Elastic waves and mechanical resonance

Choose one axial rod model

A static rod transmits force through its sections. A moving rod also has inertia: if neighbouring sections transmit different axial forces, the intervening material accelerates. The same elastic relation used in a tensile calculation can therefore generate a wave equation. Boundaries then select standing patterns, and an external periodic force can drive those patterns.

Use a straight uniform rod of length L = 1.00 m and square cross-sectional area A = 100 mm² = 1.00 × 10⁻⁴ m², with a 10 mm side. Assign Young’s modulus E = 100 GPa = 1.00 × 10¹¹ Pa, density ρ = 4000 kg/m³ and Poisson’s ratio ν = 0.25. These are model inputs, not measured properties of a named material. Let x label the reference position along the rod and u(x,t) be axial displacement, positive along +x.

Assume homogeneous isotropic linear elasticity, small strain, traction-free lateral surfaces and wavelengths long compared with the cross-sectional dimensions. The leading rod model retains axial inertia and lets the sides respond approximately as in uniaxial stress. Lateral motion associated with Poisson contraction exists, but its inertial corrections are omitted here. A small amplitude and a long wavelength are separate requirements.

Stress, strain, and Hooke’s law supplies the uniaxial relation. The classical wave equation and Boundary conditions and standing waves are useful return paths for travelling versus standing patterns. Coupled oscillators and normal modes and Driven oscillations and resonance explain the finite-oscillator ideas reused below. This lesson defines its own boundaries, mode coordinate and damping model; follow the force balance and graphs first, then check the integrals.

Turn a force imbalance into a wave equation

The leading axial strain is ε_xx = u_x, where the subscript x means a partial derivative. The axial stress is σ_xx = E u_x. Integrating over the uniform section gives the signed axial force N = Aσ_xx = EA u_x, measured in N. Here N is a whole-section force; it is not the force per unit cut length used for a membrane.

Retain a short segment from x to x + dx. Its right face receives axial force +N(x + dx,t), and its left face receives −N(x,t). This opposite-face sign follows from traction acting on the two opposite normals. The net force is N(x + dx,t) − N(x,t) ≈ N_x dx. The segment mass is ρA dx, so Newton’s law gives ρA u_tt = (EA u_x)_x when no distributed external axial force is present.

For constant E, A and ρ, the area cancels: u_tt = c_rod² u_xx, with c_rod = √(E/ρ). A uniform axial strain has constant N and produces no interior force imbalance. A varying strain produces acceleration. Also, E/ρ has unit (N/m²)/(kg/m³) = m²/s², as required for a squared speed.

For the assigned rod, c_rod = √[(1.00 × 10¹¹)/(4000)] = 5000 m/s. In this nondispersive reduced model, a travelling sinusoid with angular wavenumber β has angular frequency ω = c_rod β. Its phase speed is 5000 m/s independently of its amplitude. The speed of an individual piece of material is u_t and can be much smaller.

Distinguish a thin rod from bulk P and S waves

A longitudinal motion does not by itself identify the modulus controlling its speed. In the thin rod, free lateral surfaces allow transverse elastic strain, leaving leading axial stress E u_x. An infinite-medium longitudinal plane wave has a different strain pattern: displacement depends on the propagation coordinate and points along it, while the two transverse normal strains vanish. Its effective longitudinal modulus is K + 4G/3 rather than E.

For a homogeneous isotropic bulk solid, a P wave has displacement parallel to propagation and speed c_P = √[(K + 4G/3)/ρ]. A bulk S wave has displacement perpendicular to propagation and speed c_S = √(G/ρ). G is the shear modulus and K is the bulk modulus. These are unbounded-medium plane-wave results. They do not impose the rod’s free-side boundary conditions.

Poisson’s ratio is ν = −ε_transverse/ε_axial in a uniaxial test with free lateral surfaces. For isotropic linear elasticity, G = E/[2(1 + ν)] and K = E/[3(1 − 2ν)]. The assigned ν = 0.25 gives G = 40 GPa and K = 200/3 GPa. Thus c_P = 1000√30 m/s ≈ 5477.23 m/s and c_S = 1000√10 m/s ≈ 3162.28 m/s. The same supplied material parameters give three speeds because the models retain different deformation patterns.

Changing the rod’s cross-sectional area at fixed material parameters leaves this leading c_rod unchanged. It does change whole-section force, mass and modal stiffness. Changing boundary conditions changes the permitted standing patterns. These distinctions prevent a single “elastic wave speed” from being used without specifying geometry and motion.

