How to determine the period of the following Lissajous figure?

Question

How do I determine the period of the following Lissajous figure?

$$ x(t) =\cos(2t)-\sin(t)\\ y(t)=\cos(t-\frac{\pi}{3}) $$

Highly appreciated,

Bowser

Answer 1

The period of cosine and sine functions are $2\pi$.

The period of $\cos(2t)$ is $2\pi / 2 = \pi$ (because $2t$ runs through the interval twice as fast as plain "$t$" does).

So the period of your $y$ is $2\pi$ and the period of your $x$ is the least common multiple of $\pi$ (from cos(2t)) and $2\pi$ (from sin(2t)), which is $2\pi$.

The period of your figure is the least common multiple of these: $2\pi$.

Answer 2

If $x(A)=x(B),$

$\cos2A-\sin A=\cos2B-\sin B$

Now $\cos2A-\cos2B=-2(\sin^2A-\sin^2B)=-2(\sin A-\sin B)(\sin A+\sin B)$

$$\iff-2\{\sin A-\sin B\}\{\sin A+\sin B\}-\{\sin A-\sin B\}=0$$

$$\iff(\sin A-\sin B)\left(2\{\sin A+\sin B\}+1\right)=0$$

If $A=B+T,\sin(B+T)=\sin B,T=2n\pi$ where $n$ is any integer

Else $\dfrac14=\sin\dfrac{A+B}2\cos\dfrac{A-B}2=\sin\dfrac{2B+T}2\cos\dfrac T2$

Clearly we can't find $T$ independent of $B$ from here