Questions: Spring-Mass Systems and Mechanical Vibrations

Question 1

A car suspension is redesigned with stiffer springs (k increases) and a lighter body (m decreases). What happens to the natural frequency ω₀?

• A. It decreases, because the system has more energy stored in the springs
• B. It stays the same, because the two changes offset each other
• C. It increases, because ω₀ = √(k/m) rises when k increases and m decreases
• D. It decreases, because a stiffer spring slows down oscillation

Answer: (C)

ω₀ = √(k/m). A larger k and a smaller m both increase k/m, so ω₀ rises. Option D represents the common intuitive error — equating 'stiffer' with 'heavier' or 'slower.' Physically, a stiffer spring exerts a stronger restoring force for the same displacement, driving the mass back faster and increasing oscillation frequency. Both design changes push natural frequency upward.

Question 2

A door closer and a bouncy ball-on-a-spring have the same natural frequency ω₀. The door returns to closed position quickly without swinging back; the ball oscillates many times before settling. Which damping condition does the door closer exemplify?

• A. Overdamping — the highest damping always produces the fastest response
• B. Underdamping — the door oscillates too fast to be visible
• C. Critical damping — the system returns to equilibrium as fast as possible without oscillating
• D. Zero damping — friction is minimized in the door mechanism

Answer: (C)

Critical damping (c² = 4mk) is the threshold condition where the system returns to equilibrium as quickly as possible without any oscillation. Option A is the most common misconception: overdamping uses more damping but returns more slowly than critical damping. The door closer is the textbook example of critical damping design — fast enough to close promptly, with no bounce-back.

Question 3

Adding more damping to a spring-mass system generally makes it return to equilibrium faster.

• T. True
• F. False

Answer: False

Critical damping produces the fastest return to equilibrium. Adding more damping pushes the system into the overdamped regime, where return is actually slower. The relationship is not monotonic: underdamped systems oscillate before settling, critical damping is the fastest non-oscillatory return, and overdamped systems are sluggish. 'More damping = faster settling' is only true when moving from underdamped toward critical — not beyond it.

Question 4

The natural frequency of an undamped spring-mass system decreases as the mass m increases.

• T. True
• F. False

Answer: True

ω₀ = √(k/m). A larger mass means more inertia, so the same spring force produces smaller acceleration, leading to slower oscillation. Doubling m halves k/m, so ω₀ decreases by a factor of √2. More massive systems oscillate more slowly at the same spring stiffness.

Question 5

What determines which of the three damping regimes a spring-mass system is in, and why does this distinction matter for engineering applications?

Model answer: The discriminant c² − 4mk determines the regime: negative gives underdamping (oscillatory decay), zero gives critical damping (fastest non-oscillatory return), positive gives overdamping (slow non-oscillatory return). For engineering, the regime determines qualitative behavior: a pendulum clock needs underdamping to keep oscillating; a car shock absorber wants near-critical damping to suppress bouncing; an aircraft control surface must not overshoot its target position.

The three regimes correspond directly to the three cases of characteristic roots (complex, repeated real, distinct real). Understanding which regime you're in tells you the qualitative shape of the solution before computing anything. Engineers tune the ratio c²/(4mk) — the damping ratio — to achieve the desired response for each application.