CalcElfCalcElf › Spring SHM: k = 400 N/m, m = 4 kg, A = 0.2 m
Spring SHM: k = 400 N/m, m = 4 kg, A = 0.2 m
A spring oscillator on a smooth horizontal surface has a spring of stiffness 400 N/m, a mass of 4 kg and an amplitude of 0.2 m, and moves with simple harmonic motion. Find the period of the motion and the maximum speed of the mass as it passes the equilibrium position.
Answer
Period ≈ 0.6283 s, maximum speed 2 m/s
Step-by-step
- ω = √(k/m) = √(400/4) = 10A spring oscillator's angular frequency is set by stiffness and mass
- T = 2π/ω = 2π/10 ≈ 0.6283The period is independent of amplitude (a property of simple harmonic motion)
- v_max = Aω = 0.2×10 = 2Speed is greatest at the equilibrium position
Why it works
The angular frequency of a spring oscillator is fixed by stiffness and mass together, ω = √(k/m), so T = 2π/ω; the speed is greatest at the equilibrium position, where v_max = Aω.
Common mistakes
- Writing the period as 2π√(k/m), with k and m the wrong way round.
- Believing that a larger amplitude means a longer period.
Related
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