#period independence

Forced Oscillations and Resonance

Why does the period of a spring-mass system remain unaffected by changes in gravitational field strength?

**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. The period T = 2π √((m/k)) depends only on mass and spring constant, not gravity, which affects pendulums but not spring systems. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The restoring force is independent of gravity follows, reflecting SHM dependence on amplitude A, ω and s

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

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Simple Pendulum and Angular SHM

Which of the following best explains why the period of a simple pendulum is independent of its mass?

**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. The period T = 2π √((L/g)) depends on length and gravity. Mass cancels out in the equation of motion ( a = -(g/L) θ ), as both force and inertia scale with mass. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The restoring force and inertia both depend on

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

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Periodic Motion and SHM Basic Concepts

Which aspect of SHM ensures that the time period is independent of the initial conditions?

**Periodic motion** repeats after fixed period T, x(t+T)=x(t), while oscillatory motion involves to-and-fro about equilibrium. SHM is special periodic motion where restoring force proportional to displacement, F = -k x, acceleration a = -ω² x, leading to sinusoidal displacement x = A cos(ωt + φ). The period T = (2π/ω) depends only on ω = √((k/m)) , a constant derived from system properties, not initial displacement or velocity. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The constant angular frequency follows, reflecting SHM dependence on amplitude A

Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts

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Simple Pendulum and Angular SHM

In a simple pendulum executing SHM, what ensures that the period is independent of the initial angular displacement?

**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. For small angles, the approximation sin θ ≈ θ holds, making the period T = 2π √((L/g)) independent of amplitude, a feature of linear systems. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The small angle approximation 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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