Engineering Acoustics/Simple Oscillation

Solving for the Position Equation

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,

$f=ma=m\frac{d^{2}x}{d{t}^2}$

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

${\omega }_0^2=\frac{s}{m}$

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

$x=C_1{e}^{j{\omega }_{0}t}+C_2{e}^{-j{\omega }_{0}t}$

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

$x(t)=x_{0}cos({\omega }_{0}t)+(u_0/{\omega }_0)sin({\omega }_{0}t)$

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Alternate Position Equation Forms

If A₁ and A₂ are of the form

$A_1=Acos(\varphi )$ $A_2=Asin(\varphi )$

Then the position equation can be written

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

By applying the initial conditions (x(0)=x₀, u(0)=u₀) it is found that

$x_0=Acos(\varphi )$

$\frac{u_0}{{\omega }_0}=Asin(\varphi )$

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

$\varphi =ta{n}^{-1}(\frac{u_0}{x_0{\omega }_0})$

The position equation can also be written as the Real part of the imaginary position equation

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

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

$x(t)=A{e}^{j({\omega }_{0}t-\varphi )}$

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