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 $\mathbf{m}{(\mathbf{y}\mathbf{\text{'}})}^2\mathbf{+}{\mathbf{ky}}^2\mathbf{=}\mathbf{constant}$.

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.role="math" localid="1655363942650" $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$.

Mass–spring oscillator equation is given as:

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

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:

role="math" localid="1655364278131" $\mathrm{y}"=-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{y}\dots \left(2\right)$

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