2024-04-12 Vibrations

… missed first 30 min

Longiudinal vs Transverse vibration of a linear molecule:

diagrams here

These are independent and can be solved separately

• 3 variables but only 2 longitudinal modes

figure here

We can eliminate translation by requiring the center-of-mass to be stationary during vibration

$$m{x}_1+M{x}_2+m{x}_3=0$$ $$x_2=-\frac{m}{M}(x_1+x_3)$$ $$T=\frac{1}{2}m{\dot{x}}_1^2+\frac{1}{2}M{\dot{x}}_2^2+\frac{1}{2}m{\dot{x}}_3^2$$ $$\cdots =\frac{1}{2}m({\dot{x}}_1^2+{\dot{x}}_3^2)+\frac{1}{2}\frac{m^2}{M}({\dot{x}}_1^2+{\dot{x}}_3^2+2{\dot{x}}_1{\dot{x}}_3)$$

We can decouple the system by switching coordinates. We must eliminate the ${\dot{x}}_1{\dot{x}}_3$ term.

Generalized coordinates:

$$q_1=x_1+x_3$$ $$q_2=x_3-x_1$$ $$x_3=\frac{1}{2}(q_1+q_2),x_1=\frac{1}{2}(q_1-q_2),x_2=-\frac{m}{M}{q}_1$$

Find the kinetic energy of this system

$$T=\frac{1}{2}m{\dot{x}}_1^2+\frac{1}{2}M{\dot{x}}_2^2+\frac{1}{2}m{\dot{x}}_3^2$$ $$=\frac{1}{2}m\frac{1}{4}({\dot{q}}_1-{\dot{q}}_2{)}^2+\frac{1}{2}M\frac{m^2}{M^2}{\dot{q}}_1^2+\frac{1}{2}m\frac{1}{4}({\dot{q}}_1+{\dot{q}}_2{)}^2$$ $$=\frac{1}{2}m\frac{1}{4}({\dot{q}}_1^2+{\dot{q}}_2^2-\cancel{2{\dot{q}}_1{\dot{q}}_2})+\frac{1}{2}\frac{m^2}{M}{\dot{q}}_1^2+\frac{1}{2}m\frac{1}{4}({\dot{q}}_1^2++{\dot{q}}_2^2+\cancel{2{\dot{q}}_1{\dot{q}}_2})$$

Our coupling term cancels out. Kinetic energy finally given by

$$T=\frac{1}{4}m{\dot{q}}_2^2+\frac{mM+2m^2}{4M}{\dot{q}}_1^2$$

Finding potential:

$$u=\frac{1}{2}{\kappa }_1(x_2-x_1{)}^2+\frac{1}{2}{\kappa }_1(x_3-x_2{)}^2$$ $$=\frac{1}{2}{\kappa }_1(x_2^2+x_1^2-2x_1{x}_2)+\frac{1}{2}{\kappa }_1(x_3^2+x_2^2-2x_2{x}_3)$$ $$=\frac{1}{2}{\kappa }_1\left(\frac{m^2}{M^2}{q}_1^2+\frac{1}{4}(q_1^2+q_2^2-q_1{q}_2)+2\left(\frac{1}{2}\right)(q_1-q_2)\frac{m}{M}{q}_1\right)\dots$$ $$+\frac{1}{2}{\kappa }_1\left(\frac{1}{4}(q_1^2+q_2^2+2q_1{q}_2)+\frac{m^2}{M^2}{q}_1^{2}2\frac{1}{2}(q_1+q_2)\frac{m}{M}{q}_1\right)$$ $$={\kappa }_1\frac{m^2}{M^2}{q}_1^2+\frac{1}{4}{\kappa }_1(q_1^2+q_2^2)+{\kappa }_1{q}_1^2\frac{m}{M}$$ $$\frac{1}{4}{\kappa }_1{q}_2^2+{\kappa }_1{q}_1^2\left[\frac{4m^2}{4m^2}+\frac{m^2}{4m^2}+\frac{4mM}{4m^2}\right]$$ $$=\frac{1}{4}{\kappa }_1{q}_2^2+{\kappa }_1{q}_1^2\left[\frac{4m^2+4mM+M^2}{4m^2}\right]$$ $$u={\left(\frac{2m+M}{2M}\right)}^2{\kappa }_1{q}_1^2+\frac{1}{4}{\kappa }_1{q}_2^2$$ $$\sum _k(A_{jk}-{\omega }^2{m}_{jk})q_k=0$$ $$A_{jk}=\frac{{\partial }^{2}u}{\partial q_j\partial q_k}$$ $$q_k(t)=q_k{e}^{i(\omega t-\delta )}$$ $$A_{11}=\frac{{\partial }^{2}u}{\partial q_1^2}=2\left(\frac{2m+M}{2m}\right){\kappa }_1$$ $$A_{22}=\frac{{\partial }^{2}u}{\partial q_2^2}=\frac{1}{2}{\kappa }_1$$ $$A_{12}=A_{21}=0$$ $$T=\frac{1}{2}\sum _{jk}{m}_{jk}{\dot{q}}_j{\dot{q}}_k$$ $$\frac{1}{2}{m}_{11}{\dot{q}}_1^2=\frac{mM+2m^2}{4m}{\dot{q}}_1^2$$ $$m_{11}=\frac{mM+2m^2}{2m}$$ $$\frac{1}{2}{m}_{22}{\dot{q}}_2^2=\frac{1}{4}m{\dot{q}}_2^2$$ $$m_{22}=\frac{m}{2}$$ $$m_{12}=m_{21}=0$$ $$A_{jk}=\left[\begin{array}{cc}2{\left(\frac{2m+M}{2m}\right)}^2{\kappa }_1& 0\\ 0& \frac{1}{2}{\kappa }_1\end{array}\right]$$ $$m_{jk}=\left[\begin{array}{cc}\frac{mM+2m^2}{2m}& 0\\ 0& \frac{m}{2}\end{array}\right]$$

