Question 3.11: The single degree of freedom mass–spring system shown in Fig...
The single degree of freedom mass–spring system shown in Fig. 3.17 has mass m = 5 kg, stiffness coefficient of the spring k=5×10^{3}N/m, coefficient of dry friction μ = 0.1, initial displacement x_{0} = 0.03 m, and initial velocity \dot{x_{0}} = 0. Determine the number of cycles of oscillation of the mass before it comes to rest.
The friction force F_{f} is defined by Eq. 3.109 as F_{f} = μN where N is the normal reaction force given by N = mg. Therefore, the force F_{f} is
F_{f} = μmg = (0.1)(5)(9.81) = 4.905 N
The motion will stop if the amplitude of the cycle is such that the spring force is less than or equal to the friction force, that is, kX_{f} ≤ 4.905, or
X_{f} ≤ \frac{4.905}{k} = \frac{4.905}{5 × 10³} = 0.981 × 10^{−3} m
The amplitude loss per half-cycle is
2F_{f}/k = \frac{2(4.905)}{5 × 10³} = 1.962 × 10^{−3} m
The number of half-cycles n completed before the mass comes to rest can be obtained from the following equation
x_{0} − n (\frac{2F_{f}}{k}) ≤ 0.981 × 10^{−3}
where x_{0} = 0.03. It follows that
0.03 − n(1.962 × 10^{−3}) ≤ 0.981 × 10^{−3}
The smallest n that satisfies this inequality is n = 15 half-cycles, and the number of cycles completed before the mass comes to rest is 7.5.