Physics · Oscillations of a spring: restoring force and force constant
A mass of is suspended from a spring of stiffness . It is set oscillating and it
A mass of \( 50 \mathrm{kg} \) is suspended from a spring of stiffness \( 10 \mathrm{kN} / \mathrm{m} \). It is set oscillating and it is observed that two successive oscillations have amplitudes of \( 10 \mathrm{mm} \) and \( 1 \mathrm{mm} \) Determine the damping ratio.
- A. 0.315
- B. 0.328
- C. 0.344
- D. 0.353
Step-by-step solution
The logarithmic decrement δ = ln(amplitude ratio) = ln(10/1) ≈ 2.3026. For underdamped vibration, δ = 2πζ/√(1-ζ^2). Solving for ζ gives ζ = δ/√(4π²+δ²) = 2.3026/√(39.4784+5.3019) = 2.3026/6.6918 ≈ 0.344.
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