Prove that this function is periodic.

Question

Let $x(t)=\sqrt{3}$ sin $\left(\pi t+\pi \right)-$ cos$\left(\pi t+\pi \right)$.

I'm asked to prove that this function represents an "harmonic oscillator" (I'm not sure if this is the correct term in English, but it's referring to a weight attached to a sling). I suppose that what I'm being asked is to prove that this function is periodic?

I tought about showing that for a given $t$ in the domain of $x$, the corresponding $x(t)$ repeats itself in such way that $t + kn = x(t)$ -- $k$ being a constant and $n \in \mathbb{Z}$.

Is this the most appropriate way to prove this?

Answer 1

Observe that

$$\begin{cases}\sin(\pi t+\pi)=\sin \pi t\cos\pi+\sin\pi\cos\pi t=-\sin\pi t\\{}\\ \cos(\pi t+\pi)=\cos\pi t\cos \pi-\sin\pi t\sin\pi=-\cos\pi t\end{cases}$$

Thus your function is in fact

$$f(t)=-(\sqrt 3\,\sin\pi t-\cos\pi t) $$

and this clearly has a period of $\;2\;$:

$$f(t+2)=-\left(\sqrt3\,\sin(\pi t+2\pi)-\cos(\pi t+2\pi)\right)=-\left(\sqrt3\,\cos\pi t-\cos\pi t)\right)=f(t)$$

Answer 2

Note that given a sinusoid of the form

$$A\cos{(2\pi f t + \phi)}$$

its period is $$T = \frac{1}{f}$$

Given this, note that the period of each of your sinusoids are equal, meaning that their sum is also periodic.

Answer 3

Sinusoidal functions of the form $\cos(2\pi f t+\phi)$ have fundamental period of $T_0=1/f$. The first sinusoidal repeat itself in $T_1=2$ seconds, and the second one also repeats itself in $T_2$ seconds. So, the overall expression, repeats itself in $T_0=2$ seconds again. So we have $x(t)=x(t+2)$ for all $t$.

Answer 4

$$x(t)=2\sin\left(\pi t+\pi-\dfrac\pi6\right)$$

Now if $x(t)=x(T+t)$ where $T>0$

$$\sin\left(\pi(t+T)+\pi-\dfrac\pi6\right)=\sin\left(\pi t+\pi-\dfrac\pi6\right)$$

$\implies\pi(t+T)+\pi-\dfrac\pi6=n\pi+(-1)^n\left(\pi t+\pi-\dfrac\pi6\right)$ where $n$ is any integer

$\iff(t+T)+1-\dfrac16=n+(-1)^n\left(t+1-\dfrac16\right)$

If $n$ is even $=2m$(say)

$(t+T)+1-\dfrac16=2m+\left(t+1-\dfrac16\right)$

$\iff T=2m$

Clearly, the smallest positive value of $t$ is $2$

Observe that if $n$ is odd $=2m+1$(say), $T$ is not constant, hence can not be period.