#mechanical energy

Forced Oscillations and Resonance

Which condition ensures that the total mechanical energy in an SHM system remains conserved during the motion?

**Forced oscillations** result when external periodic driving force F = F₀ cos(ω_d t) acts on oscillator, steady-state frequency equals driving frequency ω_d, amplitude A = F₀/√((k - m ω_d²)² + (b ω_d)²) depends on proximity to natural frequency ω₀ = √(k/m). Resonance when ω_d ≈ ω₀, amplitude maximum. Total mechanical energy (kinetic + potential) is conserved in SHM when no external dissipative forces (e.g., friction) act, allowing energy to transform without loss. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Absence of dissipative forces follows, ref

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

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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 \( m = 0.4 \, \text{kg}, k = 160 \, \text{N/m} \). 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

In an ideal SHM system, what occurs to the total mechanical energy as the particle moves from the mean position to an ex

**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. Total mechanical energy in ideal SHM (no friction) is conserved, remaining constant as kinetic energy converts to potential energy during the motion. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It remains constant follows, reflecting SHM dependence on amplitude A, ω and

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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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.5 \, \text{kg}, k = 200 \, \text{N/m} \). If the amplitude is \( 4 \, \text{cm} \), wh

**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 = 200 N/m . E = 0.5 × 200 × (0.04)² = 0.5 × 200 × 0.0016 = 0.16 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.16 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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Two-Dimensional Motion and Vectors

In projectile motion, what is true about the total mechanical energy, assuming no air resistance?

Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields It remains conserved. Hence option C satisfies projectile formulas.

Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Two-Dimensional Motion and Vectors

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