2024-04-23 TMECH

Final Exam information

6 problems

• Non-Relativistic Newtonian Mechanics Problem 1
  • $F=MA$
  • Free body diagrams
  • Generating newtons second law
  • Rearrange and solve integrals
  • Position and velocity as function of time
  • Drag forces?
• Central Force Problem Problem 2
  • Starting with from Lagrangian
    • See homework 7 problem 3
  • Coordinate transformation to uncouple differential equations
  • Reduced mass equation !!
• Lagrangian w/ constraints Problem 3
  • Not going to be forced to do Lagrange multiplier. Can if you want.
• Moment of inertia problem Problem 4
  • Continuous distribution

$$I_{ij}=\int \rho \left({\delta }_{ij}\sum _k{x}_{\alpha k}^2-x_{\alpha i}{x}_{\alpha j}\right)dV$$

- In Cartesian coordinates

• Coupled Oscillations Problem 5
  • Finding $A_{jk}$ matrices
  • Find determinants to find $\omega$
  • Hw 10
• Continuous Systems: Waves Problem 6
  • Hw 11
  • Plucked and struck strings
  • Maybe its a struck, plucked string
  • Lots of integrals

No Relativity, No Hamiltonian, No Non-inertial reference frame, No scattering

Equations you need

$$F=MA$$

kinematic integrals Central force L = T-L Total and reduced mass Euler-Lagrange Equation Constraints and substitution Coupled oscillation $A_{jk}$ and $M_{jk}$ and determinants

3 continuous motion equations

Plucked and struck strings

So far we have assumed no frictional forces meaning the total energy for a vibrating string must remain constant.

First

$$q(x,t)=\sum _r{\eta }_r(t)\sin \left(\frac{r\pi x}{L}\right)$$ $$=\sum _r{\beta }_r{e}^{i{\omega }_{r}t}\sin \left(\frac{r\pi x}{L}\right)$$

Kinetic energy for an element of the string:

$$dT=\frac{1}{2}\rho dx{\dot{q}}^2$$

integrate our string to find total kinetic energy

$$T=\frac{1}{2}\rho {\int }_0^L{\left(\frac{dq}{dq}\right)}^{2}dx$$ $$q(x,t)=\sum _r\left({\mu }_r\cos ({\omega }_{r}t)-{\nu }_r\sin ({\omega }_{r}t)\sin \left(\frac{r\pi x}{L}\right)\right)$$ $$\frac{dq}{dt}=\sum _r(-{\omega }_r{\mu }_r\sin ({\omega }_{r}t)-{\omega }_r{\nu }_r\cos ({\omega }_{r}t))\sin \left(\frac{r\pi x}{L}\right)$$

this will be called ${\dot{\eta }}_r$

$$T=\frac{1}{2}\rho {\int }_0^L\sum _{r,s}{\dot{\eta }}_r{\dot{\eta }}_s\sin \left(\frac{r\pi x}{L}\right)\sin \left(\frac{s\pi x}{L}\right)dx$$

We can take the first few terms out and use Kronecker delta

$$=\frac{1}{2}\rho \sum {\dot{\eta }}_r^2\frac{L}{2}$$ $$=\frac{L\rho }{4}\sum _r(-{\omega }_r{\mu }_r\sin ({\omega }_{r}t)-{\omega }_r{\nu }_r\cos ({\omega }_{r}t){)}^2$$

Potential Energy

Loaded String

$$u=\frac{1}{2}\frac{\tilde{L}}{d}\sum _j(q_{j-1}{q}_j{)}^2\cdot \frac{d}{d}$$

…

$$u=\frac{\tau }{2}\sum _r\frac{r^2{\pi }^2}{L^2}{\eta }_r^2\frac{L}{2}$$

Putting kinetic and potential together, you can prove conservation of mechanical energy

$$T+u=\frac{\rho L}{4}\sum _r{\omega }_r^2|{\beta }_r^2|$$

This is constant in time, proving energy conservation

Wave equation

1D for a vibrating string

$$\frac{{\partial }^{2}q(x,t)}{\partial x^2}=\frac{\rho }{\tau }\frac{{\partial }^{2}q(x,t)}{\partial t^2}$$ $$\rho =\frac{M}{L}$$

and $\tau$ is tension

$$\sqrt{\frac{\tau }{\rho }}=V$$

Wave equation is given by

$$\frac{\rho }{\tau }\frac{{\partial }^{2}q}{\partial t^2}=\frac{{\partial }^{2}q}{\partial x^2}$$

General solutions

$$\frac{{\partial }^2\Psi (x,t)}{\partial x^2}=\frac{\rho }{\tau }\frac{{\partial }^2\Psi (x,t)}{\partial t^2}$$

Where $\Psi$ is known as the wave function

$\Psi$ traveling wave

$$\Psi =F(x-vt)+\rho (x+vt)$$

With the right and left traveling components respectively

Separation of the wave equation

$$\Psi =\varphi (x)\chi (t)$$ $$\frac{{\partial }^2\Psi }{\partial x^2}-\frac{1}{v^2}\frac{{\partial }^2\Psi }{\partial x^2}=0$$