Spring SHM: k = 25 N/m, m = 1 kg, A = 0.4 m

A spring oscillator on a smooth horizontal surface has a spring of stiffness 25 N/m, a mass of 1 kg and an amplitude of 0.4 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) = √(25/1) = 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 = 0.4×5 = 2
   Speed is greatest at the equilibrium position

Why it works

A softer spring gives a smaller ω = √(k/m) and therefore a longer period; the maximum speed is v_max = Aω, with the whole energy of the motion stored in the spring at the turning points.

Common mistakes

• Using v_max = A/T or A·T, mixing amplitude with period, instead of v_max = Aω.
• Thinking the speed is greatest at the ends of the motion rather than at the equilibrium position.

Related

pendulum-period vertical-throw spring-shm-k200-m0-5-a0-1