Exercises — Normal modes — coupled oscillators, normal coordinates

Shape $(1,1)$ ka matlab hai ki dono masses same direction aur distance mein move karte hain — coupling spring kabhi stretch nahi hota. Yeh symmetric mode hai, jo hamesha slow hota hai. Toh $(1,1)\to 3\text{rad/s}$. Shape $(1,-1)$ ka matlab hai opposite motion — coupling spring maximally stretch/compress hota hai, jo extra stiffness deta hai, isliye yeh antisymmetric, faster mode hai: $(1,-1)\to 5\text{rad/s}$. 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 check: ${\omega }_2>{\omega }_1$ jaisa hona chahiye, kyunki antisymmetric mode stiffer hota hai.

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

${\eta }_1=\frac{1}{\sqrt{2}}(0.06+(-0.02))=\frac{0.04}{\sqrt{2}}\approx 0.02828\text{m}.$ ${\eta }_2=\frac{1}{\sqrt{2}}(0.06-(-0.02))=\frac{0.08}{\sqrt{2}}\approx 0.05657\text{m}.$ Dono nonzero hain ⇒ is instant mein dono modes ka mix hai. Yeh bilkul theek hai — real motion ek superposition hoti hai.

(a) Parent ke worked result se, $x_1(t)=a\cos \left(\frac{{\omega }_2-{\omega }_1}{2}t\right)\cos \left(\frac{{\omega }_1+{\omega }_2}{2}t\right).$ 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 fully mass 2 par tab hoti hai jab envelope $\cos ({\omega }_{\text{env}}t)$ pehli baar zero ho, yaani ${\omega }_{\text{env}}t=\pi /2$: $T_{\text{transfer}}=\frac{\pi /2}{{\omega }_{\text{env}}}=\frac{\pi }{{\omega }_2-{\omega }_1}=\frac{\pi }{4-\sqrt{10}}\approx 3.751\text{s}.$ (Poora slosh-out-and-back period hai $2T_{\text{transfer}}\approx 7.50\text{s}$.)

${\omega }_1=\sqrt{k/m}$, ${\omega }_2=\sqrt{(k+2k_c)/m}=\sqrt{k/m}\sqrt{1+2k_c/k}$. Small $2k_c/k$ ke liye, $\sqrt{1+u}\approx 1+u/2$ use karo: ${\omega }_2\approx \sqrt{\frac{k}{m}}(1+\frac{k_c}{k})={\omega }_1+\sqrt{\frac{k}{m}}\cdot \frac{k_c}{k}={\omega }_1+\frac{k_c}{\sqrt{mk}}.$ Isliye ${\omega }_2-{\omega }_1\approx \frac{k_c}{\sqrt{mk}}$. Kyunki $T_{\text{transfer}}=\pi /({\omega }_2-{\omega }_1)$, $k_c$ ko half karne se transfer time roughly double ho jaata hai — bahut weak springs bahut lazy energy exchange dete hain. (Dekho 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 karo: $36-12{\omega }^2-6{\omega }^2+2{\omega }^4-36=2{\omega }^4-18{\omega }^2=2{\omega }^2({\omega }^2-9)=0.$ Roots: ${\omega }^2=0$ aur ${\omega }^2=9\Rightarrow \omega =3\text{rad/s}$. $\omega =0$ ki interpretation: koi walls nahi hone se, poora system translate karne ke liye free hai — dono masses spring unstretched ke saath saath drift karte hain. Zero restoring force ⇒ zero frequency. Yeh ek rigid-body / free-translation mode hai, vibration nahi. Nonzero mode eigenvector ${\omega }^2=9$ par check karo: row 1 deta hai $(6-9)A_1-6A_2=0\Rightarrow -3A_1=6A_2\Rightarrow A_2=-\frac{1}{2}{A}_1$. Masses opposite direction mein amplitude ratio $2:1$ ke saath move karte hain — consistent with $m_1{A}_1+m_2{A}_2=1\cdot A_1+2\cdot (-\frac{1}{2}{A}_1)=0$, yaani vibrational mode mein centre of mass still rehta hai.

$\mathbf{M}{\mathbf{A}}_1=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}2\\ -1\end{array}\right)=\left(\begin{array}{c}2\\ -2\end{array}\right).$ ${\mathbf{A}}_0^T(\mathbf{M}{\mathbf{A}}_1)=(1)(2)+(1)(-2)=0.$ ✓ Note karo ki plain dot product ${\mathbf{A}}_0\cdot {\mathbf{A}}_1=(1)(2)+(1)(-1)=1\ne 0$ — woh ordinary-orthogonal nahi hain. Sirf $\mathbf{M}$-weighted product zero hota hai. Yahi toh poori baat hai: unequal masses ke saath tumhe $\mathbf{M}$ se weight karna padega.