We have $A_{jk}$ and $m_{jk}$ matrices.

$$\det \left[\begin{array}{cc}\frac{1}{2}\left(\frac{2m+M}{m}\right){\kappa }_1-{\omega }^2\frac{mM+2m^2}{2m}& 0\\ 0& \frac{1}{2}{\kappa }_1-\frac{{\omega }^{2}m}{2}\end{array}\right]=0$$ $${\omega }_1^2=\frac{1}{2}{\left(\frac{2m+M}{m}\right)}^2{\kappa }_1\frac{2M}{m(2m+M)}$$ $${\omega }_1^2=\frac{2m+M}{mM}{\kappa }_1$$ $${\omega }_2^2=\frac{{\kappa }_1}{m}$$

Tensor is diagonalized which means that $q_1$ and $q_2$ are my normal coordinates.

$$q_1=a_{11}{\eta }_1+a_1{\eta }_2$$ $$q_2=a_{21}{\eta }_1+a_{22}{\eta }_2$$ $$\left[\begin{array}{cc}{A}_{11}-{\omega }_1^2{m}_{11}& 0\\ 0& A_{22}-{\omega }_1^2{m}_{22}\end{array}\right]\left[\begin{array}{c}{a}_{11}\\ a_{22}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$ $$0a_{11}+a_{21}=0$$ $$0a_{11}+[]a_{21}=0$$ $$a_{21}=0$$ $$a_{11}=\text{free var.}$$ $$\left[\begin{array}{c}1\\ 0\end{array}\right]$$

Transverse Vibrations:

figure here

$$m(y_1+y_3)+M{y}_2=0$$ $$y_2=-\frac{m}{M}(y_1+y_3)$$

We will assume small oscillations, meaning that $\alpha$ is a small angle therefore

$$\alpha =\frac{\Delta y}{b}=\frac{(y_1-y_2)+(y_3-y_2)}{b}$$ $$y_1=y_3$$

and substitute $y_2$

$$\alpha =\frac{2y_1}{bM}(2m+M)$$ $$T=\frac{1}{2}m({\dot{y}}_1^2+{\dot{y}}_3^2)+\frac{1}{2}M{y}_2^2$$ $$T=\frac{m}{M}(2m+M){\dot{y}}_1^2$$ $$T=\frac{mM{b}^2}{4(2m+M)}{\dot{\alpha }}^2$$ $$u=\frac{1}{2}{\kappa }_2(b\alpha {)}^2$$ $${\omega }_3^2=\frac{2(2m+M)}{mM}{\kappa }_2$$

From the first and third normal modes EM radiation is detected because the electrical dipole moving. 2nd no radiation b/c no dipole

The Loaded Spring (12.9)

• n masses separated by distance d in equilibrium
• masses m are connected by springs or elastic strings
• Total length of string $L=(n+1)d$ and fixed at both ends

figure here

Want to treat small transverse oscillations about equlibrium

consider particles:

$$\begin{array}{ccc}j-1,& j,& j+1\end{array}$$

displaced vertically

The force on the $j^{th}$ particle (considering only nearest neighbors)

graph here