Spring SHM: k = 80 N/m, m = 0.2 kg, A = 0.2 m

A spring oscillator on a smooth horizontal surface has a spring of stiffness 80 N/m, a mass of 0.2 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.

Step-by-step

1. ω = √(k/m) = √(80/0.2) = 20
   A spring oscillator's angular frequency is set by stiffness and mass
2. T = 2π/ω = 2π/20 ≈ 0.3142
   The period is independent of amplitude (a property of simple harmonic motion)
3. v_max = Aω = 0.2×20 = 4
   Speed is greatest at the equilibrium position

Why it works

The angular frequency ω = √(k/m) depends on the ratio of stiffness to mass; here k/m = 400. The maximum speed follows from v_max = Aω, and the period from T = 2π/ω.

Common mistakes

• Forgetting to take the square root and using ω = 400 instead of 20.
• Taking a decimal mass as a mistake and rewriting 0.2 kg as 2 kg, which changes k/m.

Related

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