Figure 3.21 shows two masses m1 and m2 where m1 = 5 kg and m2 = 1 kg. The mass m1 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.

Question

Figure 3.21 shows two masses [latex]m_{1} ext{ and } m_{2}[/latex] where [latex]m_{1} = 5 \ kg ext{ and } m_2 = 1 \ kg.[/latex]
The mass [latex]m_{1} [/latex] 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 [latex] m_2 [/latex] is assumed to move with a constant velocity [latex] v_2 = −5 \ m/s.[/latex] Assuming the coefficient of restitution e = 0.9, determine the displacement equations of the two masses after impact.

Accepted Answer

The conservation of momentum and the restitution conditions are
[latex]m_{1}v^{\prime}_{1} + m_{2}v^{\prime}_{2} = m_{1}v_{1} + m_{2}v_{2}[/latex]

[latex]v^{\prime}_{2}− v^{\prime}_{1} = e(v_{1} − v_{2})[/latex]

where [latex]v_{1} = 0 ext{ and } v_{2}[/latex] = −5 m/s. Therefore, the conservation of momentum and restitution equations can be written as
[latex]5v^{\prime}_{1} + v^{\prime}_{2}[/latex] = (5)(0) + (1)(−5)

[latex]v^{\prime}_{2} − v^{\prime}_{1}[/latex] = 0.9(0 − (−5))

which yield

[latex]5v^{\prime}_{1} + v^{\prime}_{2} = −5, \quad \quad v^{\prime}_{1} − v^{\prime}_{2}[/latex] = −4.5

By adding these two equations, one obtains [latex] 6v^{\prime}_{1} = −9.5, [/latex] or
[latex]v_{1}[/latex] = −1.5833 m/s
and
[latex]v^{\prime}_{2} = v^{\prime}_{1}[/latex] + 4.5 = −1.5833 + 4.5 = 2.9167 m/s
The equations of motion of the two masses after impact can be written as

[latex]m_{1} \ddot{x_{1}} + c \dot{x_{1}} + kx_{1}[/latex] = 0
[latex]m_{2} \ddot{x_{2}}[/latex] = 0

The solutions of these two equations can be determined as
[latex]x_{1} = Xe ^{−ξωt} \sin(ω_dt + \phi)[/latex]

[latex]x_{2} = A_{1}t + A_{2}[/latex]

where [latex] X, \phi, A_1, ext{ and } A_2 [/latex] are constants that can be determined using the initial conditions [latex]x_{1 0} = 0, \dot{x}_{1 0} = v^{\prime}_{1}[/latex] = −1.5833 m/s, [latex]x_{2 0} = 0, \dot{x}_{2 0} = v^{\prime}_{2}[/latex] = 2.9167 m/s. The damping factor ξ , natural frequency ω, and the damped natural frequency [latex] ω_d [/latex] are given by

ξ = [latex]\frac{c}{C_{c}} = \frac{c}{2√km} = \frac{10}{2√(1000)(5)}[/latex] = 0.0707

ω = [latex]\sqrt{\frac{k}{m}} = \sqrt{\frac{1000}{5}}[/latex] = 14.142 rad/s

[latex]ω_{d} = ω \sqrt{1 − ξ^{2}}[/latex] = 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

[latex] x_1 = −0.1122e^{−0.9998t} \sin(14.1066t)[/latex]
[latex] x_2 = 2.9167t [/latex]

The results presented in Fig. 3.22, which shows [latex] x_1 ext{ and } x_2 [/latex] as functions of time, clearly demonstrate that the two masses do not encounter a second impact.

Question 3.13: Figure 3.21 shows two masses m1 and m2 where m1 = 5 kg and m...

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