Skip to content
Login Register

#length 0.25 m

2 public questions tagged with this topic.

Equations of SHM, Phase and Angular Frequency

A pendulum has \( L = 0.25 \, \text{m}, g = 10 \, \text{m/s}^2 \). What is its angular frequency?

**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. ω = √((g/L)) = √((10/0.25)) = √(40) ≈ 6.32 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 6.32 rad/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details
Simple Pendulum and Angular SHM

A simple pendulum of length \( 0.25 \, \text{m} \) oscillates on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its peri

**Pendulum motion** exhibits isochronism for small amplitudes, period depends only on L and g. Using g = 9.8 m/s², T calculation requires √(L/g), frequency reciprocal of period. Angular frequency directly √(g/L), e.g., L = 0.25 m gives T ≈ 1.0 s. Period: T = 2π √((L/g)) . T = 2 × 3.14 √((0.25/9.8)) = 6.28 √(0.0255) ≈ 6.28 × 0.16 ≈ 1.0 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.0 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details