Past exam of the mathematical methods II course of the University of Oxford / 2014 / 9

Take downward displacement as positive. At equilibrium, the lower spring carries $mg$ and extends by $mg/k$, while the upper spring supports both masses and extends by $2mg/k$. Write $x_1,x_2$ for the masses' downward displacements from these equilibrium positions. The extra upper-spring extension is $x_1$ and the extra lower-spring extension is $x_2-x_1$. Applying Newton's second law and cancelling the equilibrium forces givesThe gravitational terms disappear because the constant equilibrium spring forces already balance the weights. Gravity sets the equilibrium extensions, rather than the coefficients of the linear oscillation equations for ideal springs.

Set $\mathbf{x}=\mathbf{a}\cos (\omega t+\phi )$. The two normal modes satisfyThe characteristic equation is $(2-s)(1-s)-1=s^2-3s+1=0$. HenceHere $a_{\pm }$ denotes the scalar coefficient requested in the source, whereas bold $\mathbf{a}$ denotes a mode-amplitude vector. Both roots are positive.

For a mode with upper and lower displacement amplitudes $A,B$, the second equation gives $-A+(1-s)B=0$. ConsequentlyThese are signed ratios. The slow mode is in phase and the lower mass has the larger amplitude. The fast mode is out of phase and the upper mass has the larger amplitude. Taking an absolute ratio alone would discard the physically important relative phase.

Keep the same fixed laboratory equilibrium reference, and let the pivot move downward by $z(t)=z_0\cos {\omega }_{0}t$. The dynamic upper-spring extension is now $x_1-z$, soAt ${\omega }_0^2=k/m$, seek a particular solution ${\mathbf{x}}_p=\mathbf{b}\cos {\omega }_{0}t$. The amplitude equation isLet $s_{\pm }=(3\pm \sqrt{5})/2$, ${\omega }_{\pm }=\sqrt{k{s}_{\pm }/m}$ and ${\mathbf{u}}_{\pm }=(1-s_{\pm },1{)}^{\mathsf{T}}$. The unrestricted general motion isFour initial-condition constants are retained because none are specified. The driven particular response is not a resonance: ${\omega }_0$ differs from both eigenfrequencies. Its stationary upper mass is the pivot-driven two-mass spring antiresonance. Relative to the moving pivot, subtract $z(t)$ from both components.