Vibration of strings and axial bars

Wave equation for taut strings, axial bars and shafts in torsion; wave speed, natural frequencies and mode shapes for fixed and free ends, and a bar with an end mass.

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Why it matters

Strings and axial bars are the simplest continuous systems: mass and stiffness are spread out, so there are infinitely many natural frequencies instead of one or two. They introduce the wave equation, boundary conditions and mode shapes that carry straight over to torsion of wings and shafts and to beam bending. Axial (longitudinal) modes also matter directly in launch vehicles, landing-gear struts, control rods and wire-braced or cable structures.

Key ideas

Taut string. A string of mass per unit length μ under tension T, deflecting transversely by w(x, t) with small slopes, obeys T·∂²w/∂x² = μ·∂²w/∂t². This is the one-dimensional wave equation ∂²w/∂t² = c²·∂²w/∂x² with wave speed c = √(T/μ). Assumptions: tension constant and much larger than any change caused by the motion, small slopes, no bending stiffness.

Axial bar. For a uniform bar with Young's modulus E, density ρ and area A, the axial displacement u(x, t) satisfies E·A·∂²u/∂x² = ρ·A·∂²u/∂t², the same wave equation with c = √(E/ρ). This is the speed of sound in the material (about 5000 m/s for steel and aluminium). Assumptions: plane sections stay plane, lateral (Poisson) inertia neglected, wavelengths long compared with the bar's cross-section.

Torsion of shafts. Torsional vibration of a uniform circular shaft gives the same equation for the twist θ(x, t), with c = √(G/ρ). Everything below applies with G in place of E.

Separation of variables. Put w = W(x)·sin(ω·t): W″ + (ω/c)²·W = 0, so W = A·sin(β·x) + B·cos(β·x) with β = ω/c. The boundary conditions fix the allowed β values (the eigenvalues):

  • Fixed–fixed string or bar (W = 0 at both ends): sin(β·L) = 0, β·L = n·π. Modes sin(nπx/L).
  • Free–free bar (strain ∂u/∂x = 0 at both ends): also β·L = n·π, plus a rigid-body mode at zero frequency. Modes cos(nπx/L).
  • Fixed–free bar (u = 0 at x = 0, ∂u/∂x = 0 at x = L): cos(β·L) = 0, β·L = (2n − 1)·π/2. Modes sin((2n − 1)πx/(2L)).

Harmonic series. For fixed–fixed and free–free the frequencies are integer multiples of the fundamental (1 : 2 : 3); for fixed–free only odd multiples appear (1 : 3 : 5). Mode n of a fixed–fixed member has n − 1 interior nodes.

Bar with an end mass. A fixed bar of mass m_b carrying a mass M at its free end has the frequency equation β·L·tan(β·L) = m_b/M. For a heavy end mass this reduces to the SDOF result with k = E·A/L and effective mass M + m_b/3 (Rayleigh, previous topic).

Waves and modes. The modal solution is equivalent to two travelling waves bouncing between the ends; a natural frequency is one where the reflected waves reinforce. The fundamental period of a fixed–fixed member is the time for a wave to travel 2L.

Formulas

∂²w/∂t² = c²·∂²w/∂x² — wave equation.

c = √(T/μ) — string; T tension (N), μ mass per unit length (kg/m), c (m/s).

c = √(E/ρ) — axial bar; E (Pa), ρ (kg/m³). c = √(G/ρ) — torsion of a circular shaft.

f_n = n·c/(2L), n = 1, 2, 3 … — fixed–fixed string or bar, and free–free bar (plus f = 0); L length (m).

f_n = (2n − 1)·c/(4L) — fixed–free bar.

β·L·tan(β·L) = m_b/M, ω = β·c — fixed bar with end mass M (kg), bar mass m_b (kg).

k_axial = E·A/L — static axial stiffness of a bar (N/m).

Worked examples

Example 1 (standard). A steel wire (ρ = 7850 kg/m³, diameter 1 mm) of length 0.8 m is stretched between two fixed points with tension 200 N. Find the first three natural frequencies.

  1. μ = ρ·A = 7850 × π × (0.0005)² = 6.165 × 10⁻³ kg/m.
  2. c = √(T/μ) = √(200/6.165 × 10⁻³) = 180.1 m/s.
  3. f_n = n·c/(2L): f₁ = 180.1/1.6 = 112.6 Hz; f₂ = 225.1 Hz; f₃ = 337.7 Hz.

Answer: 112.6 Hz, 225.1 Hz, 337.7 Hz.

Example 2 (GATE level). An aluminium rod (E = 70 GPa, ρ = 2700 kg/m³, A = 4 × 10⁻⁴ m², L = 1.2 m) is clamped at one end. (a) Find its first two axial natural frequencies with the other end free. (b) A mass equal to the rod's own mass is now attached to the free end; find the fundamental frequency and compare with the SDOF estimate.

  1. c = √(E/ρ) = √(70 × 10⁹/2700) = 5092 m/s.
  2. (a) f_n = (2n − 1)·c/(4L): f₁ = 5092/4.8 = 1061 Hz; f₂ = 3 × 1061 = 3182 Hz.
  3. (b) m_b = ρ·A·L = 2700 × 4 × 10⁻⁴ × 1.2 = 1.296 kg = M. Solve β·L·tan(β·L) = 1: by iteration β·L = 0.8603.
  4. f₁ = β·L·c/(2π·L) = 0.8603 × 5092/(2π × 1.2) = 581 Hz.
  5. SDOF check: k = E·A/L, m_eff = M + m_b/3 = (4/3)·m_b, so ω = (c/L)·√(3/4) = 0.866·c/L, giving f = 585 Hz, 0.7 % high.

