Exercises — Normal modes — coupled oscillators, normal coordinates

The shape $(1,1)$ means both masses move the same direction and distance — the coupling spring never stretches. That is the symmetric mode, which is always the slower one. So $(1,1)\to 3\text{rad/s}$. The shape $(1,-1)$ means opposite motion — the coupling spring is maximally stretched/compressed, giving extra stiffness, so it is the antisymmetric, faster mode: $(1,-1)\to 5\text{rad/s}$. Rule of thumb: same-sign eigenvector ⇒ slow; opposite-sign ⇒ fast.

${\omega }_1^2=k/m=8/2=4\Rightarrow {\omega }_1=2\text{rad/s}$. ${\omega }_2^2=(k+2k_c)/m=(8+8)/2=8\Rightarrow {\omega }_2=2\sqrt{2}\approx 2.828\text{rad/s}$. Sanity: ${\omega }_2>{\omega }_1$ as it must, since the antisymmetric mode is stiffer.

Here $\mathbf{K}-{\omega }^2\mathbf{M}=\left(\begin{array}{cc}k+k_c-m{\omega }^2& -k_c\\ -k_c& k+k_c-m{\omega }^2\end{array}\right)=\left(\begin{array}{cc}13-{\omega }^2& -3\\ -3& 13-{\omega }^2\end{array}\right).$ Determinant zero: $(13-{\omega }^2{)}^2-9=0\Rightarrow 13-{\omega }^2=\pm 3$. Plus: $13-{\omega }^2=3\Rightarrow {\omega }^2=10$ (this is $k/m$, symmetric). Minus: $13-{\omega }^2=-3\Rightarrow {\omega }^2=16$ (this is $(k+2k_c)/m=16$, antisymmetric). So ${\omega }_1=\sqrt{10}\approx 3.162$, ${\omega }_2=4\text{rad/s}$.

${\eta }_1=\frac{1}{\sqrt{2}}(0.06+(-0.02))=\frac{0.04}{\sqrt{2}}\approx 0.02828\text{m}.$ ${\eta }_2=\frac{1}{\sqrt{2}}(0.06-(-0.02))=\frac{0.08}{\sqrt{2}}\approx 0.05657\text{m}.$ Both nonzero ⇒ this instant contains a mix of both modes. That is fine — real motion is a superposition.

(a) From the parent's worked result, $x_1(t)=a\cos \left(\frac{{\omega }_2-{\omega }_1}{2}t\right)\cos \left(\frac{{\omega }_1+{\omega }_2}{2}t\right).$ Envelope frequency ${\omega }_{\text{env}}=\frac{{\omega }_2-{\omega }_1}{2}=\frac{4-\sqrt{10}}{2}\approx 0.4189\text{rad/s}$. Carrier ${\omega }_{\text{carrier}}=\frac{{\omega }_2+{\omega }_1}{2}=\frac{4+\sqrt{10}}{2}\approx 3.5811\text{rad/s}$. (b) Energy is fully on mass 2 when the envelope $\cos ({\omega }_{\text{env}}t)$ first hits zero, i.e. ${\omega }_{\text{env}}t=\pi /2$: $T_{\text{transfer}}=\frac{\pi /2}{{\omega }_{\text{env}}}=\frac{\pi }{{\omega }_2-{\omega }_1}=\frac{\pi }{4-\sqrt{10}}\approx 3.751\text{s}.$ (The full slosh-out-and-back period is $2T_{\text{transfer}}\approx 7.50\text{s}$.)

${\omega }_1=\sqrt{k/m}$, ${\omega }_2=\sqrt{(k+2k_c)/m}=\sqrt{k/m}\sqrt{1+2k_c/k}$. For small $2k_c/k$, use $\sqrt{1+u}\approx 1+u/2$: ${\omega }_2\approx \sqrt{\frac{k}{m}}(1+\frac{k_c}{k})={\omega }_1+\sqrt{\frac{k}{m}}\cdot \frac{k_c}{k}={\omega }_1+\frac{k_c}{\sqrt{mk}}.$ Therefore ${\omega }_2-{\omega }_1\approx \frac{k_c}{\sqrt{mk}}$. Since $T_{\text{transfer}}=\pi /({\omega }_2-{\omega }_1)$, halving $k_c$ roughly doubles the transfer time — very weak springs give very lazy energy exchange. (See Beats and Superposition.)

