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Question

A pendulum of length \( 0.36 \, \text{m} \) oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its
frequency?

Options

Choose one · Correct answer highlighted

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 √((0.36/9.8)) ≈ 2 × 3.14 √(0.0367) ≈ 1.2 s . Frequency: v = (1/T) = (1/1.2) ≈ 0.833 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.833 Hz follows, reflecting SHM dependence on

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