Question 3.1.3: In vibration testing, an instrumented hammer is often used t......
In vibration testing, an instrumented hammer is often used to hit a device to excite it and to measure the impact force simultaneously. If the device being tested is a single-degreeof-freedom system, plot the response given that m=1\ kg , c=0.5\ kg / s , k=4\ N / m, and \hat{F}=0.2\ N \cdot s. It is often difficult to provide a single impact with a hammer. Sometimes a “double hit” occurs, so the exciting force may have the form
F(t)=0.2 \delta(t)+0.1 \delta(t-\tau)
Plot the response of the same system with a double hit and compare it with the response to a single impact. Assume that the initial conditions are zero.
The solution to the single unit impact and time t=0 is given by equations (3.7) and (3.8) with \omega_n=\sqrt{4}=2\ rad / s and \zeta=c /\left(2 m \omega_n\right)=0.125. Thus, with \hat{F}=0.2δ(t)
x(t)=\hat{F} h(t) (3.7)
h(t)=\frac{1}{m \omega_d} e^{-\zeta \omega_n t} \sin \omega_d t (3.8)
\begin{aligned} x_1(t) & =\frac{0.2}{(1)\left(2 \sqrt{1-(0.125)^2}\right)} e^{-(0.125)(2) t} \sin 2 \sqrt{1-(0.125)^2 t} \\ & =0.1008 e^{-0.25 t} \sin (1.984 t) m \end{aligned}
which is plotted in Figure 3.3(a). Similarly, the response to 0.18(t-\tau) is calculated from equations (3.7) and (3.9) as
h(t-\tau)=\frac{1}{m \omega_d} e^{-\zeta \omega_n(t-\tau)} \sin \omega_d(t-\tau) \quad t>\tau (3.9)
x_2(t)=0.0504 e^{-0.25(t-\tau)} \quad \sin 1.984(t-\tau) m \quad t>\tau
and x_2(t)=0 for 0<t<\tau=0.5. The force input f(t) is indicated in Figure 3.3(b). It is important to note that no contribution from x_2 occurs until time t=\tau. Using the principle of superposition for linear systems, the response to the “double impact” will be the sum of the preceding two impulse responses:
\begin{aligned} x(t) & =x_1(t)+x_2(t) \\ & =\left\{\begin{array}{l} 0.1008 e^{-0.25 t} \sin (1.984 t) \quad 0<t<\tau \\ 0.1008 e^{-0.25 t} \sin (1.984 t)+0.0504 e^{-0.25(t-\tau)} \sin 1.984( t -\tau) \quad t>\tau \end{array}\right. \end{aligned}
This is plotted in Figure 3.3(a) for the value \tau=0.5\ s.
Note that the obvious difference between the two responses is that the “doublehit” response has a “spike” at \tau=t=0.5, causing a larger amplitude. The time, \tau, represents the time delay between the two hits.
