Skip to content
Login Register

#elastic energy

8 public questions tagged with this topic.

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

View Details
Energy in SHM - Kinetic, Potential and Total

A spring system has \( m = 0.25 \, \text{kg}, k = 100 \, \text{N/m}, A = 8 \, \text{cm} \). What is the potential energy

**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. Potential energy: U = (1/2) k x² . k = 100 N/m, x = 0.04 m . U = 0.5 × 100 × (0.04)² = 0.5 × 100 × 0.0016 = 0.08 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.08 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details
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

View Details