Question 3.13: Figure 3.21 shows two masses m1 and m2 where m1 = 5 kg and m...
Figure 3.21 shows two masses m_{1} \text{ and } m_{2} where m_{1} = 5 \ kg \text{ and } m_2 = 1 \ kg.
The mass m_{1} which is initially at rest is supported by a spring and a damper.
The stiffness coefficient of the spring k is assumed to be 1000 N/m, and the damping coefficient c is assumed to be 10 N · s/m. The mass m_2 is assumed to move with a constant velocity v_2 = −5 \ m/s. Assuming the coefficient of restitution e = 0.9, determine the displacement equations of the two masses after impact.
The conservation of momentum and the restitution conditions are
m_{1}v^{\prime}_{1} + m_{2}v^{\prime}_{2} = m_{1}v_{1} + m_{2}v_{2}
where v_{1} = 0 \text{ and } v_{2} = −5 m/s. Therefore, the conservation of momentum and restitution equations can be written as
5v^{\prime}_{1} + v^{\prime}_{2} = (5)(0) + (1)(−5)
v^{\prime}_{2} − v^{\prime}_{1} = 0.9(0 − (−5))
which yield
5v^{\prime}_{1} + v^{\prime}_{2} = −5, \quad \quad v^{\prime}_{1} − v^{\prime}_{2} = −4.5
By adding these two equations, one obtains 6v^{\prime}_{1} = −9.5, or
v_{1} = −1.5833 m/s
and
v^{\prime}_{2} = v^{\prime}_{1} + 4.5 = −1.5833 + 4.5 = 2.9167 m/s
The equations of motion of the two masses after impact can be written as
m_{1} \ddot{x_{1}} + c \dot{x_{1}} + kx_{1} = 0
m_{2} \ddot{x_{2}} = 0
The solutions of these two equations can be determined as
x_{1} = Xe ^{−ξωt} \sin(ω_dt + \phi)
where X, \phi, A_1, \text{ and } A_2 are constants that can be determined using the initial conditions x_{1 0} = 0, \dot{x}_{1 0} = v^{\prime}_{1} = −1.5833 m/s, x_{2 0} = 0, \dot{x}_{2 0} = v^{\prime}_{2} = 2.9167 m/s. The damping factor ξ , natural frequency ω, and the damped natural frequency ω_d are given by
ξ = \frac{c}{C_{c}} = \frac{c}{2√km} = \frac{10}{2√(1000)(5)} = 0.0707
ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{1000}{5}} = 14.142 rad/s
ω_{d} = ω \sqrt{1 − ξ^{2}} = 14.1066 rad/s
By using the initial conditions one can show that the displacements of the two masses as functions of time can be written as
x_1 = −0.1122e^{−0.9998t} \sin(14.1066t)
x_2 = 2.9167t
The results presented in Fig. 3.22, which shows x_1 \text{ and } x_2 as functions of time, clearly demonstrate that the two masses do not encounter a second impact.
