#pendulum length

Simple Pendulum and Angular SHM

The period of a simple pendulum is \( 2 \, \text{s} \) when \( g = 9.8 \, \text{m/s}^2 \). What should be the length of

**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°. T = 2π √((L/g)) . 2 = 2π √((L/9.8)) ⇒ 1 = π √((L/9.8)) . √((L/9.8)) = (1/π) ⇒ (L/9.8) = (1/π²) ⇒ L = (9.8/π²) ≈ 1 m (using π² ≈ 9.87 ). Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA²

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Simple Pendulum and Angular SHM

What physical property of a simple pendulum primarily governs its oscillatory behavior?

**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°. The length of the pendulum determines the period ( T = 2π √((L/g)) ), controlling the frequency and thus the oscillatory behavior, more than mass or amplitude for small angles. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The

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Simple Pendulum and Angular SHM

A simple pendulum has a length of \( 0.4 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its p

**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 √((0.4/9.8)) ≈ 2 × 3.14 √(0.0408) ≈ 1.27 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.27 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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Simple Pendulum and Angular SHM

A simple pendulum has a length of \( 0.9 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its p

**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)) = 2π √((0.9/9.8)) ≈ 2 × 3.14 √(0.0918) ≈ 1.9 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.9 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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