# Engineering Acoustics/Simple Oscillation
## The Position Equation

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
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
$x(t)=x_{0}cos(\omega _{0}t)+(u_{0}/\omega _{0})sin(\omega _{0}t)\,$
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## Alternate Position Equation Forms

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
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
And if the difference of the same two equations is found, the result is that
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
* Simple Oscillator-1.2.1.a
$s_{TOTAL}=s+s{\text{ (springs are in parallel)}}$
$\omega _{0}={\sqrt {\frac {s_{TOTAL}}{m_{TOTAL}}}}={\sqrt {\frac {2s}{M}}}$
$\mathbf {f_{0}} ={\frac {\omega _{0}}{2\pi }}=\mathbf {{\frac {1}{2\pi }}{\sqrt {\frac {2s}{M}}}}$
* Simple Oscillator-1.2.1.b
$\omega _{0}={\sqrt {\frac {s_{TOTAL}}{m_{TOTAL}}}}={\sqrt {\frac {s}{2M}}}$
$\mathbf {f_{0}} ={\frac {\omega _{0}}{2\pi }}=\mathbf {{\frac {1}{2\pi }}{\sqrt {\frac {s}{2M}}}}$
* Simple Oscillator-1.2.1.c
$\mathbf {1.} {\text{ }}s(x_{1}-x_{2})=sx_{2}$
$\mathbf {2.} {\text{ }}-s(x_{1}-x_{2})=m{\frac {d^{2}x}{dt^{2}}}$
${\frac {d^{2}x_{1}}{dt^{2}}}+{\frac {s}{2m}}x_{1}=0$
$\omega _{0}={\sqrt {\frac {s}{2m}}}$
$\mathbf {f_{0}={\frac {1}{2\pi }}{\sqrt {\frac {s}{2m}}}}$
* Simple Oscillator-1.2.1.d
$\omega _{0}={\sqrt {\frac {2s}{m}}}$
$\mathbf {f_{0}={\frac {1}{2\pi }}{\sqrt {\frac {2s}{m}}}}$
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