# Show that undamped system is nonperiodic

The DE for the undamped mass-spring system with a given periodic external force can be written as $y'' +k_0^2y = A\cos{kx},$ where $k_0$ is the natural frequency of the system and $k$ is the applied frequency.

If $k \neq k_0$, the solution will be $$y(x) = \frac{A}{k_0^2 - k^2}\cos{kx}$$ Thus, if the applied frequency $k$ is close to the natural frequency $k_0$, then this particular solution represents as oscillation with the large amplitude. This is called $resonance$. If $k = k_0,$ a particular solution cannot be obtained from this solution.

Show that the particular solution is given by $$y(x) = \frac{A}{2k_0}x\sin{k_0x},$$ which is nonperiodic.

So, if $k \neq k_0$, the solution is periodic, but if $k = k_0$, i can't see why is it nonperiodic?

• It's non-periodic because $x$ isn't periodic and so $x \sin x$ isn't periodic. Commented Mar 30, 2015 at 3:27
• how do you know that $x\sin{}$ isn't periodic ? Commented Mar 30, 2015 at 3:52
• Is $x$ periodic? Commented Mar 30, 2015 at 4:08
• as far as I know when I plot $y(x) = x\sin{x}$, the function is clearly not periodic. But then, I have trouble how to prove it. Commented Mar 30, 2015 at 4:10

Recall that a function $f:\mathbb{R}\to\mathbb{R}$ is periodic with period $p$ if $f(x+p)=f(x)$ for all $x\in\mathbb{R}.$ For example $\cos(x+2\pi)=\cos x$ for all real $x$, so $\cos x$ is periodic with period $p=2\pi$.
Try applying the definition of periodic to $y(x) = \frac{A}{2k_0}x\sin{k_0x},$ and see what you conclude. Are there any values of $p$ where $y(x+p)=y(x)$ for all $x\in\mathbb{R}?$
Geometrically, the function $f$ is periodic with period $p$ if we can shift the graph of $f$ left or right by $p$ and get the same graph. Try plotting $y(x) = \frac{A}{2k_0}x\sin{k_0x}$ for some choice of $A$ and $k_0.$
• do you mean that I should take $y(x + \omega) = \frac{A}{2k_0}(x + \omega)\sin{k_0(x+\omega)}$, then plug in to original DE to see if it is equal $A\cos{kx}$. Is it correct? Commented Mar 30, 2015 at 3:51
• Close. Rather than comparing it with $A\cos kx$, you should compare $y(x+\omega)$ with $y(x) = \frac{A}{2k_0}x\sin{k_0x}.$ Commented Mar 30, 2015 at 3:54
• One more hint to add: in order for it to be periodic, the equation in the definition of periodic must be true for all $x$. Thus, to show that it's not periodic, all you need to do is find a particular value of $x$ for which the equation is not true. Commented Mar 30, 2015 at 4:15