91Ó°ÊÓ

Given displacements are $y=A\sin \mathrm{Ó\neg }t$or $y=A\sin (\mathrm{Ó\neg }t+\mathrm{Ï•})$that depends on time origin.

The energy of mass or a moving object is known as kinetic energy. The more kinetic energy an item possesses, the faster it moves. Kinetic energy is present in any moving item.

Find the average kinetic energy of the particle.

$\begin{array}{c}E=\frac{1}{2}m{y}^2\\ E=\frac{1}{2}m\frac{\mathrm{Ó\neg }}{2\mathrm{Ï€}}{A}^2{\sin }^2\left(\mathrm{Ó\neg }t\right)dt\\ E=\frac{m{A}^2}{4\mathrm{Ï€}}{∫}_0^{\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}}{\sin }^2\left(\mathrm{Ó\neg }t\right)d\left(\mathrm{Ó\neg }t\right)\\ E=\frac{m{A}^2}{4\mathrm{Ó\neg }}\end{array}$

(a) For average kinetic energy, let $\mathrm{Ó\neg }t+\mathrm{Ï•}=u$and $\mathrm{Ó\neg }dt=du$, and take the limits from $\mathrm{Ï•}$to $\mathrm{Ï•}+2\mathrm{Ï€}$.

$\begin{array}{c}E=\frac{1}{2}m{y}^2\\ E=\frac{1}{2}m\left(\frac{1}{2}{∫}_0^{\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}}{A}^2{\sin }^2(\mathrm{Ó\neg }t+\mathrm{Ï•})dt\right)\\ E=\frac{m{A}^2}{2\mathrm{Ó\neg }}\frac{1}{2\mathrm{Ï€}}{∫}_{\mathrm{Ï•}}^{\mathrm{Ï•}+2\mathrm{Ï€}}{\sin }^{2}udu\\ E=\frac{m{A}^2}{4\mathrm{Ó\neg }}\end{array}$

(b) Formula of trigonometric sine addition is,

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

Let, $u=\mathrm{Ó\neg }t$and$du=\mathrm{Ó\neg }dt$.

Find the average kinetic energy,

$\begin{array}{l}E=\frac{1}{2}m\left(\frac{1}{2\mathrm{Ï€}}{∫}_0^{\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}}{A}^2{\sin }^2(\mathrm{Ó\neg }t+\mathrm{Ï•})dt\right)\\ E=\frac{1}{2}m\left(\frac{1}{2\mathrm{Ï€}}{∫}_0^{\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}}{A}^2{(\sin \left(\mathrm{Ó\neg }t\right)\cos \mathrm{Ï•}+\cos \left(\mathrm{Ó\neg }t\right)\sin \mathrm{Ï•})}^{2}dt\right)\\ E=\frac{m{A}^2}{2\mathrm{Ó\neg }}\frac{1}{2\mathrm{Ï€}}{∫}_0^{2\mathrm{Ï€}}(\sin u\cos \mathrm{Ï•}+\cos u\sin \mathrm{Ï•})du\\ E=\frac{m{A}^2}{2\mathrm{Ó\neg }}\frac{1}{2\mathrm{Ï€}}{∫}_0^{2\mathrm{Ï€}}({\sin }^{2}u{\cos }^2\mathrm{Ï•}+{\cos }^{2}u{\sin }^2\mathrm{Ï•}+2\sin uâ‹\dots \cos uâ‹\dots \sin \mathrm{Ï•}â‹\dots \cos \mathrm{Ï•})du\end{array}$

Solve further,

$\begin{array}{l}E=\frac{m{A}^2}{2\mathrm{Ó\neg }}(\frac{1}{2}{\cos }^2\mathrm{Ï•}+\frac{1}{2}{\sin }^2\mathrm{Ï•})\\ E=\frac{m{A}^2}{4\mathrm{Ó\neg }}({\cos }^2\mathrm{Ï•}+{\sin }^2\mathrm{Ï•})\\ E=\frac{m{A}^2}{4\mathrm{Ó\neg }}\left(1\right)\\ E=\frac{m{A}^2}{4\mathrm{Ó\neg }}\end{array}$

Therefore, it is proved that the average of the kinetic energy of a particle is the same, $\frac{m{A}^2}{4\mathrm{Ó\neg }}$, for the two formulas, $y=A\sin \mathrm{Ó\neg }t$and $y=A\sin (\mathrm{Ó\neg }t+\mathrm{Ï•})$, and (a) the average kinetic energy by selecting the integration limits so that a change of variable reduces the integral to the $\sin \mathrm{Ó\neg }t$is $\frac{m{A}^2}{4\mathrm{Ó\neg }}$, and (b) by expanding $\sin (\mathrm{Ó\neg }t+\mathrm{Ï•})$by the trigonometric addition formulas and using (5.2) to write the average values is also$\frac{m{A}^2}{4\mathrm{Ó\neg }}$