Simple harmonic oscillator insight: Which mass–spring setup will oscillate with the highest frequency (fastest)?
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AA large mass on a weak spring
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BA large mass on a stiff spring
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CA small mass on a stiff spring
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DA small mass on a weak spring
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EA medium mass on a medium-stiff spring
Answer
Correct Answer: A small mass on a stiff spring
Explanation
Introduction / Context:In elementary mechanics and vibrational spectroscopy, the frequency of a harmonic oscillator depends on the restoring force and inertia. Understanding this relationship builds intuition for molecular vibrations (stiff bonds, light atoms) and for macroscopic oscillators like mass–spring systems.
Given Data / Assumptions:
- The oscillator is ideal (no damping).
- Frequency f relates to spring constant k and mass m via angular frequency ω = sqrt(k/m), so f ∝ sqrt(k/m).
- “Stiff” means higher k; “weak” means lower k. “Small” mass means lower m; “large” mass means higher m.
Concept / Approach:From ω = sqrt(k/m), a larger k increases frequency, while a larger m decreases it. Thus, the combination for the highest frequency is the smallest mass with the stiffest spring.
Step-by-Step Solution:Write relation: ω = sqrt(k/m); f = ω / (2π).Maximize k, minimize m to maximize ω.Among options, “small mass on a stiff spring” fits this requirement.Select that configuration as the fastest oscillator.
Verification / Alternative check:Check extremes: if m → 0 or k → ∞, frequency increases. Conversely, large m or weak k slows oscillations—consistent with the chosen option.
Why Other Options Are Wrong:
- Large mass + weak spring/large mass + stiff spring: Inertia dominates; frequency lower than small mass cases.
- Small mass + weak spring: Low k limits frequency despite small m.
- Medium values: Not as optimal as the extreme “small m, large k.”
Common Pitfalls:Misremembering the formula as ω ∝ k/m instead of the square-root dependence; always use sqrt(k/m).
Final Answer:A small mass on a stiff spring.