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#spring

14 public questions tagged with this topic.

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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Energy in SHM - Kinetic, Potential and Total

A spring of \( k = 300 \, \text{N/m} \) has a \( 1.5 \, \text{kg} \) mass. If \( E = 0.75 \, \text{J} \), what is the am

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Total energy: E = (1/2) k A² . 0.75 = 0.5 × 300 × A² ⇒ 0.75 = 150 A² ⇒ A² = 0.005 ⇒ A = √(0.005) ≈ 0.071 m . Applying x

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

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PHYSICS

A 1200 kg car moving at 15 m/s compresses a spring ( k = 8 × 10³N/m ). What is the maximum compression?

Given: A 1200 kg car moving at 15 m/s compresses a spring ( k = 8 × 10³N/m ). What is the maximum compression? Formula: Initial K = 1/2 × 1200 × 15² = 135000 J. Substitution & Calculation: Spring energy 1/2 k x_m² = 135000 Rightarrow 4000 x_m² = 135000 Rightarrow x_m = √33.75 approx 5.81 m . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27), Chapter: Work, Energy and Power (Latest NCERT 2026-27), Topic: Energy conservation, kinetic energy ½mv² to spring potential ½kx² and compression. The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI units and illustrative examples. Page.

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Laws of Motion and Energy Conservation

A 8kg mass falls from 5m onto a spring (k\=2500N/m). What is the maximum compression? (Take g\=10m/s2)

Potential energy mgh=8×10×5=400J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 400 J. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Laws of Motion and Energy Conservation

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