Simple harmonic oscillator insight: Which mass–spring setup will oscillate with the highest frequency (fastest)?

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.