Practice question
Question
A simple pendulum has a period of \( 1 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is
its period on the Moon (\( g = 1.63 \, \text{m/s}^2 \))?
Explanation
**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. T ∝ (1/√(g)) . (TMₒₒₙ/TEₐrth) = √((gEₐrth/gMₒₒₙ)) = √((9.8/1.63)) ≈ √(6) ≈ 2.45 . TMₒₒₙ = 1 × 2.45 ≈ 2.45 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.45 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.
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