# Physics Cup – TalTech 2021 – Pr. 3, Hint 1.

The first hint will be a long piece of theory which you can find, in principle, in textbooks, but it might be hard to collect all the required pieces, so let us present these pieces here.

Small oscillations of a system of bodies are described by a set of linear differential equations of second order. For instance, if three bodies of masses $m_1$, $m_2$, and $m_3$, respectively, are constrained to move along $x$-axis while connected with springs of stiffness $k$, we have $\ddot x_1=-k(x_1-x_2)$, $\ddot x_2=-k(2x_2-x_1-x_3)$, and $\ddot x_3=-k(x_3-x_2)$, where $x_i$ denotes the displacement of the $i$-th mass. We have no nonlinear terms because, by small amplitudes, we can neglect all quadratic and higher terms in the Taylor expansion. Such equations can be written in matrix form (which isn’t really needed to solve this problem) as $\ddot x_i=a_{ij}x_j$, where we have assumed Einstein’s summation convention and assume summation over repeated indices. For this example of three masses, assuming the masses to be all equal to $m$, we have $a_{11}=a_{33}= - k/m$, $a_{22}= - 2k/m$, $a_{12}=a_{21}= a_{23}=a_{32}= k/m$, and $a_{13}=a_{31}= 0$. Note that the matrix here is symmetric, $a_{ij}=a_{ji}$; this is a property which you may think about as following from Newton’s third law, combined with the fact that all the masses are equal. For the example above, the functions $x_{i}=x_{i}(t)$ are just the Euclidian coordinates of the point masses, but for more complicated situations, these are generalized coordinates the total number of which needs to be equal to the number of degrees of freedom. The number of degrees of freedom is the smallest number of scalar parameters needed to describe fully a dynamical system. For instance, consider three point masses connected with two rigid rods, with a hinged connection at the middle joint, and with everything constrained into a horizontal plane. In order to describe the position of each of the point masses, we would need to use their $x$ and $y$ coordinates, so that three point masses would come with six coordinates in total. However, we have two scalar constraints – the distances between neighboring points, which will reduce the number of required generalized coordinates down by two. As a result, we need four generalized coordinates; for instance, these coordinates can be $x_1=\xi$ and $x_2=\eta$ – the coordinates of the first mass, together with $x_3=\alpha$ and $x_4=\beta$ – the angles of the rods. However, we can always go to a reference frame where the center of mass of the whole system is at rest, at the origin, in which case the number of coordinates is further decreased by two (we won’t need $\xi$ and $\eta$ anymore). The key to an easy solution is a convenient choice of generalized coordinates. In some cases, you may be able to find convenient coordinates, but some scalar constraints remain still unused (i.e. your number of generalized coordinates $N$ is still greater than the number of degrees of freedom $n$). In that case, you may just consider a subspace in the $n$-dimensional space defined by your generalized coordinates.

Next, we introduce the concept of natural modes; these are the modes where all the coordinates oscillate with the same frequency. The general theory of coupled oscillators tells us that arbitrary motion of the system is a linear combination (superposition) of the natural modes. For instance, in the case of the example with three equal masses connected with springs, one of the obvious modes is when the central point mass is at rest, and the other two oscillate in opposite phase: $x_1=-x_3=x_0\cos(\omega_1 t+\varphi)$, $x_2=0$. Due to symmetry, it is clear that $x_1$ and $x_3$ remain moving symmetrically while there is no net force exerted onto the mass in the middle. Now we can easily conclude from Newton’s second law for one of the masses that $\omega_1=\sqrt {k/m}$. This motion is described by the eigenvector $\mathbf X=(1,0,-1)$ (i.e. $X_1=-X_3=1$, $X_2=0$); a natural mode can be conveniently expressed in terms of the eigenvector as $\mathbf x=\mathbf Xx_0\cos(\omega_1 t+\varphi)$. There is also the trivial mode when all the masses move together with constant speed: with eigenvector $\mathbf Y=(1,1,1)$, $\mathbf x=\mathbf Yv_0t$. This can be interpreted as a motion with an infinitely long period, i.e. with $\omega_2=0$.

We are missing one more mode which can be easily found here if we know a useful fact: for a symmetric matrix $a_{ij}=a_{ji}$, the eigenvectors corresponding to different natural frequencies are perpendicular to each other, i.e. we need to find such a vector $Z_{i}$ that $\sum_iX_iZ_{i}=0$ and $\sum_iY_iZ_{i}=0$. It is easy to see that these equalities are satisfied with $\mathbf Z=(1,-2,1)$ (keep in mind that our space of eigenvectors is three-dimensional – because we have three degrees of freedom; therefore, all the vectors which are perpendicular to the plane defined by the vectors $\mathbf X$ and $\mathbf Y$must be parallel to $\mathbf Z$). So, our missing mode is in the form $\mathbf x=\mathbf Zz_0\cos(\omega_3t+\phi)$; let us express the average kinetic and potential energies for such a mode, which must be equal to each other for sinusoidal oscillations, as $\left=\frac 14m\omega_3^2z_0^2\mathbf Z^2=\frac 32m\omega_3^2z_0^2$; $\left=\frac 12k[z_0(1+2)]^2=\frac 92kz_0^2$, hence $\omega_3^2=3k/m$.

Please submit the solution of this problem via e-mail to physcs.cup@gmail.com.

For full regulations, see the “Participate” tab.