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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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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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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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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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Simple Pendulum and Angular SHM

A pendulum of length \( 2 \, \text{m} \) oscillates with \( g = 10 \, \text{m/s}^2 \). What is its frequency?

**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 √((2/10)) = 2π √(0.2) ≈ 2.8 s . Frequency: v = (1/T) = (1/2.8) ≈ 0.357 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.357 Hz 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 \( 1.69 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its

**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 √((1.69/9.8)) ≈ 2 × 3.14 √(0.1724) ≈ 2.61 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.61 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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Equations of SHM, Phase and Angular Frequency

A pendulum has \( L = 2.25 \, \text{m}, g = 9.8 \, \text{m/s}^2 \). What is its angular frequency?

**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. ω = √((g/L)) = √((9.8/2.25)) ≈ √(4.356) ≈ 2.087 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.087 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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Simple Pendulum and Angular SHM

A simple pendulum has a frequency of \( 0.4 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length

**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 = (1/v) = (1/0.4) = 2.5 s . 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 ≈ 1.55 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.55 m

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

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Simple Pendulum and Angular SHM

A simple pendulum has a period of \( 1.4 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length?

**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)) . 1.4 = 2π √((L/9.8)) ⇒ √((L/9.8)) = (1.4/2π) ≈ 0.223 . (L/9.8) = (0.223)² ⇒ L ≈ 9.8 × 0.0497 ≈ 0.487 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.487 m

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 \( 1.44 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its

**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 √((1.44/9.8)) ≈ 2 × 3.14 √(0.1469) ≈ 2.406 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.406 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 pendulum of length \( 1.5 \, \text{m} \) oscillates with \( g = 10 \, \text{m/s}^2 \). What is its period?

**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. T = 2π √((L/g)) = 2π √((1.5/10)) = 2π √(0.15) ≈ 2 × 3.14 × 0.387 ≈ 2.43 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.43 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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