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#length calculation

2 public questions tagged with this topic.

Equations of SHM, Phase and Angular Frequency

A simple pendulum has a period of \( 2.5 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length?

**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. T = 2π √((L/g)) . 2.5 = 2π √((L/9.8)) ⇒ √((L/9.8)) = (2.5/2π) ≈ 0.398 . (L/9.8) = (0.398)² ⇒ L ≈ 9.8 × 0.158 ≈ 1.55 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.55

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Simple Pendulum and Angular SHM

A simple pendulum has a frequency of \( 0.3 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length

**Simple pendulum** for small angles approximates SHM with period T = 2π√(L/g), frequency f = (1/2π)√(g/L), angular frequency ω = √(g/L) (rad/s), independent of mass. L length from pivot to centre of mass (m), g = 9.8 m/s² acceleration due to gravity, approximation sinθ ≈ θ (rad) for θ < 10°. Period: T = (1/v) = (1/0.3) ≈ 3.333 s . T = 2π √((L/g)) ⇒ 3.333 = 2π √((L/9.8)) . √((L/9.8)) = (3.333/2π) ≈ 0.531 ⇒ (L/9.8) = (0.531)² ⇒ L ≈ 2.76 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E =

Ref: NCERT > Physics Book > Oscillations > Simple Pendulum and Angular SHM

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