Question 2.2: The system shown in Fig. 2.1, but without the applied force,......
The system shown in Fig. 2.1, but without the applied force, F, has the following properties: m = 1 kg; k = 10 000 N/m and c = 40 N/m/s. Plot the time history of z for the initial conditions: z = 0.1 m, \dot{z} = 0, at t = 0.
From Eq. (2.16),
\omega _{n} = \sqrt{\frac{k}{m} } = \sqrt{\frac{10000}{1} } = 100 rad/sFrom Eqs (2.18) and (2.19),
c_{c} = 2m\omega _{n} (2.18)
\gamma = \frac{c}{c_{c}} (2.19)
\gamma = \frac{c}{c_{c}} = \frac{c}{2 m \omega _{n}} = 40/(2 \times 1\times 100) = 0.2From Eq. (2.26), \omega _{d} = \omega _{n} \sqrt{(1-\gamma ^{2})} = 100\sqrt{(1 – 0.2^{2})} = 98.0 rad/s.
Since \gamma < 1, the response is given by Eq. (2.28):
z = e^{-\gamma \omega _{n}t}(A \cos \omega _{d}t + B \sin \omega _{d}t ) (A)
We shall also require an expression for the velocity, \dot{z}. Differentiating Eq. (A):
\dot{z} = e^{-\gamma \omega _{n}t}[(B\omega _{d} – A\gamma \omega _{n})\cos \omega _{d}t – (A\omega _{d} + B\gamma \omega _{n}) \sin \omega _{d}t] (B)
The constants A and B can be found from Eqs (A) and (B) by substituting the numerical values of \omega _{n}, \omega _{d} and \gamma , as calculated above, and the initial conditions, z = 0.1, and \dot{z} = 0 at t = 0, giving A = 0.1, and B=0.0204. Then from Eq. (A):
z = e^{-20t}(0.1 \cos 98.0 t + 0.204 \sin 98.0 t ) (C)
This is easily plotted using a spreadsheet program, as in Fig. 2.4.
