Questions: Vibrations of Single-DOF Systems

Question 1

An engineer records successive peak displacements of a vibrating structure: 10 mm, 7 mm, 4.9 mm. How should this system be classified?

• A. Overdamped — it is clearly losing energy each cycle
• B. Critically damped — the amplitude decays monotonically to zero
• C. Underdamped — it oscillates with exponentially decaying amplitude
• D. Cannot be classified without knowing the mass and stiffness separately

Answer: (C)

The presence of repeated oscillating peaks (10 → 7 → 4.9 mm) with decreasing amplitude is the signature of underdamped behavior (ζ < 1). Critically damped and overdamped systems do not oscillate — they return to equilibrium monotonically without crossing zero. The ratio of successive peaks (10/7 ≈ 7/4.9 ≈ 1.43) is consistent and can be used with the logarithmic decrement formula to extract ζ without knowing m, k, or c individually.

Question 2

The spring stiffness of a single-DOF system is doubled while the mass remains constant. What happens to the natural frequency ω_n?

• A. ω_n doubles
• B. ω_n increases by a factor of √2
• C. The period halves but ω_n is unchanged
• D. ω_n is unchanged — it depends only on mass

Answer: (B)

The natural frequency is ω_n = √(k/m). Doubling k gives ω_n = √(2k/m) = √2 · √(k/m) — an increase by a factor of √2 ≈ 1.41. A common error is to double the result instead of taking the square root. This relationship also shows why stiffer systems (larger k) and lighter systems (smaller m) vibrate faster — both increase the ratio k/m.

Question 3

A critically damped system (ζ = 1) oscillates at the damped natural frequency ω_d = ω_n√(1 − ζ²), just like an underdamped system, but with faster amplitude decay.

• T. True
• F. False

Answer: False

When ζ = 1, the formula gives ω_d = ω_n√(1 − 1²) = 0. Critical damping means no oscillation at all — the system returns to equilibrium in minimum time without any oscillatory overshoot. The formula ω_d = ω_n√(1−ζ²) applies only to underdamped systems (ζ < 1). Critical damping is the boundary between oscillatory and non-oscillatory behavior, not a faster version of underdamped response.

Question 4

The logarithmic decrement method allows you to measure the damping ratio ζ from experimental vibration data without independently knowing the mass, stiffness, or damping coefficient.

• T. True
• F. False

Answer: True

The logarithmic decrement δ = ln(x_n/x_{n+1}) = 2πζ/√(1−ζ²) depends only on ζ. From recorded peak amplitudes, δ can be computed directly from their ratio, and ζ can then be solved algebraically. The individual values of m, k, and c are not needed — only their combination through ζ matters for the decay pattern. This makes the method extremely practical for field measurements and structural health monitoring.

Question 5

Why does the damping ratio ζ (rather than the damping coefficient c) serve as the fundamental parameter for classifying a vibrating system's behavior?

Model answer: The damping coefficient c alone does not determine behavior — a large c could be underdamped if m and k are also large, or overdamped if they are small. The damping ratio ζ = c / (2mω_n) normalizes c by the critical damping coefficient c_c = 2mω_n, making ζ dimensionless. This normalization means ζ < 1 always predicts oscillatory decay, ζ = 1 always means fastest non-oscillatory return, and ζ > 1 always means sluggish non-oscillatory return — regardless of the system's physical scale. A tiny MEMS device and a suspension bridge can both have ζ = 0.1 and their transient responses will look qualitatively identical, scaled only in amplitude and frequency.

This universality is what makes the single-DOF framework so powerful. The same parameter ζ classifies the behavior of a car suspension, a building under earthquake loading, a guitar string, and a circuit's step response. The framework repeats structurally for any vibrating system; only m, k, and c take different physical forms.