91Ó°ÊÓ

The Mass–Spring Oscillator

A damped mass-spring oscillator consists of a mass m attached to a spring fixed at one end, as shown in Figure 4.1. Devise a differential equation that governs the motion of this oscillator, taking into account the forces acting on it due to the spring elasticity, damping friction, and possible external influences.

Mass–spring oscillator equation;

$\begin{array}{c}{\mathbf{F}}_{\mathrm{ext}}\mathbf{=}\left[\mathrm{inertia}\right]\mathbf{y}\mathbf{"}\mathbf{+}\left[\mathrm{damping}\right]\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left[\mathrm{stiffness}\right]\mathbf{y}\\ =\mathrm{my}"+\mathrm{by}\text{'}+\mathrm{ky}\end{array}$…… (1)

The rule for the bounded equation: Just based on stiffness we can decide whether it is bounded or not if stiffnessk > 0 then it is bounded and if k < 0 then it is unbounded.

Root finding formula:

If. ${\mathbf{b}}^2\mathbf{<}\mathbf{4}\mathbf{ac}$Then,$\mathbf{Î\pm }\mathbf{=}\mathbf{-}\frac{b}{2a}$ and.$\mathbf{β}\mathbf{=}\frac{1}{2a}\sqrt{\mathbf{4}\mathbf{ac}\mathbf{-}{\mathbf{b}}^2}$

Given that, the standard form of the second-order differential equation of mass is

.$\mathbf{my}\mathbf{"}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\mathbf{=}\mathbf{0}$

Let us assume that.$\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathrm{rt}}$ Then, two times differential with respect to t.

$\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{=}{\mathbf{re}}^{\mathrm{rt}}\\ \mathbf{y}\mathbf{"}\left(t\right)\mathbf{=}{\mathbf{r}}^2{\mathbf{e}}^{\mathrm{rt}}\end{array}$

To find the roots of the standard form substitute the derivative values in equation (1).

$\begin{array}{c}{\mathbf{mr}}^2\mathbf{+}\mathbf{br}\mathbf{+}\mathbf{k}\mathbf{=}\mathbf{0}\\ \mathbf{r}\mathbf{=}\frac{\mathbf{-}\mathbf{b}Â\pm \sqrt{{\mathbf{b}}^2\mathbf{-}\mathbf{4}\mathbf{mk}}}{2m}\end{array}$

Since the given system is underdamped. So,.${\mathbf{b}}^2\mathbf{-}\mathbf{4}\mathbf{mk}\mathbf{<}\mathbf{0}$ Then the solution is;

$\mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\frac{\mathrm{bt}}{2m}}\mathbf{sin}\left(\frac{\sqrt{\mathbf{4}\mathbf{mk}\mathbf{-}{\mathbf{b}}^2}}{2m}\right)\mathbf{t}$

.

If,$\mathbf{b}→\mathbf{0}$ then ${\mathbf{Ae}}^{\mathbf{-}\frac{\mathrm{bt}}{2m}}\mathbf{=}\mathbf{A}$and.$\frac{\sqrt{\mathbf{4}\mathbf{mk}\mathbf{-}{\mathbf{b}}^2}}{2m}\mathbf{=}\sqrt{\frac{k}{m}}$

Therefore, the natural frequency is$\mathbf{f}\mathbf{=}\frac{\sqrt{\frac{k}{m}}}{2\mathrm{Ï€}}$.