Spring SHM: k = 90 N/m, m = 0.1 kg, A = 0.2 m

A spring oscillator on a smooth horizontal surface has a spring of stiffness 90 N/m, a mass of 0.1 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) = √(90/0.1) = 30
   A spring oscillator's angular frequency is set by stiffness and mass
2. T = 2π/ω = 2π/30 ≈ 0.2094
   The period is independent of amplitude (a property of simple harmonic motion)
3. v_max = Aω = 0.2×30 = 6
   Speed is greatest at the equilibrium position

Why it works

The angular frequency of a spring oscillator is ω = √(k/m), so T = 2π/ω; here the stiffness and mass give k/m = 900. The speed is largest 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.
• Mixing the units and treating the 0.1 kg mass as a 1 kg mass, which divides ω by √10.

Related

pendulum-period spring-shm-k80-m0-2-a0-2 spring-shm-k100-m1-a0-1