Practice question
Question
A particle in SHM has an amplitude of \( 12 \, \text{cm} \) and a period of \( 1.2 \, \text{s} \). What
is its maximum velocity? (Take \( \pi = 3.14 \))
Explanation
**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. Maximum velocity: vₘₐₓ = A ω . ω = (2π/T) = (2 × 3.14/1.2) ≈ 5.23 rad/s . A = 0.12 m . vₘₐₓ = 0.12 × 5.23 ≈ 0.628 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.628 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.
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