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#spring potential energy

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Energy in SHM - Kinetic, Potential and Total

A spring system has \( m = 1.2 \, \text{kg}, k = 480 \, \text{N/m}, A = 6 \, \text{cm} \). What is the potential energy

**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. Potential energy: U = (1/2) k x² . k = 480 N/m, x = 0.03 m . U = 0.5 × 480 × (0.03)² = 0.5 × 480 × 0.0009 = 0.216 J .

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

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Energy in SHM - Kinetic, Potential and Total

A spring system has \( m = 1 \, \text{kg}, k = 100 \, \text{N/m}, A = 20 \, \text{cm} \). What is the potential energy a

**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. Potential energy: U = (1/2) k x² . At x = 0 m , U = 0 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.0 J 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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