\magnification=\magstep2

$$
\vec{F} = - k{\vec{x}}
$$

$$
m {{d^2\vec{x}}\over{dt^2}} = - k{\vec{x}}
$$

$$
m {{d^2\vec{x}}\over{dt^2}} + k{\vec{x}} = 0
$$

$$
\vec{x}(t) = \vec{A} \cos ( \omega t + \phi ) 
$$

$$
\omega = \sqrt{ {k}\over{m} }
$$

$$
\vec{F}_{\rm resist} = - b \vec{v} 
$$

$$
\vec{F} = - k{\vec{x}} - b \vec{v}
$$

$$
m {{d^2\vec{x}}\over{dt^2}} = - k{\vec{x}} - b {{d\vec{x}} \over {dt}}
$$

$$
m {{d^2\vec{x}}\over{dt^2}} + b {{d\vec{x}} \over {dt}} + k{\vec{x}} = 0
$$

$$
x(t) = A \cos ( \omega t + \phi ) e^{- t / \tau }  \qquad {\rm ???}
$$


$$
{{d x}\over{ d t }}  =  
        - {{1}\over{\tau}} A \cos ( \omega t + \phi ) e^{- t / \tau }
        - {\omega } A \sin ( \omega t + \phi ) e^{-t / \tau }
$$

$$\eqalign{
{{d^2 x}\over{ d t^2 }}  = &  
       \phantom{+\ } {{1}\over{\tau^2}} A \cos ( \omega t + \phi ) e^{- t / \tau } \cr
    & + {{\omega}\over{\tau} } A \sin ( \omega t + \phi ) e^{-t / \tau } \cr
    & + {{\omega}\over{\tau} } A \sin ( \omega t + \phi ) e^{-t / \tau } \cr
    & - {\omega^2} A \cos ( \omega t + \phi ) e^{- t / \tau } \cr
}$$


\vfill
\eject

$$
{m \over {\tau^2}} - { m \omega^2} - { {b}\over{\tau}} + k  =  0
$$

$$
{ {2 m \omega}\over{\tau}} - {b \omega} = 0
$$

$$
\tau = {{2 m} \over {b}}
$$

$$
\omega = \sqrt{ {{k}\over{m}} - {{b^2}\over{4 m^2}} }
$$

$$
{{k}\over{m}} - {{b^2}\over{4 m^2}} = 0
$$


$$
{F} = F_d \cos ( {\omega}_{d} t )- k x - b v
$$

$$
m {{d^2 x}\over{dt^2}} - F_d \cos ( {\omega}_d t ) + b {{dx} \over {dt}} + k{x} = 0
$$

$$
x(t) = A \cos ( {\omega}_d t + \phi)  
$$

$$
A = {    { F_d }   \over
        {  m  \sqrt {\left[ 
                 { ( {\omega_d^2 - {\omega}_0^2 )}^2
                    + { ( { {  b \omega_d  } \over { m } } ) }^2 ) } 
               \right] } }  
       }
$$

$$
\omega_0 = \sqrt { { k} \over {m} } 
$$

$$
U(x)  =  5 x^2      \qquad   {\rm \ Joules}
$$

$$
U(x)  =  3 x^3      \qquad   {\rm \ Joules}
$$

$$
U(x)  =  2 x^4      \qquad   {\rm \ Joules}
$$



\bye
