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# The Equation of Energy

The Equation of Energy for a body travelling in a straight line or under Simple Harmonic Motion

## The Mechanics Of Motion Of A Body Traveling In A Straight Line.

There are two integrals of the equation of Motion of a Body which have special significance because of their physical meaning. They are the equations of Energy and Momentum.

## The Equation Of Energy.

It is possible to write acceleration as
$\frac{d}{dx}\left(\frac{1}{2}v^2&space;\right)$
. The formula for Newton's Second Law ( F = ma) can now be expressed as:-

$\frac{d}{dx}\left(\frac{W\;v^2}{2g}&space;\right)&space;=&space;F$

$\therefore\;\;\;\;\;\;\frac{W\,v^2}{2g}&space;=&space;\int&space;F\,dx&space;+&space;C$

$\inline&space;\displaystyle&space;If\;\;\;\;v_{1}\;\;\;and\;\;\;v_2$ are the velocities of the body at $\inline&space;\displaystyle&space;x_{1}\;\;\;and\;\;\;x_2$ respectively.

$Then\;\;\;\;\;\frac{W\;v_2\,^2}{2g}&space;-&space;\frac{W\;v_1\,^2}{2g}&space;=&space;\int_{x_1}^{x_2}\;F\;dx$

The left hand of the equation is the increase in Kinetic Energy and the right hand side is the Work Done by the force F. This equation or its alternative form (3) are called the Equation of Energy

The Equation of Energy can also be obtained by multiplying equation (2) by $\inline&space;\displaystyle&space;\frac{dx}{dt}$ i.e. by v and then by integrating with respect to t rather than x.

$\therefore\;\;\;\;\;\;\frac{W}{g}\;\frac{dx}{dt}\frac{d^2x}{dt^2}&space;=&space;F\;\frac{dx}{dt}$

$\therefore\;\;\;\;\;\;\frac{d}{dt}\left\{\frac{W}{2g}\;\left(\frac{dx}{dt}&space;\right)^2&space;\right\}&space;=&space;F\;\frac{dx}{dt}$

$\therefore\;\;\;\;\;\frac{W}{2g}\;\left(\frac{dx}{dt}&space;\right)^2&space;=&space;\int&space;F\;\frac{dx}{dt}dt&space;+&space;C$

$=&space;\int&space;F\;dx&space;+&space;C$

Which is the same as equation (3)

## The Energy Equation For Simple Harmonic Motion.

The equation of Simple Harmonic Motion which is :-

$\ddot{x}&space;+&space;n^2x&space;=&space;0$

Can also be written in terms of v

$Hence\;\;\;\;\;\frac{d}{dx}\,\left(\frac{1}{2}\,v^2&space;\right)\;=&space;-&space;n^2\,x$

$\therefore\;\;\;\;\;\frac{1}{2}\,v^2&space;=&space;\int&space;-\,n^2\,x\,dx&space;+&space;C&space;=&space;-\,\frac{1}{2}\,n^2x^2&space;+&space;C$

This is the |b{Equation of Energy}. It is possible to find x from this equation and to show that $\inline&space;\displaystyle&space;\dot{x}&space;=&space;0\;\;and\;\;x&space;=&space;a\;\;\;at\;\;\;t&space;=&space;0$

$Since\;\;\;\;v&space;=&space;0\;\;\;at\;\;\;x&space;=&space;a\;;\;\;\;\;C&space;=&space;\frac{1}{2}n^2a^2$

$\therefore\;\;\;\;\;v^2&space;=&space;n^2\;(a^2&space;-&space;x^2)$

$\therefore\;\;\;\;\;\dot{x}&space;=&space;v&space;=&space;-\,n\;\sqrt{a^2&space;-&space;n^2}$

Note that the negative root has been taken because the body starts inwards from x = a

Thus it can be seen that:-

$\frac{dt}{dx}&space;=&space;-\frac{1}{n\sqrt{a^2&space;-&space;x^2}}$

$\therefore\;\;\;\;\;nt&space;=&space;-\,\int&space;\frac{dx}{\sqrt{a^2&space;-&space;x^2}}&space;=&space;-\,\sin^{-1}\frac{x}{a}&space;+&space;K$

But t = 0 at x = a

$\therefore\;\;\;\;\;0&space;=&space;-\,\sin^{-1}1&space;+&space;K&space;=&space;-\frac{1}{2}\pi&space;&space;+&space;K$

$\therefore\;\;\;\;\;nt&space;=&space;\frac{1}{2}\pi&space;&space;-&space;\sin^{-1}\frac{x}{a}$

$\therefore\;\;\;\;\;x&space;=&space;a\,\sin\;(\frac{1}{2}\pi&space;&space;-&space;nt)&space;=&space;a\,\cos\,nt$

## The Equation Of Momentum

The equation of Motion (2) can be written in the following two forms:-

$\frac{d}{dt}\left(\frac{W\,v}{g}&space;\right)&space;=&space;F\;\;\;\;or\;\;\;\;d\left(\frac{W\,v}{g}&space;\right)&space;=&space;F\,dt$

The first form states that the rate of increase of the Momentum $\inline&space;\displaystyle&space;\frac{W\,v}{g}$ is equal to the Force F and the second form that the increase of Momentum in a time dt is equal to the Impulse Fdt. Integrating either form produces another integral of the Equation of Motion:-

$\frac{W\,v}{g}&space;=&space;\int&space;F\,dt&space;+&space;C$

This means that if the velocity increases from $\inline&space;\displaystyle&space;v_1\;\;\;to\;\;\;v_2$ in the time from $\inline&space;\displaystyle&space;t_1\;\;\;to\;\;\;t_2$

$Then\;\;\;\;\;\;\frac{W\,v_2}{g}&space;-&space;\frac{W\,v_1}{g}&space;=&space;\int_{t_1}^{t_2}F\,dt$

This equation states that the increase of Momentum is equal to the total impulse of the force in the interval $\inline&space;\displaystyle&space;t_1\;\;\;to\;\;\;t_2$