Engineering Acoustics/Simple Oscillation

EA1.jpg

Authors · Print · License

Edit this template

Part 1: Lumped Acoustical Systems1.11.21.31.41.51.61.71.81.91.101.11

Part 2: One-Dimensional Wave Motion2.12.22.3

Part 3: Applications3.13.23.33.43.53.63.73.83.93.103.113.123.133.143.153.163.173.183.193.203.213.223.233.24

The Position EquationEdit

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^2x \over dt^2}\,

where a is the acceleration of the mass, we can get

m\frac{d^2 x}{d t^2 }= -sx

or,

\frac{d^2 x}{d t^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)=A e^{\lambda t} \,

The force equation then becomes

(\lambda^2+\omega_0^2)A e^{\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_1e^{j\omega_0 t}+C_2e^{-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)\,



Back to Main page

Alternate Position Equation FormsEdit

If A1 and A2 are of the form

A_1 = A cos( \phi)\,
A_2 = A sin( \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 = A cos(\phi)\,


\frac{u_0}{\omega_0} = A sin(\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) = A cos(\omega_0 t - \phi)\,


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

x(t) = A e^{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

  1. 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}}}



  1. 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}}}






  1. Simple Oscillator-1.2.1.c

Simple Oscillator-1.2.1.c-solution


\mathbf{1.}\text{ }
s(x_1-x_2) = sx_2


\mathbf{2.}\text{ }
-s(x_1-x_2) = m \frac{d^2x}{dt^2}


\frac{d^2x_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}}}



  1. Simple Oscillator-1.2.1.d


\omega_0=\sqrt{\frac{2s}{m}}


\mathbf{f_0 = \frac{1}{2\pi}\sqrt{\frac{2s}{m}}}










Back to Main page

Last modified on 9 August 2012, at 19:16