Skip to content
Login Register

#spring system

2 public questions tagged with this topic.

Equations of SHM, Phase and Angular Frequency

A spring system has \( k = 800 \, \text{N/m}, m = 2 \, \text{kg} \). What is the maximum speed if amplitude is \( 5 \, \

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. ω = √((k/m)) = √((800/2)) = 20 rad/s . vₘₐₓ = ω A = 20 × 0.05 = 1.0 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.0 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

View Details
Energy in SHM - Kinetic, Potential and Total

A spring system has \( m = 0.4 \, \text{kg}, k = 160 \, \text{N/m}, A = 7 \, \text{cm} \). 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 = 160 N/m, x = 0.035 m . U = 0.5 × 160 × (0.035)² = 0.5 × 160 × 0.001225 = 0.098 J .

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

View Details