Let fixed and free ends select the modes

Fix the left end axially, so u(0,t) = 0. Leave the right end free of axial traction, so N(L,t) = EA u_x(L,t) = 0. For this nonzero constant EA, the second condition becomes u_x(L,t) = 0. “Free” constrains force, not displacement. The following free vibration has no distributed drive or damping.

Try a standing mode u(x,t) = q(t)φ(x), with q_tt = −ω²q. The wave equation gives φ_xx + β²φ = 0 and ω = c_rod β. The fixed-end condition removes the cosine term, leaving φ(x) proportional to sin(βx). The free-end condition requires cos(βL) = 0. Therefore β_n = (2n − 1)π/(2L), φ_n(x) = sin(β_n x), and f_n = ω_n/(2π) = (2n − 1)c_rod/(4L), for n = 1, 2, 3, … .

The fundamental is one quarter of a sinusoidal wavelength across the rod: λ₁ = 4L. The next modes contain three and five quarter wavelengths. Their frequencies are 1250, 3750 and 6250 Hz. These odd multiples belong to this axial fixed–free problem. A fixed–free bending beam obeys a different differential equation and does not inherit this frequency sequence.

For these first three modes, wavelengths are 4.00 m, 4/3 m and 0.800 m, all long compared with the 0.010 m side. This supports the chosen geometric regime qualitatively; it is not an estimate of the full three-dimensional frequency error. Higher mode number eventually challenges the same long-wave approximation.

Read displacement and stress in one mode

Choose the first free mode with tip amplitude U = 1.00 μm: u(x,t) = U sin(πx/(2L)) cos(ω₁t), where ω₁ = 2500π rad/s. At t = 0, the tip has displacement +U and every point has zero axial velocity. A half-period later, displacement and stress reverse sign. The amplitude graphs show nonnegative peak magnitudes along the rod; they are not a record of every point being at its positive peak at every time.

Differentiate with respect to x. The stress is σ_xx(x,t) = EU[π/(2L)] cos(πx/(2L)) cos(ω₁t). At the fixed end its amplitude is EUπ/(2L) = π/20 MPa ≈ 0.157080 MPa. At the free end it is zero. Thus the displacement node at the clamp is a stress antinode, while the displacement antinode at the tip is a stress node. A large displacement at a point need not mean large local strain there.

Differentiate with respect to t instead. At the tip, the velocity amplitude is ω₁U = 2.5π mm/s ≈ 7.85398 mm/s. This material velocity is about 0.00785 m/s, while c_rod is 5000 m/s. A travelling-wave phase speed, a particle velocity and a standing mode’s frequency are different quantities. The standing pattern does not move bodily down the rod.

The two plotted ordinates have their own labelled units: axial displacement amplitude in μm and stress amplitude in MPa. Vertical displacement on the graph is a data coordinate. The physical material motion is along the rod axis; the curve is not a bent rod outline.

Give one mode a mass and a stiffness

Write u(x,t) = q(t)φ₁(x), with φ₁(x) = sin(πx/(2L)). Because φ₁(L) = 1, q is the physical tip displacement in this one-mode motion. The kinetic energy is ½∫[0,L]ρA u_t² dx = ½m₁q_t², where m₁ = ∫[0,L]ρAφ₁² dx. Since the integral of sin²(πx/(2L)) is L/2, m₁ = ρAL/2 = 0.200 kg.

The axial elastic energy is ½∫[0,L]EA u_x² dx = ½K₁q². Its modal stiffness is K₁ = ∫[0,L]EA(φ₁,x)² dx = EAπ²/(8L) = 1.25 × 10⁶π² N/m ≈ 1.23370 × 10⁷ N/m. Their ratio gives K₁/m₁ = ω₁², agreeing with the wave-equation result. Here β₁ is a spatial wavenumber in m⁻¹; K₁ is a stiffness in N/m.

The whole rod mass is ρAL = 0.400 kg. The effective mass is smaller because different points move at different speeds in this tip-normalised mode. No material was removed. If the mode shape were multiplied by a constant, q would be divided by that constant and the modal mass and stiffness would both change accordingly, leaving the physical field and natural frequency unchanged.

For the free first mode with q initially U and q_t initially zero, energy is ½K₁U² = 0.625π² μJ ≈ 6.16850 μJ. It alternates between elastic and kinetic energy while the lossless model conserves their sum. This gives the bridge from a continuous displacement field to a single oscillator coordinate.

