Spring SHM: k = 200 N/m, m = 0.5 kg

A spring oscillator on a smooth horizontal surface has a spring of stiffness 200 N/m, a mass of 0.5 kg and an amplitude of 0.1 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) = √(200/0.5) = 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 = 0.1×20 = 2
   Speed is greatest at the equilibrium position

Why it works

Stiff spring and light mass both raise ω = √(k/m), which shortens the period T = 2π/ω and raises the maximum speed Aω.

Common mistakes

• Forgetting the square root when working out ω from k and m.
• Writing the period as 1/ω, which drops the factor 2π.

Related

pendulum-period vertical-throw spring-shm-k100-m1-a0-1