Question 3.4.4: Calculate the response of an underdamped spring–mass system ......
Calculate the response of an underdamped spring–mass system to a unit impulse. Assume zero initial conditions.
The equation of motion is
m \ddot{x}+c \dot{x}+k x=\delta(t)
Taking the Laplace transform of both sides of this expression using the results of Example 3.4.1 and entry 1 in Table 3.1 yields
\left(m s^2+c s+k\right) X(s)=1
Solving for X(s) yields
X(s)=\frac{1 / m}{s^2+2 \zeta \omega_n s+\omega_n^2}
Assuming that \zeta<1 and consulting entry 8 of Table 3.1 yields
x(t)=\frac{1 / m}{\omega_n \sqrt{1-\zeta^2}} e^{-\zeta \omega_n t} \sin \left(\omega_n \sqrt{1-\zeta^2} t\right)=\frac{1}{m \omega_d} e^{-\zeta \omega_n t} \sin \omega_d t
in agreement with equation (3.6).
x(t)=\frac{\hat{F} e^{-\zeta \omega_n t}}{m \omega_d} \sin \omega_d t (3.6)
| Table 3.1 Common Laplace Transforms for Zero Initial Conditions^a | ||
| F(s) | f(t) | |
| 1. | 1 | δ(0) unit impulse |
| 2. | 1/s | 1, unit step Φ(t) |
| 3. | \frac{1}{s+a} | e^{-a t} |
| 4. | \frac{1}{(s+a)(s+b)} | \frac{1}{b-a}\left(e^{-a t}-e^{-b t}\right) |
| 5. | \frac{\omega_n}{s^2+\omega_n^2} | \sin \omega_n t |
| 6. | \frac{s}{s^2+\omega_n^2} | \cos \omega_n t |
| 7. | \frac{1}{s\left(s^2+\omega_n^2\right)} | \frac{1}{\omega_n^2}\left(1-\cos \omega_n t\right) |
| 8. | \frac{1}{s^2+2 \zeta \omega_n s+\omega_n^2} | \frac{1}{\omega_d} e^{-\zeta \omega_n t} \sin \omega_d t, \zeta<1, \omega_d=\omega_n \sqrt{1-\zeta^2} |
| 9. | \frac{\omega_n^2}{s\left(s^2+2 \zeta \omega_n s+\omega_n^2\right)} | 1-\frac{\omega_n}{\omega_d} e^{-\zeta \omega_n t} \sin \left(\omega_d t+\phi\right), \phi=\cos ^{-1} \zeta, \zeta<1 |
| 10. | e^{-a s} | δ(t – a) |
| 11. | F(s – a) | e^{a t} f(t) \geq 0 |
| 12. | e^{-a s} F(s) | f(t – a)Φ(t – a) |
^aA more complete table appears in Appendix B. Here the Heaviside step function or unit step function is denoted by Φ. Other notations for this function include μ and H.