Project a distributed force onto the mode

To drive the same fixed–free problem, apply a distributed axial force per reference length b(x,t), measured in N/m. The field equation becomes ρA u_tt = (EA u_x)_x + b. We retain zero end traction at x = L. Applying a nonzero tip force instead would change that boundary condition.

Prescribe b(x,t) = (2F₁/L)φ₁(x) cos(Ωt), with F₁ = 1.00 N. Multiplying the field equation by φ₁ and integrating over the rod gives the generalised force ∫[0,L]φ₁b dx. Because ∫φ₁² dx = L/2, that force is F₁ cos(Ωt). F₁ is the modal force amplitude, not the spatial integral of b. The shape of the distributed drive is part of the assigned model.

The selected drive couples only to the first mode in the ideal undamped axial model, because the different sine modes are orthogonal on [0,L]. For example, ∫[0,L]φ₁φ₂ dx = 0. A force proportional to φ₂ would have zero first-mode overlap, even if its time frequency equalled ω₁. It could still excite another mode. Matching a natural frequency is insufficient when the applied force has no coupling to that mode.

We now retain this first-mode coordinate and add a prescribed viscous modal damping coefficient C₁. The equation is m₁q_tt + C₁q_t + K₁q = F₁ cos(Ωt). This damping is a separate phenomenological input; it was not derived from the lossless Hooke relation. The comparison below assigns two damping ratios while keeping the rod, drive and boundary conditions fixed.

Calculate the damped steady amplitude

Define ζ = C₁/(2m₁ω₁), the positive dimensionless modal damping ratio, and r = Ω/ω₁, the drive-to-natural frequency ratio. Assign ζ = 0.02 or 0.05. These give C₁ = 20π N s/m ≈ 62.8319 N s/m and C₁ = 50π N s/m ≈ 157.080 N s/m. They are supplied comparison values, not measured material damping.

After transients decay, the steady displacement is q_ss(t) = Q cos(Ωt − δ). Substitution gives Q = (F₁/K₁)/√[(1 − r²)² + (2ζr)²], with phase lag δ = atan2(2ζr, 1 − r²). The two-argument angle chooses the correct quadrant: for positive damping it runs from 0 toward π as r increases. The static reference amplitude is F₁/K₁ = 0.0810569 μm.

At the natural frequency, r = 1, the displacement lags the force by π/2 and Q = F₁/(2ζK₁). Thus Q = 2.02642 μm for ζ = 0.02, and Q = 0.810569 μm for ζ = 0.05. Increasing the assigned damping ratio by a factor 2.5 reduces this particular amplitude by the same factor. The table and figure compare the steady harmonic response, not the early transient or a sweep with a prescribed rate.

For these responses the first-mode spatial shape remains φ₁: u_ss(x,t) = q_ss(t)φ₁(x). Only its modal amplitude and phase change with drive frequency. Arbitrary force shapes and other initial data generally require more modes, while a damping model with mode coupling requires more than this independent scalar equation.

Separate the displacement peak from the natural frequency

For fixed F₁, minimising the squared response denominator gives r_peak = √(1 − 2ζ²), provided 0 < ζ < 1/√2. The positive-frequency displacement peak is slightly below ω₁. For ζ = 0.02 it occurs at about 1249.50 Hz with Q_peak ≈ 2.02683 μm; for ζ = 0.05 it occurs at about 1246.87 Hz with Q_peak ≈ 0.811585 μm. The natural frequency remains 1250 Hz. A coarse plot may hide the small difference, so the figure labels it explicitly.

At r = 1, q_ss = Q sin(Ωt), so velocity q_t = ΩQ cos(Ωt) is in phase with the force F₁ cos(Ωt). The mean injected power is then ½F₁ΩQ. More generally, steady mean damping loss is ½C₁Ω²Q², equal to mean injected power after transients have decayed. For ζ = 0.02 at r = 1, this is about 0.00795775 W. This is an energy-transfer statement, distinct from the location of the displacement maximum.

If ζ = 0 and Ω = ω₁, the formula has no finite steady harmonic amplitude. With zero initial displacement and velocity, the ideal solution is q(t) = [F₁t/(2m₁ω₁)]sin(ω₁t). Its envelope grows linearly while the model remains applicable. At t = 0.010 s the envelope is 10/π μm ≈ 3.18310 μm; the instantaneous displacement also contains the sine factor. This ideal growth does not predict unlimited motion in a real rod.

