Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 7

For a zero-natural-length spring, extension is the full distance between its endpoints, so the squared spring lengths are quadratic functions of the Cartesian coordinates. The Lagrangian isThe gravitational potential energy has positive derivative in the upward coordinate, giving downward forces. Applying the Euler-Lagrange equations to all four coordinates givesThe two spring forces are equal and opposite on the two masses for the connecting spring, as required by Newton's third law.

At equilibrium, the two horizontal equations imply ${\hat{X}}_2={\hat{X}}_1$ and then $k_1{\hat{X}}_1=0$, so both vanish. Adding the two vertical equations gives $k_1{\hat{Y}}_1=-(m_1+m_2)g$; the second separately gives $k_2({\hat{Y}}_2-{\hat{Y}}_1)=-m_{2}g$. HenceRetain the imposed constraint $X_2=0$ and set $X_1=x_1$, $Y_1={\hat{Y}}_1+y_1$, $Y_2={\hat{Y}}_2+y_2$. Subtracting the equilibrium equations cancels all constant gravitational terms. For $q=(x_1,y_1,y_2{)}^T$, the small-oscillation mass and stiffness matrices areThe horizontal reaction enforcing the constraint replaces the removed $X_2$ equation; it is not necessary to demand that the original unconstrained horizontal force on mass 2 vanish during motion. Because the zero-natural-length spring potentials are exactly quadratic, these displacement equations are actually exact within the ideal spring model, not merely first-order approximations.

Keep the constraint from part (b). For equal masses and spring constants, the generalized eigenvalue problem for small oscillations reduces toThe horizontal coordinate decouples, givingIn this normal mode, mass 1 oscillates horizontally, mass 2 has its horizontal coordinate fixed, and neither mass has a vertical displacement.

For the vertical block, write $\lambda =m{\omega }^2/k$. Its characteristic polynomial is $(2-\lambda )(1-\lambda )-1={\lambda }^2-3\lambda +1$. Thus ${\lambda }_{\pm }=(3\pm \sqrt{5})/2$, and the first vertical equation gives $y_2/y_1=2-{\lambda }_{\pm }$. If $\phi =(1+\sqrt{5})/2$ is the golden ratio, the two vertical modes areIn the lower-frequency mode, the masses move vertically in phase, with the lower mass having the larger amplitude. In the higher-frequency mode they move oppositely, with lower-to-upper amplitude ratio $1/\phi$. Their Euclidean inner product is zero because $1+\phi (-1/\phi )=0$; since $M=mI$, this is also mass-matrix orthogonality of normal modes.

The three frequencies satisfy ${\omega }_{-}<{\omega }_x<{\omega }_{+}$. Each mode has a scalar coordinate $A\cos \omega t+B\sin \omega t$, and their superposition gives the general constrained motion. No zero-frequency mode remains because the first spring anchors the system. Gravity fixes the equilibrium extensions but drops out of these frequencies for zero-natural-length springs.