Practice question
Question
A particle in SHM has an amplitude of \( 8 \, \text{cm} \) and a period of \( 0.4 \, \text{s} \). What
is its maximum velocity? (Take \( \pi = 3.14 \))
Explanation
**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Maximum velocity: vₘₐₓ = A ω . ω = (2π/T) = (2 × 3.14/0.4) = 15.7 rad/s . A = 0.08 m . vₘₐₓ = 0.08 × 15.7 = 1.256 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.256 m/s follows, reflecting SHM dependence on
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