#pendulum period

Forced Oscillations and Resonance

A simple pendulum has a length of \( 2.25 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its

**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.

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Simple Pendulum and Angular SHM

A simple pendulum has a period of \( 3 \, \text{s} \) on Earth. What will be its period on a planet where \( g = 2.45 \,

**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 = 2π √((L/g)) . T ∝ (1/√(g)) . (Tₚlₐₙₑt/TEₐrth) = √((gEₐrth/gₚlₐₙₑt)) = √((9.8/2.45)) = √(4) = 2 . Tₚlₐₙₑt = 2 × 3 = 6 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 6 s

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

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Simple Pendulum and Angular SHM

A simple pendulum has a length of \( 0.81 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its

**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)) = 2π √((0.81/9.8)) ≈ 2 × 3.14 √(0.0827) ≈ 1.805 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.805 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

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Simple Pendulum and Angular SHM

A simple pendulum has a length of \( 1.6 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its p

**Angular SHM** of pendulum results from restoring torque τ = -m g L sinθ ≈ -m g L θ for small θ, analogous to linear SHM with ω = √(g/L). This relation allows period determination from length and local gravity, illustrating gravitational influence on oscillation. Period: T = 2π √((L/g)) = 2π √((1.6/9.8)) ≈ 2 × 3.14 √(0.1633) ≈ 2.54 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.54 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

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