Practice question
Question
A spring of \( k = 400 \, \text{N/m} \) is attached to a \( 1 \, \text{kg} \) mass. What is the angular
frequency?
Explanation
**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. ω = √((k/m)) = √((400/1)) = 20 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 rad/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.
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