Practice question

**Oscillation of mass attached to spring** follows Hooke's law F = -k x, SHM with ω = √(k/m). Effective stiffness for two identical springs in parallel doubles, k_eff = 2k, increasing frequency by √2, while series halves stiffness to k/2, lowering frequency. Energy E = ½ k_eff A². Total energy: E = (1/2) k A² = (1/2) m ω² A² = 0.5 × 1 × 5² × 2² = 50 J . Potential energy: U = (1/2) m ω² x² = 0.5 × 1 × 25 × 1² = 12.5 J . Kinetic energy: K = E - U = 50 - 12.5 = 37.5