Single Mass-Spring Oscillators
For a mass m attached to an ideal Hookean spring of effective stiffness k, oscillations about equilibrium have omega = sqrt(k/m), independent of amplitude in the linear regime.
Why this shows up in the exam
Horizontal and vertical spring blocks · Inferring stiffness from static extension · Atomic bond oscillator estimates
Learn the idea
A linear spring sets frequency through stiffness and mass; gravity shifts vertical equilibrium but not the ideal frequency. The spring decides how hard the system pulls back and the mass decides how reluctant it is to accelerate, giving the square-root frequency rule.
🧠 Memory hook: Stiffer is faster; heavier is slower.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- omega = sqrt(k/m) — mass-spring angular frequency
- T = 2 pi sqrt(m/k) — mass-spring period
- x_eq = mg/k — vertical equilibrium extension for one hanging mass
How to approach it
- 1Shift coordinates to equilibrium
- 2Identify the moving mass and effective stiffness
- 3Use ratios before numerical substitution
Common slip-ups that cost marks
- •Including gravity in frequency after shifting equilibrium
- •Using natural length as the oscillation origin
- •Confusing spring constant with angular frequency
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A mass of 1 kg is attached to a spring of force constant 100 N/m. Find the angular frequency of small oscillations.
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