Find the response of the single degree of freedom system shown in Fig. 5.9 to the rectangular impulsive force shown in Fig. 5.8, where m = 10 kg, k = 9,000 N/m, c = 18N · s/m, and F0 = 10,000 N. The force is assumed to act at time t = 0 and the impact interval is assumed to be 0.005 s.

Question

Find the response of the single degree of freedom system shown in Fig. 5.9 to the rectangular impulsive force shown in Fig. 5.8, where m = 10 kg, k = 9,000 N/m, c = 18N · s/m, and [latex]F_0[/latex] = 10,000 N. The force is assumed to act at time t = 0 and the impact interval is assumed to be 0.005 s.

Accepted Answer

The linear impulse I is given by
I = [latex]\int_ {t_{1}}^{t_{2}}F(t) dt = \int_0^{0.005} 10,000 dt = 10,000(0.005) =[/latex] 50 N · s
The natural frequency of the system ω is given by
ω = [latex]\sqrt{\frac{k}{m}} =\sqrt{\frac{9000}{10}}= [/latex] 30 rad/s
The critical damping coefficient [latex]C_c[/latex] is

[latex]C_c[/latex] = 2mω = 2(10)(30) = 600

The damping factor ξ is
[latex]ξ = \frac{c}{C_c} = \frac{18}{600} = 0.03[/latex]
The damped natural frequency [latex]ω_d[/latex] is

[latex]ω_d = ω\sqrt{1 − ξ^2} = 30\sqrt{1 − (0.03)^2} = 29.986[/latex] rad/s

The system response to the impulsive force is then given by
x(t) = [latex]\frac{I}{mω_d} e^{−ξωt} \sin ω_dt\[0.5 cm] = \frac{50} {(10)(29.986)}e^{-(0.03)(30)t} \sin 29.986t\[0.5 cm] = 0.1667e^{−0.9t} \sin 29.986t[/latex]

Question 5.5: Find the response of the single degree of freedom system sho...

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