I need to calculate the period, I need help to verify my answer

Question

I need to determine if $x(t) = 9\cos(2t) + 4\sin(\pi t)$ is periodic. If it is periodic I need to find the period. this what I have done \begin{align*} T_0 &= 2\pi/w\\ T_1 &= 2\pi/2 = \pi \\ T_2 &= 2\pi/\pi = 2\\ T &= T_1/T_2\\ T &= \pi/2 \end{align*} I am a little confused here, can someone check and tell me if I have the right answer, thanks

Answer 1

$\cos(2t)$ is periodic with period $\pi$; i.e. it repeats after $\pi, 2\pi, 3\pi$, etc. $\sin(\pi t)$ is is periodic with period $2$; i.e. it repeats after $2,4,6$, etc.

The period of the combined function will be the least common integer multiple of $\pi$ and $2$, the smallest number appearing in both lists above. Unfortunately, there is no such common multiple.

Contrast this with the case of $\cos(2t)$ and $\sin(3t)$. The latter repeats after $2\pi/3, 4\pi/3, 6\pi/3,\ldots$. Note that $2\pi$ appears on both lists, so is the minimum period of all nontrivial linear combinations of these two functions.

Answer 2

Your $x(t)$ is not periodic. If we assume that $T$ is a period of $x(t)$, then it is also a period of $x''(t)=-36\cos(2t)-4\pi^2\sin(\pi t)$ and also of $4x(t)+x''(t)=(16-4\pi^2)\sin(\pi t)$ and of $\pi^2x(t)+x''(t)=(9\pi^2 -36)\cos(2t)$. Since $\pi^2$ is irrational, the coefficients $16-4\pi^2$ and $9\pi^2-36$ are nonzero, and this would imply that $T$ is both a multiple of $2$ (the periods of $\sin(\pi t)$) and of $\pi$ (the periods of $\cos(2t)$). Again, since $\pi$ is irrational, there is no common multiple of $\pi$ and $2$ (apart from $0$).