Q4.9-15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

How can one deduce the value of the damping constant b by observing the motion of an underdamped system? Assume that the mass m is known. 

Step-by-Step Solution

Answer

Therefore, the damping constant b by observing the motion of an under-damped system is.$\mathbf{b}\mathbf{=}\pm \sqrt{\mathbf{4}\mathbf{mk}\mathbf{-}{\left(\frac{4\mathrm{m\pi }}{T}\right)}^2}$

Step 1: General form

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: 

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

    

The rule for the bounded equation: Just based on stiffness we can decide whether it is bounded or not if stiffness k > 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$\mathbf{\alpha }\mathbf{=}\mathbf{-}\frac{b}{2a}$, and$\mathbf{\beta }\mathbf{=}\frac{1}{2a}\sqrt{\mathbf{4}\mathbf{ac}\mathbf{-}{\mathbf{b}}^2}$.

 

Step 2: Evaluate the equation

Referring to Problem 14: The solution for underdamped oscillation is;

 

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

 

Then find the time period T.

 $\begin{array}{c}\mathbf{T}\mathbf{=}\frac{2\pi }{\frac{\sqrt{\mathbf{4}\mathbf{mk}\mathbf{-}{\mathbf{b}}^2}}{2m}}\\ \mathbf{=}\frac{4\mathrm{m\pi }}{\sqrt{\mathbf{4}\mathbf{mk}\mathbf{-}{\mathbf{b}}^2}}\\ \sqrt{\mathbf{4}\mathbf{mk}\mathbf{-}{\mathbf{b}}^2}\mathbf{=}\frac{4\mathrm{m\pi }}{T}\\ \mathbf{4}\mathbf{mk}\mathbf{-}{\mathbf{b}}^2\mathbf{=}{\left(\frac{4\mathrm{m\pi }}{T}\right)}^2\\ {\mathbf{b}}^2\mathbf{=}\mathbf{4}\mathbf{mk}\mathbf{-}{\left(\frac{4\mathrm{m\pi }}{T}\right)}^2\\ \mathbf{b}\mathbf{=}\pm \sqrt{\mathbf{4}\mathbf{mk}\mathbf{-}{\left(\frac{4\mathrm{m\pi }}{T}\right)}^2}\end{array}$

 

Therefore with the values of k, m, and T we can calculate it further to know the value of damping coefficient b.