A vertical spring stretches by 2 mm when a mass of 200 g is attached and allowed to come to equilibrium. From this equilibrium position, the mass is pulled downward by an additional 2 mm and released. Taking g = 10 m·s⁻², which pair gives the frequency of oscillation and the maximum elastic potential energy stored in the spring?
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Question

A vertical spring stretches by 2 mm when a mass of 200 g is attached and allowed to come to equilibrium. From this equilibrium position, the mass is pulled downward by an additional 2 mm and released. Taking g = 10 m·s⁻², which pair gives the frequency of oscillation and the maximum elastic potential energy stored in the spring?

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G1 Physics · Accepted Answer

At equilibrium, kx₀ = mg. Given, m = 0.20 kg, x₀ = 2 mm = 2 × 10⁻³ m. Therefore, k = mg ÷ x₀ = 0.20 × 10 ÷ (2 × 10⁻³) = 1000 N·m⁻¹. The oscillation frequency is f = (1 ÷ 2π)√(k ÷ m) = (1 ÷ 2π)√(1000 ÷ 0.20) = √5000 ÷ (2π) = 5√50 ÷ π Hz. The mass is pulled 2 mm below equilibrium. Since the equilibrium extension is already 2 mm, the maximum extension from the natural length is x(max) = 4 mm = 4 × 10⁻³ m. Maximum spring energy: U(max) = (1 ÷ 2)kx(max)² = (1 ÷ 2)(1000)(4 × 10⁻³)² = 8 × 10⁻³ J. Corr…