# Why is amplitude going to infinity in forced damped oscillator at resonance?

I'm trying to find the amplitude of steady state response of the following differential equation:

$$\ddot{x}+2p\dot x + {\omega_0}^2x=\cos(\omega t)$$

A particular solution is $$x_p=\Re{\dfrac{e^{i\omega t}}{\omega_0^2 - \omega^2 + i2p\omega}}$$

The amplitude at steady state is then $$A=\dfrac{1}{\sqrt{(\omega_0^2 - \omega^2)^2 + (2p\omega)^2}}$$

The denominator has minimum value when $$\omega^2 =\omega_0^2 - 2p^2$$: $$A=\dfrac{1}{2p\sqrt{\omega_0^2-p^2}}$$

This expression seems to suggest that the amplitude goes to infinity as $$p$$ approaches $$\omega_0$$.

But amplitude has to be finite(from other examples of LRC tank circuit etc).
Pretty sure I'm wrong but not able to see where. Any help?

• @Qmechanic Ah right.. $p=\omega_0$ gives a single real root for homogeneous part: $\ddot{x}+2p \dot{x}+\omega_0^2x=0$. I also notice that $p=\omega_0$ is $0$ input frequency AND $p^2 > w_0^2/2$ corresponds to negative input frequency $\omega$ .. but that's it... not so sure what it means :( Feb 8 at 14:12
• The denominator is a sum of two positive quantities: $(\omega^2-\omega_0^2)^2$ and $(2p\omega)^2$: thus it cannot go to $0$ unless $p$ has an imaginary part. Feb 8 at 15:40
• Linked. Show your work. The linked question has answers writing the denominator in a form minimized by inspection and thus preventing your algebra error. Feb 8 at 15:41

You algebra is wrong. If $$\omega^2-\omega_0^2=-2p^2$$ you get $$(-2p^2)^2+(2 p\omega )^2= (2p^2)^2+(2 p\omega )^2$$= under the square root and, being the sum of two postive numbers, this can never be zero.