Question 3.9: A damped single degree of freedom mass–spring system has mas...
A damped single degree of freedom mass–spring system has mass m = 5 kg and stiffness coefficient k = 500 N/m. From the experimental measurements, it was observed that the amplitude of vibration diminishes from 0.02 to 0.012 m in six cycles. Determine the damping coefficient c.
In this example, n of Eq. 3.94 is equal to 6. Let
ln \frac{x_{i}}{x_{i+n}} = nξωτ_{d} = nδ (3.94)
x_{i} = 0.02, x_{i+6} = 0.012
The logarithmic decrement δ is determined from
nδ = 6δ = ln \frac{0.02}{0.012} = 0.51083
that is,
δ = 0.085138
The damping factor ξ can be determined from Eq. 3.95 as
ξ = \frac{δ}{\sqrt{(2π)² + δ²}} = \frac{0.085138}{\sqrt{(2π)² + (0.085138)²}} = 0.013549
The damping coefficient c is defined as c = ξC_{c}, where the critical damping coefficient C_{c} is
C_{c} = 2 \sqrt{km} = 2\sqrt{(500)(5)} = 100N · s/m
Therefore,
c = (0.013549)(100) = 1.3549N · s/m
The specific energy loss is
\frac{ΔU}{U_{i}} = 1 − e^{−2δ} = 1 − e^{−2(0.085138)} = 0.1566