#period

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

Two identical springs (\( k = 100 \, \text{N/m} \)) are attached to a \( 2.0 \, \text{kg} \) mass as in Fig. 13.14. What

**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. Effective kₑff = 2k = 2 × 100 = 200 N/m . T = 2π √((m/kₑff)) = 2π √((2/200)) = 2π √(0.01) = 2π × 0.1 ≈ 0.628 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.628 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 oscillates with \( T = 0.4 \, \text{s} \) when \( m = 0.2 \, \text{kg} \). What is the spring const

**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. T = 2π √((m/k)) . 0.4 = 2π √((0.2/k)) ⇒ (0.4/2π) = √((0.2/k)) . (0.0637)² = (0.2/k) ⇒ k = (0.2/0.00406) ≈ 49.26 N/m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 49.26 N/m 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 happens to the period of a spring-mass system if both the mass and spring constant are doubled?

**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. Period T = 2π √((m/k)) . If m' = 2m and k' = 2k , then T' = 2π √((2m/2k)) = 2π √((m/k)) = T , so the period remains unchanged. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It remains unchanged follows, reflecting SHM dependence on

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A mass oscillates with \( T = 0.4 \, \text{s} \) when attached to a spring of \( k = 100 \, \text{N/m} \). What is the m

**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. T = 2π √((m/k)) . 0.4 = 2π √((m/100)) ⇒ (0.4/2π) = √((m/100)) . ((0.4/6.28))² = (m/100) ⇒ m = 100 × (0.0637)² ≈ 0.405 kg . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.405 kg 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 oscillates with \( T = 0.6 \, \text{s} \) when \( m = 0.9 \, \text{kg} \). What is the spring const

**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. T = 2π √((m/k)) . 0.6 = 2π √((0.9/k)) ⇒ (0.6/2π) = √((0.9/k)) . (0.0955)² = (0.9/k) ⇒ k = (0.9/0.00912) ≈ 98.68 N/m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 98.68 N/m 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 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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Equations of SHM, Phase and Angular Frequency

A particle in SHM has \( a = -9 x \) (in SI units). What is its period?

**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°. For SHM, a = -ω² x . Given a = -9 x , ω² = 9 ⇒ ω = 3 rad/s . T = (2π/ω) = (2π/3) ≈ 2.09 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.09 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

What effect does tripling the spring constant have on the period of a spring-mass system if the mass is also tripled?

**Oscillation of mass attached to spring** follows Hooke's law F = -k x, SHM with ω = √(k/m). Effective stiffness for two identical springs in parallel doubles, k_eff = 2k, increasing frequency by √2, while series halves stiffness to k/2, lowering frequency. Energy E = ½ k_eff A². Period T = 2π √((m/k)) . If k' = 3k and m' = 3m , then T' = 2π √((3m/3k)) = 2π √((m/k)) = T , so the period remains unchanged. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It remains unchanged

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

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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 particle in SHM has an amplitude of \( 8 \, \text{cm} \) and a period of \( 0.4 \, \text{s} \). What is its maximum ve

**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°. Maximum velocity: vₘₐₓ = A ω . ω = (2π/T) = (2 × 3.14/0.4) = 15.7 rad/s . A = 0.08 m . vₘₐₓ = 0.08 × 15.7 = 1.256 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.256 m/s follows, reflecting SHM dependence on

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

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Energy in SHM - Kinetic, Potential and Total

A particle in SHM has an amplitude of \( 9 \, \text{cm} \) and a period of \( 0.8 \, \text{s} \). What is its maximum ve

**Energy distribution** shows maximum kinetic at equilibrium and maximum potential at extremes, sum constant. This relation enables calculation of amplitude, velocity at any displacement via v = ±√(2(E-U)/m), and understanding of energy storage in oscillating system for NEET problems. Maximum velocity: vₘₐₓ = A ω . ω = (2π/T) = (2 × 3.14/0.8) = 7.85 rad/s . A = 0.09 m . vₘₐₓ = 0.09 × 7.85 ≈ 0.7065 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.7065 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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