Potential $V=\frac{1}{2}{k}_c(x_1-x_2{)}^2$, kinetic $T=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2{\dot{x}}_2^2$. $\mathbf{K}=\left(\begin{array}{cc}{k}_c& -k_c\\ -k_c& k_c\end{array}\right)=\left(\begin{array}{cc}6& -6\\ -6& 6\end{array}\right),\mathbf{M}=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right).$ $\det \left(\begin{array}{cc}6-{\omega }^2& -6\\ -6& 6-2{\omega }^2\end{array}\right)=(6-{\omega }^2)(6-2{\omega }^2)-36=0.$ Expand: $36-12{\omega }^2-6{\omega }^2+2{\omega }^4-36=2{\omega }^4-18{\omega }^2=2{\omega }^2({\omega }^2-9)=0.$ Roots: ${\omega }^2=0$ and ${\omega }^2=9\Rightarrow \omega =3\text{rad/s}$. Interpretation of $\omega =0$: with no walls, the whole system is free to translate — both masses drift together with the spring unstretched. Zero restoring force ⇒ zero frequency. This is a rigid-body / free-translation mode, not a vibration. Check the nonzero mode eigenvector at ${\omega }^2=9$: row 1 gives $(6-9)A_1-6A_2=0\Rightarrow -3A_1=6A_2\Rightarrow A_2=-\frac{1}{2}{A}_1$. The masses move oppositely with amplitude ratio $2:1$ — consistent with $m_1{A}_1+m_2{A}_2=1\cdot A_1+2\cdot (-\frac{1}{2}{A}_1)=0$, i.e. the centre of mass stays still in the vibrational mode.

$\mathbf{M}{\mathbf{A}}_1=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}2\\ -1\end{array}\right)=\left(\begin{array}{c}2\\ -2\end{array}\right).$ ${\mathbf{A}}_0^T(\mathbf{M}{\mathbf{A}}_1)=(1)(2)+(1)(-2)=0.$ ✓ Note that the plain dot product ${\mathbf{A}}_0\cdot {\mathbf{A}}_1=(1)(2)+(1)(-1)=1\ne 0$ — they are not ordinary-orthogonal. Only the $\mathbf{M}$-weighted product vanishes. This is the whole point: with unequal masses you must weight by $\mathbf{M}$.

Displacements $x_1,x_2,x_3$. Each mass feels its two neighbouring springs. Equations of motion: $m{\ddot{x}}_1=-k{x}_1-k(x_1-x_2),m{\ddot{x}}_2=-k(x_2-x_1)-k(x_2-x_3),m{\ddot{x}}_3=-k(x_3-x_2)-k{x}_3.$ Stiffness matrix (factor $k$): $\mathbf{K}=k\left(\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right),\mathbf{M}=m1.$ Let $\lambda =m{\omega }^2/k$. Secular equation $\det (\mathbf{K}-{\omega }^2\mathbf{M})=0$ becomes, with $d=2-\lambda$: $D=\det \left(\begin{array}{ccc}d& -1& 0\\ -1& d& -1\\ 0& -1& d\end{array}\right).$ Why expand along the top row? it has a zero in the corner, so one of the three cofactor terms dies and the algebra is shortest. Expanding along row 1: $D=d\cdot \det \left(\begin{array}{cc}d& -1\\ -1& d\end{array}\right)-(-1)\cdot \det \left(\begin{array}{cc}-1& -1\\ 0& d\end{array}\right)+0.$ The two $2\times 2$ determinants are $d^2-1$ and $(-1)(d)-(-1)(0)=-d$. So $D=d(d^2-1)+1\cdot (-d)=d^3-d-d=d^3-2d=0.$ Factor: $d(d^2-2)=0\Rightarrow d=0,\pm \sqrt{2}$. Then $\lambda =2-d$: ${\lambda }_1=2-\sqrt{2}\approx 0.586,{\lambda }_2=2,{\lambda }_3=2+\sqrt{2}\approx 3.414,$ i.e. ${\omega }^2=(2-\sqrt{2})\frac{k}{m},2\frac{k}{m},(2+\sqrt{2})\frac{k}{m}$. The lowest mode $(2-\sqrt{2})$ is the "all-together" gentlest sway; the highest $(2+\sqrt{2})$ is the zig-zag where neighbours fight hardest — the seed idea behind a phonon dispersion curve (see Phonons and Lattice Vibrations).

