PX153 - B4 - application - first look at smh

• as an example, let's look at a mass on a spring
• the force on the mass $\overrightarrow{F}=-kx\hat{\overrightarrow{x}}$
• using PX155 - A2 - newton's second law, we can derive the equation of motion:

or,

• let ${\omega }_0=\sqrt{\frac{k}{m}}$
• now we need to solve the 2nd order differential equation:$$\frac{d^2x}{dt^2}=-\omega_0^2x$$
• try solution: $x(t)=A{e}^{\lambda t}$$$\frac{dx}{dt}=A\lambda e^{\lambda t}=\lambda x$$$$\frac{d^{2}x}{d{t}^2}=A{\lambda }^2{e}^{\lambda t}={\lambda }^{2}x$$
• from this,

- for this to be generally true, $-{\omega }_0^2={\lambda }^2⟹\lambda =\pm i{\omega }_0$
- both roots are equally valid, hence the general solution is the sum of both:
$$x(t)=Ce^{i\omega_0t}+De^{-i\omega_0t}$$

• aside: helpful to consider dimensions

• eg: what are the dimensions of ${\omega }_0$?

  • $e^{-i{\omega }_{0}t}$ so, ${\omega }_{0}t$ must be dimensionless
  • $[{\omega }_0]=s^{-1}$

• suppose the initial condition are $x(0)=x_0$ and the value of $v(0)=\frac{dx(0)}{dt}=0$

  • we then have:
  $$x(0)=C{e}^0+D{e}^0=C+D=x_0$$
  • and:
    $v(0)=\frac{dx}{dt}=i{\omega }_{0}C{e}^0-i{\omega }_{0}D{e}^0=0$
    $⟹C-D=0$
    $\therefore C=D=\frac{1}{2}{x}_0$
    $⟹x(t)=\frac{1}{2}{x}_0(e^{i{\omega }_{0}t}+e^{-i{\omega }_{0}t})$

• if instead, the initial conditions are $x(0)=0$ and $v(0)=v_0$

  $$x(t)=C{e}^{i{\omega }_{0}t}+D{e}^{-i{\omega }_{0}t}$$

  $⟹0=C{e}^0+D{e}^0⟹0=C+D$
  and $v=\frac{dx}{dt}=i{\omega }_{0}C{e}^0-i{\omega }_{0}D{e}^0=v_0$
  $⟹C-D=\frac{v_0}{i{\omega }_0}$
  $⟹2C=\frac{v_0}{i{\omega }_0}⟹C=\frac{v_0}{2i{\omega }_0}$
  $⟹D=-C=-\frac{v_0}{2i{\omega }_0}$
  giving,
  \1\n\2\n