Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the energy integral lemma to show that motions of the free undamped mass-spring oscillator $\mathbf{my}\mathbf{"}\mathbf{+}\mathbf{ky}\mathbf{=}\mathbf{0}$ obey  $\mathbf{m}{(\mathbf{y}\mathbf{\text{'}})}^2\mathbf{+}{\mathbf{ky}}^2\mathbf{=}\mathbf{constant}$.

Step-by-Step Solution

Answer

Therefore, the given statement is true. That is,$\mathbf{m}{(\mathbf{y}\mathbf{\text{'}})}^2\mathbf{+}{\mathbf{ky}}^2\mathbf{=}\mathbf{constant}$ is true.

Step 1: The Energy Integral Lemma

Let $\mathrm{y}\left(\mathrm{t}\right)$ be a solution to the differential equation $\mathrm{y}"=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t.

Let F(y) is an indefinite integral of $\mathrm{f}\left(\mathrm{y}\right)$i.e. $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity $\mathrm{E}\left(\mathrm{t}\right)=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$ is constant; i.e. $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$ .

Mass–spring oscillator equation is given as:

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

Step 2: Evaluate the given equation

Given that, $\mathrm{my}"+\mathrm{ky}=0\dots \left(1\right)$

 

To prove: $\mathrm{m}{(\mathrm{y}\text{'})}^2+{\mathrm{ky}}^2=\mathrm{constant}$.

 

Equation (1) can be written as:

$\mathrm{y}"=-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{y} \dots \left(2\right)$

Rewrite the equation (2) into $\mathrm{F}\left(\mathrm{y}\right)$ form.

$\begin{array}{c}\mathrm{y}"=-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{y}\\ =\frac{\mathrm{d}}{\mathrm{dy}}\left(-\frac{\mathrm{k}}{2\mathrm{m}}{\mathrm{y}}^2\right)\end{array}$

Therefore, $\mathrm{F}\left(\mathrm{y}\right)=-\frac{\mathrm{k}}{2\mathrm{m}}{\mathrm{y}}^2$ . Then,

$\begin{array}{c}\mathrm{E}\left(\mathrm{t}\right)=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)\\ =\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2+\frac{\mathrm{k}}{2\mathrm{m}}{\mathrm{y}}^2\\ 2\mathrm{mE}=\mathrm{my}\text{'}{\left(\mathrm{t}\right)}^2+{\mathrm{ky}}^2\end{array}$

Thus, it is proved that $\mathrm{m}{(\mathrm{y}\text{'})}^2+{\mathrm{ky}}^2=\mathrm{constant}$.