Practice question
Question
A spring of \( k = 250 \, \text{N/m} \) has a \( 2.5 \, \text{kg} \) mass. If \( E = 1.25 \, \text{J}
\), what is the amplitude?
Explanation
**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. 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 cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.1 m follows,
Discussion
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