A spring of \( k = 250 \, \text{N/m} \) has a \( 1 \, \text{kg} \) mass. If \( E = 1.25 \, \text{J} \), what is the ampl
**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 . 1.25 = 0.5 × 250 × A² ⇒ 1.25 = 125 A² ⇒ A² = 0.01 ⇒ A = 0.1 m . Applying x = A
Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total