Past exam of the mathematical methods II course of the University of Oxford / 2021 resit / 11

Use signed arc length displacements $x_1,x_2$ in the common clockwise sense shown by both arrows in the source diagram. Near the diametrically opposite equilibrium, one spring changes length by $x_2-x_1$ and the other by $x_1-x_2$. Their combined elastic potential energy isThe kinetic energy is $K=(m_1{\dot{x}}_1^2+q{m}_1{\dot{x}}_2^2)/2$. The tangential components of Newton's second law, or the derivatives $-\partial U/\partial x_i$, giveBoth springs contribute to each restoring force. The hoop radius drops out because the coordinates are arc lengths; angular coordinates $x_i/r$ give the same equations after cancelling the common factor $r^2$ in the energies.

To find the normal modes of two arc-spring-coupled beads on a hoop, take $x_i=a_i{e}^{i\Omega t}$ for a normal mode. The resulting generalized eigenvalue problem isIts determinant isThus the angular frequencies areThe ordinary nonzero frequency is $\omega /(2\pi )$. The zero mode has mode shape $(1,1)$ and represents a common rotation with no change in spring length. The oscillatory mode shape is $(1,-1/q)$.

Define a common displacement coordinate $X=(x_1+q{x}_2)/(1+q)$ and a relative displacement $\delta =x_1-x_2$. Adding and subtracting the equations of motion givesThe general real motions of the two normal modes are thereforeHere $C,D,A,B$ are independent real constants. The zero mode includes uniform drift as well as a static common displacement; writing only a constant times $\cos 0t$ would miss the independent linear-in-time solution. Their superposition isIn the oscillatory normal mode, $m_1{x}_1+m_2{x}_2=0$ and the two amplitudes have ratio $1/q$, with opposite phases. Equivalently the oscillatory motion has zero total angular momentum about the hoop center.

As $q\to \infty$ with $m_1,k$ fixed, $\omega \to \sqrt{2k/m_1}$ and the heavier particle's oscillatory amplitude tends to zero for fixed lighter-particle amplitude. The lighter particle then oscillates against two effectively anchored springs with combined spring constant $2k$. A common uniform rotation remains an allowed zero mode; the heavy particle is stationary in the limit only if the initial common drift is zero or tends to zero, as it does for bounded total angular momentum.

For $q=2$, the common coordinate has $X(0)=(x_1(0)+2x_2(0))/3=0$ and $\dot{X}(0)=\pi \omega r/20$. The relative coordinate has $\delta (0)=3\pi r/20$ and $\dot{\delta }(0)=0$. ConsequentlySince $x_1=X+2\delta /3$ and $x_2=X-\delta /3$, the full superposition isThe heavier particle's velocity isIt first vanishes when $\sin \omega t=-1$, soThis velocity is nonnegative and only touches zero: the heavier particle does not reverse its direction at that instant. Both the uniform zero mode and the oscillatory normal mode are needed to obtain this result.