Mechanical Vibrations

Why it matters

Mechanical vibrations are crucial in engineering because they affect the performance, safety, and longevity of mechanical systems. Understanding vibrations helps in designing systems that can withstand oscillatory forces, reducing noise and wear, and preventing structural failures.

Key ideas

• Types of Vibrations: Vibrations can be classified into free and forced vibrations. Free vibrations occur without external force after an initial disturbance, while forced vibrations occur due to continuous external forces.
• Damping: Damping is the mechanism by which vibrational energy is gradually lost to the surroundings, reducing the amplitude of oscillations over time.
• Natural Frequency: The frequency at which a system naturally oscillates when disturbed is its natural frequency. It is crucial for avoiding resonance, which can cause excessive vibrations.
• Resonance: An undamped oscillator driven harmonically at ω_n has a growing response. Damping limits the steady response. For a viscously damped oscillator under a constant-amplitude harmonic force, the displacement-amplitude peak occurs at ω_n√(1 − 2ζ²) when ζ < 1/√2, so the peak need not be exactly at ω_n.
• Degrees of Freedom: The number of independent ways in which a mechanical system can move.

Linear single-degree-of-freedom model

The equation m ẍ + c ẋ + kx = F(t) assumes a linear spring, viscous damping, and displacement from static equilibrium. Free vibration has F(t) = 0. Define ω_n = √(k/m), ζ = c/(2√(mk)), and ω_d = ω_n√(1 − ζ²).

For 0 ≤ ζ < 1, x(t) = e^(−ζω_n t)[x₀ cos(ω_d t) + ((v₀ + ζω_n x₀)/ω_d) sin(ω_d t)]. Critical and overdamped systems (ζ ≥ 1) require nonoscillatory solution forms.

Formulas

• f_n = (1 / (2π)) * √(k / m)
  • f_n: Natural frequency (Hz)
  • k: Stiffness of the system (N/m)
  • m: Mass of the system (kg)
• ζ = c / (2 * √(m * k))
  • ζ: Damping ratio (dimensionless)
  • c: Damping coefficient (Ns/m)
• x(t) = X * e^(-ζω_nt) * cos(ω_dt + φ)
  • x(t): Displacement at time t (m)
  • X: Envelope constant set by both initial conditions (m); x(0) = X cos φ
  • ω_n: Natural angular frequency (rad/s)
  • ω_d: Damped angular frequency (rad/s)
  • φ: Phase angle (rad)

Worked example

A linear spring–mass–damper has m = 2 kg, k = 200 N/m, and c = 5 N·s/m. Displacement is measured from static equilibrium. At t = 0 it is released from x₀ = 0.1 m with zero velocity. Find x at 1 s during free vibration.

ω_n = √(k/m) = 10 rad/s, so f_n = 10/(2π) = 1.5915 Hz.

The critical damping coefficient is c_c = 2√(mk) = 40 N·s/m. Thus ζ = 5/40 = 0.125, and ω_d = 10√(1 − 0.125²) = 9.92157 rad/s.

For these initial conditions:

x(t) = e^(−1.25t)[0.1 cos(9.92157t) + (0.125/9.92157) sin(9.92157t)] m.

Using radians, x(1) = −0.02691 m, or about −26.91 mm. The negative sign identifies the side of equilibrium. The envelope has decayed; displacement is not the same as the envelope amplitude.

Common mistakes

• Confusing natural frequency with angular frequency.
• Ignoring damping effects in systems where they are significant.
• Miscalculating the phase angle or initial conditions.

For GATE ME

Questions often involve calculating natural frequencies, damping ratios, and analyzing the response of systems to vibrations. Practice problems on resonance conditions and the effects of damping.

Quick check

1. What is the effect of increasing the damping coefficient on the system's vibrations?
2. How does resonance affect a mechanical system?
3. Define natural frequency.

Answers: 1. Reduces amplitude and increases damping. 2. Causes large amplitude oscillations. 3. Frequency at which a system naturally oscillates when disturbed.