I'm heaving trouble solving the following undamped forced vibration problem using Laplace transforms:
$$\ddot{q}(t) + \omega_n^2 q(t) = \cos(\omega t).$$
I will show what I have done so far, and I'd appreciate any insights.
Taking the Laplace transform of both sides and applying the derivative identities yields,
$$\mathcal{L}\{\ddot{q}(t)\} + \mathcal{L}\{\omega_n^2 q(t)\} = \mathcal{L}\{\cos(\omega t)\},$$
$$s^2 \mathcal{L}\{{q}(t)\} – s q(0) – \dot{q}(0) + \omega_n^2 \mathcal{L}\{q(t)\} = \frac{s}{s^2 + \omega^2},$$
I asume the initial conditions are zero, then group and solve for $\mathcal{L}\{{q}(t)\}$,
$$\mathcal{L}\{{q}(t)\} = \frac{1}{s^2 + \omega_n^2} \frac{s}{s^2 + \omega^2} .$$
Multiplying and dividing by $\omega_n$ allows recognizing the convolution product,
$$ \mathcal{L}\{{q}(t)\} = \frac{1}{\omega_n} \mathcal{L}\{\sin(\omega_n t)\} \mathcal{L}\{\cos(\omega t)\}.$$
And the time-domain solution would be
$$
q(t) = \frac{1}{\omega_n} \int_{0}^{t} \cos(\omega (t – \tau))\sin(\omega_n \tau) \,d \tau.
$$
Now here is my problem: I don't know if made a mistake up to this point, but assuming I haven't when I try to evaluate the convolution integral the results are not what I expected, namely I expected,
$$
\lim_{\omega \rightarrow \omega_n} \lim_{t \rightarrow \infty} q(t) \rightarrow \infty,
$$
but I don't get that from the convolution integral.
Any thought?
Best Answer
I had evaluated the convolution integral incorrectly. The answer is
$$ q(t) = \frac{1}{\omega_n} \int_{0}^{t} \cos(\omega(t-\tau)) \sin(\omega_n \tau) \,d \tau = \frac{\cos(\omega t) - \cos(\omega_n t)}{\omega_n^2 - \omega^2} $$
The limit as $\omega \rightarrow \omega_n$ can be obtained L'Hopital's rule,
$$\lim_{\omega \rightarrow \omega_n} q(t) = \frac{t \sin(\omega_n t)}{2 \omega_n}$$
Clearly the magnitude of the response approaches infinity as $t \rightarrow \infty$.
Thanks to Ian for his comments.