Q4.9-13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Show that for the underdamped system of Example 3, the times when the solution curve y(t) in (33) touches the exponential curves $\pm \sqrt{\frac{7}{12}}{\mathbf{e}}^{-2t}$are not the same values of t for which the function y(t) attains its relative extrema.

 

Step-by-Step Solution

Answer

Therefore, the exponential curve  $\pm \sqrt{\frac{7}{12}}{\mathbf{e}}^{-2t}$does not have the same values of t and the function y(t) does not attain the maximum value at$\mathbf{t}\mathbf{=}\frac{\frac{\left(2n+1\right)\mathbf{\pi }}{2}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{85}}{\mathbf{2}\sqrt{3}}$.

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;

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

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 Example 3: (33)   $\mathbf{y}\left(t\right)\mathbf{=}\sqrt{\frac{7}{12}}{\mathbf{e}}^{-2t}\mathbf{sin}\left(\mathbf{2}\sqrt{3}\mathbf{t}\mathbf{+}\varphi \right)$…… (2)

And maximum displacement occurs at.$\mathbf{t}\mathbf{=}\frac{1}{\mathbf{2}\sqrt{3}}\mathbf{arctan}\frac{\sqrt{3}}{5}\Rightarrow t\approx \mathbf{0}\mathbf{.}\mathbf{096}$

Given that,$\pm \sqrt{\frac{7}{12}}{\mathbf{e}}^{-2t}$ .

 

Let us take.$y=\pm \sqrt{\frac{7}{12}}{\mathbf{e}}^{-2t}$ Then find the value of t.

 

Substitute the y value in equation (2).

 

$\begin{array}{c}\pm \sqrt{\frac{7}{12}}{\mathbf{e}}^{-2t}\mathbf{=}\sqrt{\frac{7}{12}}{\mathbf{e}}^{-2t}\mathbf{sin}\left(\mathbf{2}\sqrt{3}\mathbf{t}\mathbf{+}\varphi \right)\\ \mathbf{sin}\left(\mathbf{2}\sqrt{3}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{85}\right)\mathbf{=}\pm 1\\ \mathbf{2}\sqrt{3}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{85}\mathbf{=}\frac{\left(2n+1\right)\mathbf{\pi }}{2}\\ \mathbf{2}\sqrt{3}\mathbf{t}\mathbf{=}\frac{\left(2n+1\right)\mathbf{\pi }}{2}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{85}\\ \mathbf{t}\mathbf{=}\frac{\frac{\left(2n+1\right)\mathbf{\pi }}{2}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{85}}{\mathbf{2}\sqrt{3}}\end{array}$

 

Since,$\mathbf{t}\mathbf{=}\frac{\frac{\left(2n+1\right)\mathbf{\pi }}{2}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{85}}{\mathbf{2}\sqrt{3}}\ne \frac{1}{\mathbf{2}\sqrt{3}}\mathbf{arctan}\frac{\sqrt{3}}{5}$.

Therefore, the exponential curve$\pm \sqrt{\frac{7}{12}}{\mathbf{e}}^{-2t}$ does not have the same values of t and the function y(t) does not attain the maximum value at$\mathbf{t}\mathbf{=}\frac{\frac{\left(2n+1\right)\mathbf{\pi }}{2}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{85}}{\mathbf{2}\sqrt{3}}$.