# Engineering Acoustics/Simple Oscillation

 Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24

## 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 C1 and C2 are constants given by the initial conditions of the system

If the position of the mass at t = 0 is denoted as x0, then

$C_{1}+C_{2}=x_{0}\,$

and if the velocity of the mass at t = 0 is denoted as u0, 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

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

## Alternate Position Equation Forms

If A1 and A2 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)=x0, u(0)=u0) 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 = 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

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

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

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

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

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

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

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

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

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

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

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

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