# Solving forced undamped vibration using Laplace transforms

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?

• You should anticipate that the amplitude of the oscillation grows over time, not that the limit is $+\infty$. – Ian Sep 27 '15 at 18:24
• In fact being explicit your solution is apparently $\frac{\cos(\omega t)-\cos(\omega_n t)}{\omega_n^2-\omega^2}$, or something like this. For $t \gg 1/|\omega_n-\omega|$, the numerator will be (sometimes) on the order of $1$ while the denominator will remain small. – Ian Sep 27 '15 at 18:31
• You can qualitatively see what is going on when you look at $\omega=\omega_n$, then the solution is $t$ times a sinusoid (maybe plus some periodic terms). So it does not blow up monotonically but the amplitude still grows. For $\omega$ close to $\omega_n$ the amplitude grows but not without bound. – Ian Sep 27 '15 at 18:32
• Ian, you actually answered the question. I evaluated the convolution integral incorrectly. And indeed the solution tends to infinity as $\omega \rightarrow \omega_n$. – Indrid Cold Sep 27 '15 at 18:49
• ...No, it doesn't. Again, look at the truly resonant case: $y''+y=\cos(y),y(0)=0,y'(0)=0$ has solution $y=\frac{1}{2} t \sin(t)$. On a fixed finite time interval, the solution where we replace $\cos(y)$ with $\cos(ay)$ for $a \approx 1$ will be similar to this. – Ian Sep 27 '15 at 18:51

## 1 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.