Practice question
Question
A spring system has \( m = 0.3 \, \text{kg}, k = 120 \, \text{N/m}, A = 8 \, \text{cm} \). What is the
potential energy at \( x = 4 \, \text{cm} \)?
Explanation
**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. Potential energy: U = (1/2) k x² . k = 120 N/m, x = 0.04 m . U = 0.5 × 120 × (0.04)² = 0.5 × 120 × 0.0016 = 0.096 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E =
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