Question Detail
Question
A 2-kg object oscillates on a spring with a force constant k = 400 N/m and initial amplitude of 3 cm. If the energy decreases by 1% per period, what is the damping constant 'b'?
Correct Answer
Option B is the correct answer.
Detailed Explanation
From Worked Example 20 (page 34), we first find the period T = 2π√(m/k) = 2π√(2 kg / 400 N/m) ≈ 0.444 s. The fractional energy loss per period is ΔE/E = 0.01. The relationship between energy loss, damping constant 'b', and period T is ΔE/E = bT/m. Solving for b: b = (ΔE/E) * (m/T) = 0.01 * (2 kg / 0.444 s) ≈ 0.045 kg/s.
Hint
First, calculate the period T of the oscillation using the mass and spring constant. Then use the given fractional energy decrease per period (ΔE/E) and the formula ΔE/E = bT/m to solve for the damping constant 'b'.