What is the difference between $x$-component of the period and $y$-component of the period in a position vector

Question

The position vector $r(t)$ of the front of a toy train at time $t$ seconds on a closed track is given by: $$r(t) = \sin \!\left(\frac{t}{3}\right) i + \frac{1}{2}\sin\! \left(\frac{2t}{3}\right)j, \quad t\ge 0.$$ where displacement components are measured in meters.

c) Find the exact time, in seconds, that it takes the train to complete one circuit of the track.

Well my initial working was to find the period $\frac{2\pi }{b}$, but there is $x$ and $y$ components of the period. What do they each represent? What is the period for the train to complete one loop?

The question then follows up with another question:

e) Write down a definite integral, which gives the distance travelled by the train when it completes exactly one circuit of the track.

So I did:

$$\int_{t_0}^{t_1}\left|v\left(t\right)\right|\:dt.$$

where $t_2$ is the start of period and $t1$ is the end of the period. Once again what period should I use? the $x$-component of the period or $y$? How do they differentiate form each other? What information does each of them carries?

Answer 1

c) The required time $T$ is the smallest common period of $x$-coordinate of $r$ (which is $6\pi$ seconds) and $y$-coordinate of $r$ (which is $3\pi$ seconds), that is $6\pi$ seconds.

e) This situation can happen, for instance, when the train travels from $0$ to $T$ seconds. Thus the required distance is $$\int_0^T |r’(t)|dt=\int_0^T\frac 13\sqrt{\cos^2\left(\frac t3\right)+ 2\cos^2\left(\frac {2t}3\right)}dt.$$

Answer 2

This is actually a superposition of two simple harmonic motions in two directions x and y. The figures that the trajectories of these motions form are termed as Lissajous figures. However, my demonstration doesn't require any knowledge of it.

I am going to use the word vertical for the SHM along y-axis and horizontal for x-axis.

The time period of the vertical SHM is $3\pi$ seconds and that of the horizontal SHM is $6\pi$ seconds. That means the train completes two rounds in the y-direction in the same time when it completes only one round in the x-direction. (its trajectory looks like this)

So, the LCM of time periods of x and y direction,i.e. $6\pi$ seconds, is the time period of the whole motion. This was the first part of your question.

Now, to solve to solve the second part, we need to calculate the path length of the curve, which is 4 times of the length in the first quadrant.

I will leave it to the reader to figure out that the function $$y=\sin(2\sin^{-1}x)$$ is the same as the path described in the question.

The infinitesimal arc length is given by $$\left(ds\right)=\sqrt{\left(dx\right)^{2}+\left(dy\right)^{2}}$$

Hence total length of the path: $$l=4\int_{ }^{ }\sqrt{\left(dy\right)^{2}+\left(dx\right)^{2}}$$

Also, $$dy=\frac{2\cos\left(2\sin^{-1}x\right)}{\sqrt{1-x^{2}}}dx$$

Substituting the values, we have: $$l=4\int_{0}^{1}\sqrt{1+\frac{4\cos^{2}\left(2\sin^{-1}x\right)}{1-x^{2}}}dx$$

This was in terms of $x$. You can also do it in terms of $t$ as mentioned in the answer by @Alex Ravsky.

Hope this helps :)