Subfield: Mechanics


Question: A tesseract is a 4 dimensional example of cube. It can be drawn in 3 dimensions by drawing two cubes and connecting their vertices together as shown in the picture below:

![](https://cdn.mathpix.com/cropped/2023_12_21_4d38ea6c92285d4052d0g-1.jpg?height=643&width=659&top_left_y=188&top_left_x=733)

Now for the 3D equivalent. The lines connecting the vertices are replaced with ideal springs of constant $k=10 \mathrm{~N} / \mathrm{m}$ (in blue in the figure). Now, suppose the setup is placed in zero-gravity and the outer cube is fixed in place with a sidelength of $b=2 \mathrm{~m}$. The geometric center of the inner cube is placed in the geometric center of the outer cube, and the inner cube has a side-length $a=1 \mathrm{~m}$ and mass $m=1.5 \mathrm{~kg}$. The inner cube is slightly displaced from equilibrium. Consider the period of oscillations

- $T_{1}$ : when the springs have a relaxed length of 0 ;
- $T_{2}$ : when the springs are initially relaxed before the inner cube is displaced.

What is $T_{1}+T_{2}$ ?

Solution:
First let us prove that there is a net external torque of $\vec{\tau}=0$ on the cube for small displacements which means the inner cube behaves like a point mass. Consider a simple case when the cube is pushed to one side.
![](https://cdn.mathpix.com/cropped/2023_12_21_4d38ea6c92285d4052d0g-1.jpg?height=612&width=615&top_left_y=1724&top_left_x=755)

If we label the vertices of the cube from 1 to 4 clockwise, where 1 is the top left side, it is apparent that sides 1 and 2 provide a positive torque while sides 3 and 4 provide a negative torque. As the displacement is small, the angles created are small enough such that $\sin \theta \approx \theta$. As force is



proportional to the extension of the spring as $F \propto x$, we can write that

$$
\tau \propto \theta\left(r_{+}+r_{-}-r_{+}-r_{-}\right) \propto 0
$$

If torque is zero when the cube is displaced in the $x$-direction, then by symmetry, the torque is zero when the cube is displaced in the $y$-direction. Superposing both solutions implies that torque as a function of displacements in the $x$ and $y$ directions $\alpha \hat{x}+\beta \hat{y}$ is

$$
\tau(\alpha x+\beta y)=\tau(\alpha x)+\tau(\beta y)=\alpha \tau(x)+\beta \tau(y)=0
$$

1.Label the vertices of the outer cube as $1,2, \ldots, 8$ and the vectors that point to these vertices from the inner cube as $\vec{r}_{1}, \vec{r}_{2}, \ldots, \vec{r}_{2}$. Consider when the inner cube deviates from equilibrium with a vector $\vec{r}$. The force as a function of $\vec{r}$ is
$$
\begin{aligned}
F(\vec{r}) & =k\left[\left(\vec{r}_{1}-\vec{r}\right)+\left(\vec{r}_{2}-\vec{r}\right)+\ldots\left(\vec{r}_{8}-\vec{r}\right)\right] \\
& =k\left(\sum_{i=1}^{8} \vec{r}_{i}-8 \vec{r}\right) \\
& =-8 k \vec{r}
\end{aligned}
$$

This implies the period of oscillations is

$$
T_{1}=2 \pi \sqrt{\frac{m}{8 k}}
$$

2.Let the center of the inner cube be $(0,0,0)$. Consider the coordinates $(a / 2, a / 2, a / 2)$ and $(b / 2, b / 2, b / 2)$ which correspond to the vertex of the inner and larger cube respectively. Consider moving the cube in the $x$-direction. From defining $y=b / 2-a / 2$, the compressional/extension of each spring $\pm \Delta \ell$ is then
$$
\begin{aligned}
\Delta \ell & = \pm \sqrt{(x+y)^{2}+2 y^{2}}-\sqrt{3} y \\
& = \pm \sqrt{3} y \sqrt{1+\frac{2 x}{3 y}+\mathcal{O}\left(x^{2}\right)}-\sqrt{3} y \\
& \approx \pm \sqrt{3} y \frac{x}{3 y} \\
& = \pm \frac{x}{\sqrt{3}}
\end{aligned}
$$

The total energy in all springs together are then

$$
E=8 \times \frac{1}{2} k\left(\frac{x}{\sqrt{3}}\right)^{2} \Longrightarrow F=-\frac{8 k}{3} x \Longrightarrow T_{2}=2 \pi \sqrt{\frac{3 m}{8 k}}
$$

Hence, our total answer is

$$
T_{1}+T_{2}=2 \pi(1+\sqrt{3}) \sqrt{\frac{m}{8 k}}
$$

Final answer: 2.35 s