Practice question
Question
A simple pendulum has a length of \( 2.25 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2
\). What is its period?
Explanation
**Forced oscillations** result when external periodic driving force F = F₀ cos(ω_d t) acts on oscillator, steady-state frequency equals driving frequency ω_d, amplitude A = F₀/√((k - m ω_d²)² + (b ω_d)²) depends on proximity to natural frequency ω₀ = √(k/m). Resonance when ω_d ≈ ω₀, amplitude maximum. Period: T = 2π √((L/g)) = 2π √((2.25/9.8)) ≈ 2 × 3.14 √(0.2296) ≈ 3.01 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.01 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.
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