Step 1 — write the total energy in normal coordinates. With $m=1$, the kinetic energy is $T=\frac{1}{2}{\dot{x}}_1^2+\frac{1}{2}{\dot{x}}_2^2$. Substituting $x_1=\frac{1}{\sqrt{2}}({\eta }_1+{\eta }_2)$, $x_2=\frac{1}{\sqrt{2}}({\eta }_1-{\eta }_2)$ and expanding, the cross terms ${\dot{\eta }}_1{\dot{\eta }}_2$ cancel (the $\frac{1}{\sqrt{2}}$ orthonormal transform is exactly what makes this happen), leaving $T=\frac{1}{2}{\dot{\eta }}_1^2+\frac{1}{2}{\dot{\eta }}_2^2$. The same substitution in $V$ gives $V=\frac{1}{2}{\omega }_1^2{\eta }_1^2+\frac{1}{2}{\omega }_2^2{\eta }_2^2$ with no cross term. Hence $E=\underset{E_1}{\underbrace{(\frac{1}{2}{\dot{\eta }}_1^2+\frac{1}{2}{\omega }_1^2{\eta }_1^2)}}+\underset{E_2}{\underbrace{(\frac{1}{2}{\dot{\eta }}_2^2+\frac{1}{2}{\omega }_2^2{\eta }_2^2)}}.$ Step 2 — each $E_a$ is separately conserved. Because there is no cross term, $E_1$ depends only on ${\eta }_1$ and $E_2$ only on ${\eta }_2$. Each ${\eta }_a$ obeys the SHM equation ${\ddot{\eta }}_a=-{\omega }_a^2{\eta }_a$, and a single SHM conserves its own energy $\frac{1}{2}{\dot{\eta }}_a^2+\frac{1}{2}{\omega }_a^2{\eta }_a^2$. So $E_1$ and $E_2$ are each constant in time — energy sloshes between the masses but never between the modes. Step 3 — evaluate at $t=0$. From the initial data, ${\eta }_1(0)=\frac{a}{\sqrt{2}}$, ${\eta }_2(0)=\frac{a}{\sqrt{2}}$, and ${\dot{\eta }}_a(0)=0$. Then $E_1=\frac{1}{2}{\omega }_1^2(\frac{a}{\sqrt{2}}{)}^2=\frac{{\omega }_1^2{a}^2}{4}=\frac{10a^2}{4}=2.5a^2,E_2=\frac{{\omega }_2^2{a}^2}{4}=\frac{16a^2}{4}=4a^2.$ Step 4 — check the total. $E_1+E_2=2.5a^2+4a^2=6.5a^2$. Cross-check directly from the physical state at $t=0$: only the wall/coupling springs are stretched with $x_1=a,x_2=0$, giving $V(0)=\frac{1}{2}k{a}^2+\frac{1}{2}{k}_c{a}^2=\frac{1}{2}(10)a^2+\frac{1}{2}(3)a^2=6.5a^2$, and $T(0)=0$. The two agree — total energy is $6.5a^2$, constant for all time. ✓ Conclusion — the split is NOT equal. Even though the two modes start with equal amplitude (${\eta }_1(0)={\eta }_2(0)=\frac{a}{\sqrt{2}}$), their energies differ: $E_1:E_2=2.5:4=10:16={\omega }_1^2:{\omega }_2^2$. The stiffer, faster mode carries more energy for the same displacement. "Equal amplitude" is not "equal energy."

Secular determinant gives the ${\omega }^2$; eigenvectors give the mode shapes; normal coordinates decouple the motion; energy is conserved within each mode while it sloshes between the masses.