#period calculation

Forced Oscillations and Resonance

A particle in SHM has \( a = -25 x \) (in SI units). What is its period?

**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. For SHM, a = -ω² x . Given a = -25 x , ω² = 25 ⇒ ω = 5 rad/s . Period: T = (2π/ω) = (2π/5) ≈ 1.256 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.256 s

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

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Equations of SHM, Phase and Angular Frequency

A pendulum oscillates with a period of \( 1.5 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its lengt

**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)) . 1.5 = 2π √((L/9.8)) ⇒ (1.5/2π) = √((L/9.8)) . ((1.5/2 × 3.14))² = (L/9.8) ⇒ L = 9.8 × (0.238)² ≈ 0.56 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.56

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

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Equations of SHM, Phase and Angular Frequency

Two identical springs (\( k = 50 \, \text{N/m} \)) are attached to a \( 0.5 \, \text{kg} \) mass as in Fig. 13.14. What

**SHM representation** using sine or cosine equivalent with phase offset, ω relates to system parameters like mass and stiffness. Understanding ω and φ permits prediction of position at any time and comparison of two SHM via phase difference Δφ = φ₂ - φ₁. Effective kₑff = 2k = 2 × 50 = 100 N/m . T = 2π √((m/kₑff)) = 2π √((0.5/100)) = 2π √(0.005) ≈ 0.44 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.44 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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 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

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

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 \( 0.4 \, \text{m} \) 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.

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

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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

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