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, $$f = ma = m{d^2x \over dt^2}\,$$ 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 $$x(t) = x_0 cos(\omega_0 t) + (u_0 /\omega_0 )sin(\omega_0 t)\,$$

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Alternate Position Equation Forms
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 (ej&phi; = cos&phi; + jsin&phi;), x(t) is of the form $$x(t) = A e^{j(\omega_0 t - \phi)}\,$$

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