The Position Equation [ edit ]
This section shows how to form the equation describing the position of a mass on a spring.

For a simple oscillator consisting of a mass m attached to one end of a spring with a spring constant s , the restoring force, f , can be expressed by the equation

$f=-sx\,$
where x is the displacement of the mass from its rest position. Substituting the expression for f into the linear momentum equation,

$f=ma=m{d^{2}x \over dt^{2}}\,$
where a is the acceleration of the mass, we can get

$m{\frac {d^{2}x}{dt^{2}}}=-sx$
or,

${\frac {d^{2}x}{dt^{2}}}+{\frac {s}{m}}x=0$
Note that the frequency of oscillation $\omega _{0}$ is given by

$\omega _{0}^{2}={s \over m}\,$
To solve the equation, we can assume

$x(t)=Ae^{\lambda t}\,$
The force equation then becomes

$(\lambda ^{2}+\omega _{0}^{2})Ae^{\lambda t}=0,$
Giving the equation

$\lambda ^{2}+\omega _{0}^{2}=0,$
Solving for $\lambda$

$\lambda =\pm j\omega _{0}\,$
This gives the equation of x to be

$x=C_{1}e^{j\omega _{0}t}+C_{2}e^{-j\omega _{0}t}\,$
Note that

$j=(-1)^{1/2}\,$
and that C_{1} and C_{2} are constants given by the initial conditions of the system

If the position of the mass at t = 0 is denoted as x_{0} , then

$C_{1}+C_{2}=x_{0}\,$
and if the velocity of the mass at t = 0 is denoted as u_{0} , then

$-j(u_{0}/\omega _{0})=C_{1}-C_{2}\,$
Solving the two boundary condition equations gives

$C_{1}={\frac {1}{2}}(x_{0}-j(u_{0}/\omega _{0}))$

$C_{2}={\frac {1}{2}}(x_{0}+j(u_{0}/\omega _{0}))$

The position is then given by

$x(t)=x_{0}cos(\omega _{0}t)+(u_{0}/\omega _{0})sin(\omega _{0}t)\,$

This equation can also be found by assuming that x is of the form

$x(t)=A_{1}cos(\omega _{0}t)+A_{2}sin(\omega _{0}t)\,$
And by applying the same initial conditions,

$A_{1}=x_{0}\,$

$A_{2}={\frac {u_{0}}{\omega _{0}}}\,$

This gives rise to the same position equation

$x(t)=x_{0}cos(\omega _{0}t)+(u_{0}/\omega _{0})sin(\omega _{0}t)\,$
Back to Main page

Alternate Position Equation Forms [ edit ]
If A_{1} and A_{2} are of the form

$A_{1}=Acos(\phi )\,$
$A_{2}=Asin(\phi )\,$

Then the position equation can be written

$x(t)=Acos(\omega _{0}t-\phi )\,$

By applying the initial conditions (x(0)=x_{0} , u(0)=u_{0} ) it is found that

$x_{0}=Acos(\phi )\,$

${\frac {u_{0}}{\omega _{0}}}=Asin(\phi )\,$

If these two equations are squared and summed, then it is found that

$A={\sqrt {x_{0}^{2}+({\frac {u_{0}}{\omega _{0}}})^{2}}}\,$

And if the difference of the same two equations is found, the result is that

$\phi =tan^{-1}({\frac {u_{0}}{x_{0}\omega _{0}}})\,$
The position equation can also be written as the Real part of the imaginary position equation

$\mathbf {Re} [x(t)]=x(t)=Acos(\omega _{0}t-\phi )\,$

Due to euler's rule (e^{jφ} = cosφ + jsinφ), x (t) is of the form

$x(t)=Ae^{j(\omega _{0}t-\phi )}\,$
Example 1.1
GIVEN: Two springs of stiffness, $s$ , and two bodies of mass, $M$

FIND: The natural frequencies of the systems sketched below

$s_{TOTAL}=s+s{\text{ (springs are in parallel)}}$

$\omega _{0}={\sqrt {\frac {s_{TOTAL}}{m_{TOTAL}}}}={\sqrt {\frac {2s}{M}}}$

$\mathbf {f_{0}} ={\frac {\omega _{0}}{2\pi }}=\mathbf {{\frac {1}{2\pi }}{\sqrt {\frac {2s}{M}}}}$

$\omega _{0}={\sqrt {\frac {s_{TOTAL}}{m_{TOTAL}}}}={\sqrt {\frac {s}{2M}}}$

$\mathbf {f_{0}} ={\frac {\omega _{0}}{2\pi }}=\mathbf {{\frac {1}{2\pi }}{\sqrt {\frac {s}{2M}}}}$

$\mathbf {1.} {\text{ }}s(x_{1}-x_{2})=sx_{2}$

$\mathbf {2.} {\text{ }}-s(x_{1}-x_{2})=m{\frac {d^{2}x}{dt^{2}}}$

${\frac {d^{2}x_{1}}{dt^{2}}}+{\frac {s}{2m}}x_{1}=0$

$\omega _{0}={\sqrt {\frac {s}{2m}}}$

$\mathbf {f_{0}={\frac {1}{2\pi }}{\sqrt {\frac {s}{2m}}}}$

$\omega _{0}={\sqrt {\frac {2s}{m}}}$

$\mathbf {f_{0}={\frac {1}{2\pi }}{\sqrt {\frac {2s}{m}}}}$

Back to Main page