Spring SHM: k = 50 N/m, m = 2 kg, A = 0.2 m

A spring oscillator on a smooth horizontal surface has a spring of stiffness 50 N/m, a mass of 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) = √(50/2) = 5
   A spring oscillator's angular frequency is set by stiffness and mass
2. T = 2π/ω = 2π/5 ≈ 1.2566
   The period is independent of amplitude (a property of simple harmonic motion)
3. v_max = Aω = 0.2×5 = 1
   Speed is greatest at the equilibrium position

Why it works

A spring oscillator has ω = √(k/m) and T = 2π/ω, so the period grows when the mass grows. Its greatest speed occurs as it sweeps through the equilibrium position, where all the energy is kinetic.

Common mistakes

• Writing T = 2π√(k/m) with k and m swapped.
• Thinking the maximum speed occurs at the extreme positions instead of at the equilibrium position.

Related

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