#displacement

Damped Oscillations

A spring system has \( m = 1.0 \, \text{kg}, k = 400 \, \text{N/m}, A = 6 \, \text{cm} \). What is the potential energy

**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. Potential energy: U = (1/2) k x² . k = 400 N/m, x = 0.03 m . U = 0.5 × 400 × (0.03)² = 0.5 × 400 × 0.0009 = 0.18 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.18 J follows, reflecting SHM

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A mass of \( 2.0 \, \text{kg} \) on a spring with \( k = 800 \, \text{N/m} \) has \( A = 5 \, \text{cm} \). What is the

**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. Total energy: E = (1/2) k A² = 0.5 × 800 × (0.05)² = 1 J . Potential energy: U = (1/2) k x² = 0.5 × 800 × (0.025)² = 0.25 J . Kinetic energy: K = E - U = 1 - 0.25 = 0.75 J . Applying x = A cos(ωt + φ), v = -ωA

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A mass oscillates with \( v = -8 \cos (2t) \) (in m/s). What is its displacement function?

**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. Velocity: v = -ω A sin (ω t) , but given v = -8 cos (2t) . ω = 2 s⁻¹, vₘₐₓ = ω A = 8 ⇒ A = (8/2) = 4 m . Since v = -A ω sin (ω t) , adjust phase: x = 4 sin (2t) . Applying x = A cos(ωt + φ), v = -ωA

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A mass of \( 4 \, \text{kg} \) is attached to a spring with \( k = 1600 \, \text{N/m} \) and displaced by \( 5 \, \text{

**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. Total energy: E = (1/2) k A² . A = 0.05 m, k = 1600 N/m . E = 0.5 × 1600 × (0.05)² = 0.5 × 1600 × 0.0025 = 2 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.0 J

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A particle’s displacement in SHM is \( x = 6 \sin (3t) \) (in cm). What is the 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 cm = 0.06 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 0.06 = 9 × 0.06 = 0.54 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.54 m/s² follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.8 \, \text{kg}, k = 320 \, \text{N/m} \). If displaced by \( 5 \, \text{cm} \), what i

**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. Total energy: E = (1/2) k A² . A = 0.05 m, k = 320 N/m . E = 0.5 × 320 × (0.05)² = 0.5 × 320 × 0.0025 = 0.4 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.4 J 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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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.6 \, \text{kg}, k = 240 \, \text{N/m} \). If displaced by \( 7 \, \text{cm} \), what i

**Spring-mass system** has period T = 2π√(m/k), frequency f = (1/2π)√(k/m), ω = √(k/m), where k spring constant (N/m) and m mass (kg). For parallel combination, k_eff = k₁ + k₂, series gives 1/k_eff = 1/k₁ + 1/k₂, affecting ω = √(k_eff/m) and T = 2π√(m/k_eff). Total energy: E = (1/2) k A² . A = 0.07 m, k = 240 N/m . E = 0.5 × 240 × (0.07)² = 0.5 × 240 × 0.0049 = 0.588 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.588 J

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

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.25 \, \text{kg}, k = 100 \, \text{N/m} \). If displaced by \( 6 \, \text{cm} \), what

**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. Total energy: E = (1/2) k A² . A = 0.06 m, k = 100 N/m . E = 0.5 × 100 × (0.06)² = 0.5 × 100 × 0.0036 = 0.18 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.18 J 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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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 4 \, \text{kg}, k = 400 \, \text{N/m} \). If displaced by \( 20 \, \text{cm} \), what is

**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. Potential energy: U = (1/2) k x² . At x = 10 cm = 0.1 m : U = (1/2) × 400 × (0.1)² = 0.5 × 400 × 0.01 = 2 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result

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

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.4 \, \text{kg}, k = 160 \, \text{N/m} \). If displaced by \( 6 \, \text{cm} \), what i

**Spring-mass system** has period T = 2π√(m/k), frequency f = (1/2π)√(k/m), ω = √(k/m), where k spring constant (N/m) and m mass (kg). For parallel combination, k_eff = k₁ + k₂, series gives 1/k_eff = 1/k₁ + 1/k₂, affecting ω = √(k_eff/m) and T = 2π√(m/k_eff). Total energy: E = (1/2) k A² . A = 0.06 m, k = 160 N/m . E = 0.5 × 160 × (0.06)² = 0.5 × 160 × 0.0036 = 0.288 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.288 J

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

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Energy in SHM - Kinetic, Potential and Total

A particle in SHM has \( x = 3 \cos (5t) \) (in m). What is its kinetic energy at \( x = 1.5 \, \text{m} \) if \( m = 2

**Conservation of mechanical energy** in undamped SHM implies total energy proportional to amplitude squared A² and spring constant k. E = ½ k A² allows amplitude determination from known E and k via A = √(2E/k), with k = m ω² linking dynamical and energetic descriptions for spring-mass system. Total energy: E = (1/2) m ω² A² = 0.5 × 2 × 5² × 3² = 225 J . Potential energy: U = (1/2) m ω² x² = 0.5 × 2 × 25 × (1.5)² = 56.25 J . Kinetic energy: K = E - U = 225 - 56.25 = 168.75 J . Applying

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

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.9 \, \text{kg}, k = 360 \, \text{N/m} \). If displaced by \( 4 \, \text{cm} \), what i

**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. Total energy: E = (1/2) k A² . A = 0.04 m, k = 360 N/m . E = 0.5 × 360 × (0.04)² = 0.5 × 360 × 0.0016 = 0.288 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.288 J 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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