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#angular frequency

31 public questions tagged with this topic.

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 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 spring-mass system has \( m = 2 \, \text{kg}, k = 800 \, \text{N/m} \). 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 = √((k/m)) = √((800/2)) = √(400) = 20 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 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

A particle’s x-projection from circular motion is \( x = 6 \cos (3t) \) (in m). What is its maximum acceleration?

**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. Maximum acceleration: aₘₐₓ = ω² A . A = 6 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 6 = 9 × 6 = 54 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 54 m/s² 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 spring-mass system has \( m = 1.25 \, \text{kg}, k = 500 \, \text{N/m} \). 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 = √((k/m)) = √((500/1.25)) = √(400) = 20 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 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 spring-mass system has \( m = 1.6 \, \text{kg}, k = 640 \, \text{N/m} \). 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 = √((k/m)) = √((640/1.6)) = √(400) = 20 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 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 pendulum has \( L = 0.25 \, \text{m}, g = 10 \, \text{m/s}^2 \). What is its angular frequency?

**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. ω = √((g/L)) = √((10/0.25)) = √(40) ≈ 6.32 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 6.32 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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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 2.5 \, \text{kg}, k = 1000 \, \text{N/m} \). What is its angular frequency?

**Mass-spring dynamics** show T depends on mass and stiffness, independent of amplitude for ideal spring. Given T and m, k = 4π² m/T² extracted, and energy E = ½ k A² connects amplitude to total mechanical energy, illustrating isochronism. ω = √((k/m)) = √((1000/2.5)) = √(400) = 20 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 rad/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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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

In SHM, which quantity is directly proportional to the square of the 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°. Acceleration a = -ω² x has a maximum magnitude of ω² A , making it proportional to ω² , unlike velocity ( ω A ) or displacement. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Acceleration 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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