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#spring constant 1000 N/m

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 2.5 \, \text{kg}, k = 1000 \, \text{N/m} \). What is its angular frequency?

**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. ω = √((k/m)) = √((1000/2.5)) = √(400) = 20 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 rad/s 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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Energy in SHM - Kinetic, Potential and Total

A spring of \( k = 1000 \, \text{N/m} \) has a \( 2 \, \text{kg} \) mass. If \( E = 5 \, \text{J} \), what is the amplit

**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 . 5 = 0.5 × 1000 × A² ⇒ 5 = 500 A² ⇒ A² = 0.01 ⇒ A = 0.1 m . Applying x = A

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

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