 2, x′(0) = −1. Show that the solution remains positive for all t > 0.x″ + 8x′ + 12x = 0, x(0) = 0, x′(0) = 3: solve and determine when the maximum displacement occurs.0.8 kg mass hangs from an unknown spring, stretching it 0.04 m. Find k. If this mass is displaced 0.1 m from equilibrium and released, find the period and amplitude of oscillation.L has period T = 2π√(L/g). What length gives a period of exactly 1 second? (Use g = 9.8 m/s².) This is the "seconds pendulum."400 kg and spring constant k = 40000 N/m. What damping coefficient produces critical damping? If the actual b = 3000 N·s/m, is the system underdamped or overdamped?x(t) = 0.3 cos 4t m with m = 1 kg and k = 16 N/m, verify that E = (1/2)m(x′)² + (1/2)kx² is constant by evaluating it at t = 0, t = π/16, and t = π/8.m = 1 kg, k = 9 N/m. System A: b = 2 N·s/m; System B: b = 8 N·s/m. Classify each. For System A, find ω_d and compare to ω₀.L = 0.2 H, R = 2 Ω, C = 0.05 F, and no applied voltage. The charge satisfies Lq″ + Rq′ + (1/C)q = 0. Classify this circuit as underdamped, critically damped, or overdamped. Find the oscillation frequency if applicable.m = 0.25 kg and k = 4 N/m. A sinusoidal force F₀ cos(ωt) is applied. At what forcing frequency ω does resonance occur? What is the period of the resonant force?Each problem contains an error. Identify it and write the correction.
The exponent is wrong. The repeated root for critical damping is r = −b/(2m) = −4/(2·1) = −2, not −b/m = −4. The correct solution is x(t) = (C₁ + C₂ t)e^{−2t}.
The conceptual error is stating that ω₀ rad/s equals ω₀ Hz. ω₀ = 3 rad/s gives f = 3/(2π) ≈ 0.48 Hz and T = 2π/3 ≈ 2.09 s. The computation of T was accidentally correct via the check formula, but the intermediate claim "3 rad/s = 3/(2π) Hz" (missing the division) reflects the confusion — though the actual 3/(2π) result was stated correctly.
The algebra is correct and the equation mx″ = −kx is right. But the equilibrium in this coordinate system is at x = 0 only if x is measured from the equilibrium position (where the spring is stretched by δ). If x is measured from the natural length, then x = 0 is the natural length, and the equilibrium is at x = δ, not at x = 0. The setup here is fine if x is understood as displacement from the natural length plus δ — which is the same as displacement from equilibrium. The student's conclusion is correct but the reasoning about what "x = 0" means needs to be stated carefully.
The product rule was omitted. The correct derivative is x′(t) = −e^{−t}(C₁ cos 3t + C₂ sin 3t) + e^{−t}(−3C₁ sin 3t + 3C₂ cos 3t). At t = 0: x′(0) = −C₁ + 3C₂ = −1 + 3C₂ = 0, giving C₂ = 1/3, not C₂ = 0.
2 kg mass is attached to a spring with k = 18 N/m and a damper with b = 6 N·s/m. Write the governing differential equation, compute the discriminant, and classify the damping.b would make the system in Problem 1 critically damped?x(0) = 0 and x′(0) = 3 m/s, solve the initial value problem completely, showing all steps including the product rule.Ae^{−λt} cos(ω_d t − φ). State the amplitude at t = 0 and at t = 1 second.m = 0.5 kg and k = 8 N/m. The mass starts at x(0) = −0.2 m with x′(0) = 0.8 m/s. Solve the differential equation. Find the amplitude and period.m = 1 kg, k = 4 N/m, b = 4 N·s/m has x(0) = −1 m and x′(0) = 0. Solve the initial value problem. Does the mass cross x = 0 at any time t > 0? Justify analytically.x(t) = A cos(ω₀ t), at what positions is the kinetic energy maximum? At what positions is the potential energy maximum? Explain using the formulas for KE and PE.k = 400 N/m supports a mass that produces a static deflection of 0.0196 m. (Use g = 9.8 m/s².) Find the mass, the natural frequency in rad/s and in Hz, and the period.b = 40 N·s/m is added to the system in Problem 10, classify the motion. What value of b would produce critical damping?ω₀ = 5 rad/s. An external force F₀ cos(ωt) is applied. For what value of ω does resonance occur? Explain in physical terms why forcing at this frequency is dangerous.t? What does this factor imply about the amplitude over time?Chapter 7 developed the complete theory of free vibrations — homogeneous equations with zero on the right-hand side — and introduced the physical picture of forced oscillations. Chapter 8 will provide the mathematical machinery needed to compute particular solutions systematically: the method of undetermined coefficients for polynomial, exponential, sine, and cosine forcing; the modification rule for the resonance case; and variation of parameters as a more general technique. Students who understand the physical interpretation from Chapter 7 will find that Chapter 8 makes the transient-plus-steady-state picture mathematically precise. Every forced oscillation problem, every driven circuit, every engineering system receiving a sinusoidal input will become fully solvable.