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Forced Oscillations and Resonance

A pendulum has \( L = 1.4 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. ω = √((g/L)) = √((9.8/1.4)) ≈ √(7) ≈ 2.65 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.65 rad/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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Forced Oscillations and Resonance

A pendulum of length \( 0.36 \, \text{m} \) oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its frequency?

**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 √((0.36/9.8)) ≈ 2 × 3.14 √(0.0367) ≈ 1.2 s . Frequency: v = (1/T) = (1/1.2) ≈ 0.833 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.833 Hz follows, reflecting SHM dependence on

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

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Forced Oscillations and Resonance

A pendulum has \( L = 0.6 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. ω = √((g/L)) = √((9.8/0.6)) ≈ √(16.33) ≈ 4.04 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 4.04 rad/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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Forced Oscillations and Resonance

A simple pendulum has a period of \( 1 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its period on th

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. T ∝ (1/√(g)) . (TMₒₒₙ/TEₐrth) = √((gEₐrth/gMₒₒₙ)) = √((9.8/1.63)) ≈ √(6) ≈ 2.45 . TMₒₒₙ = 1 × 2.45 ≈ 2.45 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.45 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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Damped Oscillations

A pendulum has \( L = 1.96 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. ω = √((g/L)) = √((9.8/1.96)) = √(5) ≈ 2.24 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.24 rad/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A pendulum has \( L = 0.5 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. ω = √((g/L)) = √((9.8/0.5)) = √(19.6) ≈ 4.43 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 4.43 rad/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

What is the effect on the frequency of a simple pendulum if it is taken to a planet where gravity is one-fourth that of

**Damped oscillations** occur when resistive forces dissipate energy, amplitude decays exponentially as A(t) = A₀ e^(-b t/2m), b damping coefficient (kg/s), frequency slightly reduced ω' = √(ω₀² - (b/2m)²). Damping arises from friction or viscosity, energy loss per cycle proportional to velocity squared, motion eventually stops. Frequency v = (1/2π) √((g/L)) . If g' = (g/4) , then v' = (1/2π) √((g/4/L)) = (1/2) v , halving the frequency. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It halves follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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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

A simple pendulum has a frequency of \( 0.25 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its lengt

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Period: T = (1/v) = (1/0.25) = 4 s . T = 2π √((L/g)) ⇒ 4 = 2π √((L/9.8)) . √((L/9.8)) = (4/2π) ≈ 0.637 ⇒ (L/9.8) = (0.637)² ⇒ L ≈ 3.98 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.98 m follows, reflecting

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

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Equations of SHM, Phase and Angular Frequency

A pendulum has \( L = 1.2 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. ω = √((g/L)) = √((9.8/1.2)) ≈ √(8.17) ≈ 2.86 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.86 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

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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 pendulum oscillates with \( \theta_{\text{max}} = 0.2 \, \text{rad}, L = 2 \, \text{m}, g = 10 \, \text{m/s}^2 \). Wha

**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°. ω = √((g/L)) = √((10/2)) = √(5) ≈ 2.24 rad/s . Arc length amplitude: A = L θₘₐₓ = 2 × 0.2 = 0.4 m . vₘₐₓ = ω A = 2.24 × 0.4 ≈ 0.896 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x

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

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