Answer: (a) 1061 Hz and 3182 Hz; (b) 581 Hz (SDOF estimate 585 Hz).

Common mistakes

  • Using √(T/μ) as a frequency; it is a wave speed, and must be divided by 2L (or 4L).
  • Mixing up fixed–free and fixed–fixed: the fundamental of a cantilevered bar is half that of the same bar fixed at both ends.
  • Taking μ as mass per unit volume, or mixing grams and millimetres.
  • Forgetting the zero-frequency rigid-body mode of a free–free bar.
  • Expecting even harmonics in a fixed–free bar; only odd multiples occur.
  • Applying the string formula to a stiff wire with significant bending stiffness.

For GATE AE

Expect frequencies of strings and bars from tension, mass, modulus and density, frequency ratios between boundary conditions, mode shapes and node counts, the effect of changing tension or length, and wave-speed questions. Torsional vibration of a uniform shaft follows the same formulas with G. Practise converting rod dimensions into μ and checking an answer with an SDOF estimate using k = EA/L.

Quick check

  1. If the tension in a string is quadrupled, what happens to its frequencies?
  2. What is the frequency ratio f₂/f₁ for a fixed–free bar?
  3. Steel bar, c = 5048 m/s, L = 0.5 m, fixed–free: what is f₁?
  4. How many interior nodes does mode 3 of a fixed–fixed string have?
  5. What is the lowest frequency of a free–free bar?

Answers: 1. they double; 2. 3; 3. 5048/2 = 2524 Hz; 4. two; 5. zero (rigid-body mode), the first elastic mode being c/(2L).

Interview questions

All Structural Dynamics and Aeroelasticity interview questions

Try answering each one aloud before you open it.

  1. 1.What is the fundamental frequency of a vibrating string?Concept

    The fundamental frequency of a vibrating string is the lowest frequency at which the string vibrates. It is determined by the length of the string, the tension in the string, and the mass per unit length of the string. Mathematically, it can be expressed as f₁ = (1/2L) * √(T/μ), where L is the length of the string, T is the tension, and μ is the mass per unit length.

  2. 2.Explain the concept of mode shapes in the context of vibrating strings.Concept

    Mode shapes refer to the specific patterns of vibration that a string can exhibit at its natural frequencies. Each mode shape corresponds to a particular natural frequency, and the shape is characterized by nodes (points of zero displacement) and antinodes (points of maximum displacement). The fundamental mode shape has no nodes other than the endpoints, while higher modes have additional nodes.

  3. 3.How does the tension in a string affect its vibration frequency?Application

    The tension in a string directly affects its vibration frequency. As the tension increases, the frequency of vibration also increases. This is because higher tension results in a greater restoring force, leading to faster oscillations. The relationship is given by the formula f = (1/2L) * √(T/μ), indicating that frequency is proportional to the square root of the tension.

  4. 4.What happens to the natural frequency of an axial bar if its length is doubled?Application

    If the length of an axial bar is doubled, its natural frequency decreases. This is because the natural frequency is inversely proportional to the length of the bar. Specifically, for a bar with fixed ends, the fundamental frequency is given by f₁ = (1/2L) * √(E/ρ), where E is the modulus of elasticity and ρ is the density. Doubling the length results in halving the frequency.

  5. 5.Explain the significance of boundary conditions in the vibration analysis of strings and bars.Concept

    Boundary conditions are crucial in vibration analysis as they define how the ends of the string or bar are constrained. These conditions affect the natural frequencies and mode shapes of the system. Common boundary conditions include fixed, free, and pinned ends. The choice of boundary conditions determines the possible vibration modes and influences the dynamic response of the structure.

  6. 6.Calculate the fundamental frequency of a string that is 1 meter long, has a tension of 100 N, and a mass per unit length of 0.01 kg/m.Numerical

    To calculate the fundamental frequency, use the formula f₁ = (1/2L) * √(T/μ). Here, L = 1 m, T = 100 N, and μ = 0.01 kg/m. Plugging in the values, f₁ = (1/2*1) * √(100/0.01) = 0.5 * √10000 = 0.5 * 100 = 50 Hz.

  7. 7.What is the effect of increasing the mass per unit length on the vibration frequency of a string?Application

    Increasing the mass per unit length of a string decreases its vibration frequency. This is because the frequency is inversely proportional to the square root of the mass per unit length, as given by the formula f = (1/2L) * √(T/μ). A higher mass per unit length results in a lower frequency, as the string becomes heavier and oscillates more slowly.

  8. 8.Describe how damping affects the vibration of axial bars.Concept

    Damping in axial bars reduces the amplitude of vibrations over time, leading to a gradual dissipation of energy. It is an important factor in controlling vibrations and preventing resonance, which can cause structural damage. Damping can be introduced through material properties or external devices, and it helps in stabilizing the structure by minimizing oscillations.

  9. 9.A steel bar with a modulus of elasticity of 200 GPa and a density of 7850 kg/m³ is 2 meters long. Calculate its fundamental frequency assuming fixed ends.Numerical

    For a bar with fixed ends, the fundamental frequency is given by f₁ = (1/2L) * √(E/ρ). Here, L = 2 m, E = 200 GPa = 200 x 10⁹ N/m², and ρ = 7850 kg/m³. Plugging in the values, f₁ = (1/2*2) * √(200 x 10⁹ / 7850) = 0.25 * √(25477707.64) ≈ 0.25 * 5047.5 ≈ 1261.88 Hz.

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