Different response quantities can have different peak conditions, and a phrase such as “at resonance” should identify what is being maximised. Here the plotted resonance curve is displacement amplitude under a force drive of fixed modal amplitude.

Keep the wave and resonance models within their scope

The derivation concerns axial motion of a slender free-side rod and bulk plane P/S waves in a homogeneous isotropic solid. Bending waves in a beam are another family. In the ideal unforced Euler–Bernoulli model, ρA w_tt + EI w_xxxx = 0 gives ω² = [EI/(ρA)]β⁴ for a travelling sinusoid, so its phase speed depends on wavelength. The constant axial c_rod and odd fixed–free axial spectrum should not be transferred to bending, plates or shells.

A model check has several parts: verify the force and displacement boundary conditions, identify the retained deformation pattern, compare the wavelength with the short geometric dimensions, and keep strains within the linear regime. Changes in material anisotropy, lateral inertia, elastic heterogeneity, large motion or support response can change the equations. The present calculations do not supply a failure limit or an accuracy percentage for those omitted effects.

The useful chain of reasoning is concrete. Cut-face force imbalance produces local acceleration. Geometry and elastic constraints set the restoring modulus. Boundaries select normal modes. The force’s spatial overlap selects which modes it drives. Supplied damping then fixes the finite steady response of the chosen oscillator coordinate.

Worked examples

Try it yourself

1. The same assigned material forms an unbounded isotropic solid. A plane S wave propagates along x with displacement along y. Which speed applies: 5000, 5477.23 or 3162.28 m/s?

   Show the answer and reasoning

   c_S = √(G/ρ) = 1000√10 m/s ≈ 3162.28 m/s.

   The motion is a bulk shear plane wave. The 5000 m/s value belongs to the free-side axial rod, and 5477.23 m/s belongs to the bulk P plane wave.

2. Does φ(x) = sin(πx/L) satisfy both axial boundary conditions of the fixed-left, free-right rod? If not, identify the failed condition.

   Show the answer and reasoning

   No. It satisfies φ(0) = 0 but fails φ_x(L) = 0.

   φ_x = (π/L)cos(πx/L), so φ_x(L) = −π/L. Its right-end displacement is zero, but the axial stress there is nonzero; a free end constrains traction.

3. Replace φ₁ by ψ₁ = 2φ₁ and write the same field as u = s(t)ψ₁. What are s in terms of q, the new effective mass and the new stiffness?

   Show the answer and reasoning

   s = q/2; m_ψ = 0.800 kg; K_ψ = 4K₁.

   Both modal integrals contain the square of the shape or its derivative, so multiplying the shape by two multiplies both coefficients by four. The smaller coordinate keeps physical energy and displacement unchanged; K_ψ/m_ψ = K₁/m₁.

4. At drive frequency 1250 Hz and modal force amplitude 1.00 N, change ζ from 0.02 to 0.05. What happens to the steady tip amplitude?

   Show the answer and reasoning

   It decreases by a factor 2.5, from about 2.02642 μm to 0.810569 μm.

   At r = 1, Q = F₁/(2ζK₁). The comparison holds F₁, K₁ and the natural frequency fixed; it is specific to the supplied viscous modal model.

5. Start the ideal undamped rod at rest and drive it with b(x,t) = (2F₁/L)φ₂(x)cos(ω₁t), where φ₂ = sin(3πx/(2L)). Is the first mode driven at its natural frequency?

   Show the answer and reasoning

   Its drive frequency matches ω₁, but its first-mode generalised force is zero.

   ∫φ₁b dx = (2F₁/L)cos(ω₁t)∫φ₁φ₂ dx = 0 by orthogonality. The first coordinate stays zero from zero initial data. The second mode can still respond off its own natural frequency; the total field need not remain zero.

What to carry forward

• A difference between opposite cut-face forces accelerates rod material; the leading free-side axial wave speed is √(E/ρ).
• Bulk P and S speeds use different restoring moduli and deformation patterns from the thin axial rod.
• A free axial end has zero stress, allowing a displacement antinode; fixed–free axial frequencies are odd multiples of the fundamental.
• Modal mass, stiffness and force depend on shape normalisation; the physical field and natural frequency do not.
• Resonance requires spatial force coupling. In the selected force-driven mode, positive damping gives a finite harmonic response at the natural frequency and lets transients decay. The displacement peak can differ from the natural frequency.