Displacements $x_1,x_2,x_3$. Har mass apne do neighbouring springs feel karta hai. 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 $k$): $\mathbf{K}=k\left(\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right),\mathbf{M}=m1.$ $\lambda =m{\omega }^2/k$ lo. Secular equation $\det (\mathbf{K}-{\omega }^2\mathbf{M})=0$ ban jaata hai, $d=2-\lambda$ ke saath: $D=\det \left(\begin{array}{ccc}d& -1& 0\\ -1& d& -1\\ 0& -1& d\end{array}\right).$ Top row ke along expand kyun karein? Kyunki corner mein ek zero hai, toh teen cofactor terms mein se ek mar jaata hai aur algebra sabse chhoti hoti hai. Row 1 ke along expand karte hain: $D=d\cdot \det \left(\begin{array}{cc}d& -1\\ -1& d\end{array}\right)-(-1)\cdot \det \left(\begin{array}{cc}-1& -1\\ 0& d\end{array}\right)+0.$ Do $2\times 2$ determinants hain $d^2-1$ aur $(-1)(d)-(-1)(0)=-d$. Toh $D=d(d^2-1)+1\cdot (-d)=d^3-d-d=d^3-2d=0.$ Factor karo: $d(d^2-2)=0\Rightarrow d=0,\pm \sqrt{2}$. Phir $\lambda =2-d$: ${\lambda }_1=2-\sqrt{2}\approx 0.586,{\lambda }_2=2,{\lambda }_3=2+\sqrt{2}\approx 3.414,$ yaani ${\omega }^2=(2-\sqrt{2})\frac{k}{m},2\frac{k}{m},(2+\sqrt{2})\frac{k}{m}$. Sabse neeche wala mode $(2-\sqrt{2})$ hai "all-together" gentlest sway; sabse upar wala $(2+\sqrt{2})$ hai zig-zag jahan neighbours sabse zyada fight karte hain — yeh phonon dispersion curve ke peeche ka seed idea hai (dekho Phonons and Lattice Vibrations).

Step 1 — total energy normal coordinates mein likho. $m=1$ ke saath, kinetic energy hai $T=\frac{1}{2}{\dot{x}}_1^2+\frac{1}{2}{\dot{x}}_2^2$. $x_1=\frac{1}{\sqrt{2}}({\eta }_1+{\eta }_2)$, $x_2=\frac{1}{\sqrt{2}}({\eta }_1-{\eta }_2)$ substitute karke aur expand karne par, cross terms ${\dot{\eta }}_1{\dot{\eta }}_2$ cancel ho jaate hain (woh $\frac{1}{\sqrt{2}}$ orthonormal transform exactly yahi karne ke liye hai), bacha rehta hai $T=\frac{1}{2}{\dot{\eta }}_1^2+\frac{1}{2}{\dot{\eta }}_2^2$. Wahi substitution $V$ mein karne se milta hai $V=\frac{1}{2}{\omega }_1^2{\eta }_1^2+\frac{1}{2}{\omega }_2^2{\eta }_2^2$ bina kisi cross term ke. Isliye $E=\underset{E_1}{\underbrace{(\frac{1}{2}{\dot{\eta }}_1^2+\frac{1}{2}{\omega }_1^2{\eta }_1^2)}}+\underset{E_2}{\underbrace{(\frac{1}{2}{\dot{\eta }}_2^2+\frac{1}{2}{\omega }_2^2{\eta }_2^2)}}.$ Step 2 — har $E_a$ separately conserved hai. Kyunki koi cross term nahi hai, $E_1$ sirf ${\eta }_1$ par depend karta hai aur $E_2$ sirf ${\eta }_2$ par. Har ${\eta }_a$ SHM equation ${\ddot{\eta }}_a=-{\omega }_a^2{\eta }_a$ follow karta hai, aur ek single SHM apni energy $\frac{1}{2}{\dot{\eta }}_a^2+\frac{1}{2}{\omega }_a^2{\eta }_a^2$ conserve karta hai. Toh $E_1$ aur $E_2$ dono time mein constant hain — energy masses ke beech slosh karti hai lekin modes ke beech kabhi nahi. Step 3 — $t=0$ par evaluate karo. Initial data se, ${\eta }_1(0)=\frac{a}{\sqrt{2}}$, ${\eta }_2(0)=\frac{a}{\sqrt{2}}$, aur ${\dot{\eta }}_a(0)=0$. Toh $E_1=\frac{1}{2}{\omega }_1^2(\frac{a}{\sqrt{2}}{)}^2=\frac{{\omega }_1^2{a}^2}{4}=\frac{10a^2}{4}=2.5a^2,E_2=\frac{{\omega }_2^2{a}^2}{4}=\frac{16a^2}{4}=4a^2.$ Step 4 — total check karo. $E_1+E_2=2.5a^2+4a^2=6.5a^2$. Directly physical state se cross-check karo $t=0$ par: sirf wall/coupling springs stretch hain $x_1=a,x_2=0$ ke saath, jo deta hai $V(0)=\frac{1}{2}k{a}^2+\frac{1}{2}{k}_c{a}^2=\frac{1}{2}(10)a^2+\frac{1}{2}(3)a^2=6.5a^2$, aur $T(0)=0$. Dono agree karte hain — total energy hai $6.5a^2$, har time ke liye constant. ✓ Conclusion — split EQUAL nahi hai. Haalaanki dono modes equal amplitude se start karte hain (${\eta }_1(0)={\eta }_2(0)=\frac{a}{\sqrt{2}}$), unki energies alag hain: $E_1:E_2=2.5:4=10:16={\omega }_1^2:{\omega }_2^2$. Stiffer, faster mode same displacement ke liye zyada energy carry karta hai. "Equal amplitude" ka matlab "equal energy" nahi hai.

Secular determinant se ${\omega }^2$ milte hain; eigenvectors se mode shapes milte hain; normal coordinates motion ko decouple karte hain; energy har mode ke andar conserved hai jabki woh masses ke beech slosh karti hai.