Underdamped vibrations — in a single-degree-of-freedom viscously damped system, how does the amplitude of free vibration vary with time when the system is underdamped?
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Adecreases linearly with time
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Bincreases linearly with time
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Cdecreases exponentially with time
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Dincreases exponentially with time
Answer
Correct Answer: decreases exponentially with time
Explanation
Introduction / Context: Real mechanical systems usually possess some damping. When damping is light (underdamped), oscillations persist but their envelope decays. Recognizing the correct decay law is essential for estimating settling time and vibration severity.
Given Data / Assumptions:
- Single mass–spring–damper with damping ratio 0 < ζ < 1.
- No external forcing (free vibration).
- Undamped natural frequency ωn=√(k/m).
Concept / Approach: The displacement solution is x(t)=X0e−ζωntcos(ωdt+φ), where ωd=ωn√(1−ζ²). The amplitude envelope X(t)=X0e−ζωnt decays exponentially with time.
Step-by-Step Solution:
1) Write characteristic equation for mẍ + cẋ + kx=0.2) For 0<ζ<1, complex roots yield oscillatory motion.3) Envelope is exponential: e−ζωnt → amplitude decreases exponentially.Verification / Alternative check: Log-decrement δ=ln(xi/xi+1)=2πζ/√(1−ζ²) confirms an exponential decay trend across cycles.
Why Other Options Are Wrong:Linear increase/decrease — decay is not linear in time.Increases exponentially — contradicts dissipative energy loss.
Common Pitfalls: Confusing forced resonance amplitude growth with free-decay behavior; mixing underdamped decay with critically damped non-oscillatory return.
Final Answer: decreases exponentially with time.