Often, the force $F$ depends only on position $y(t).$

We write

By the Fundamental Theorem of Calculus, if $F$ is a continuous function, it has an anti-derivative $\phi(y)$ . Hence, $\displaystyle \frac{d\phi}{dy}=F(y).$

We multiply both sides by $\displaystyle \frac{dy}{dt}$

Applying the chain rule, the LHS becomes

while the RHS is

Equation $m\dfrac{d^2 y}{dt^2}=F(y,t)$ may then be written as

Equation $m\dfrac{d^2 y}{dt^2}=F(y,t)$ may then be written as

$E$ is the total energy, or simply the energy of the system. It is constant in time, or in physics parlance, conserved.

To simplify the equation, let's introduce a function $V(y)=-\phi(y).$ Energy can now be expressed as the sum of two terms, rather than the difference.

The first term, $\displaystyle \frac{1}{2}m\left(\frac{dy}{dt}\right)^2$ is a function of velocity (mass is a constant). It is the contribution to total energy due to the object's motion, and is called kinetic energy (K.E). Note that $\left(\frac{dy}{dt}\right)^2 \ge 0,$ so kinetic energy is non-negative.

The second term, $\displaystyle V(y(t))$ is a function of position. It's called potential energy (P.E.), because it has the potential to be converted into kinetic energy.

There are several advantages to adopting this formalism:

1. Newton's second law of motion is a second order ODE. The above technique allows any system with a continuous, position-dependent force to be described as a separable, first order ODE. Any such system is solvable in principle.
2. Energy $E$ is seen as an integration constant, calculated from initial position and velocity.
3. Given knowledge of the energy, position can be calculated from velocity, and vice-versa.

We can re-state the result in another way.

If we write

Then we have

This calculation shows energy cannot be created or destroyed, but can be changed from one form to another. 🤯

Since kinetic energy is non-negative, it follows that total energy can never be less than potential energy. We can visualise this for a physical system by graphing $V$ against $y$ using a potential energy diagram:

Potential energy diagrams provide a visual description of a system's dynamics. Picture a ball rolling on the hilly terrain imposed by the shape of $V(y).$

If the ball is placed where $V$ has an upward slope ($V^\prime(y)>0$), your intuition tells you it should be pushed to the left.

Indeed, since the force is $\displaystyle F(y)=-V^\prime(y)\lt 0,$ your intuition is correct. For the same reason, if $V$ has a downward slope, it is pushed to the right.

For $a\lt y\lt b,$ $V(y)\lt E.$ This means kinetic energy is positive, and hence motion can take place here. The same is true when $y>c$.

On the other hand, the system does not have enough energy to occupy positions $y\lt a$, or $b\lt y\lt c.$

Positions $a,$ $b,$ and $c$ are called turning points, because an object in this system reverses its direction here. Since $V(y)=E$ at these points, kinetic energy must be zero, and so an object must be temporarily at rest.

A system with the same potential energy function $V(y)$ but a different total energy $E^\star$ will behave differently. In the graph below, motion is only possible for $y>e.$

Example: Show that equilibrium solutions to equation $m\dfrac{d^2y}{dt}= F\left(y(t)\right)$ can only occur where kinetic energy is $0,$ and only at critical points of $V$.

An object at point $d$ in the graph above, with total energy $E^\star,$ represents an equilibrium solution.

There is a negative sign on the right of this equation because gravity pulls down towards the earth, but $y$ is measured as distance up from the the earth. Now also recall that $V,$ the potential energy has been defined as the negative of the integral of the force with respect to distance.

Setting $c$ equal to $0$, the ODE ($\#$) may be written;

As $y$ decreases, the P.E. of the particle decreases and the K.E. increases so that the total energy is conserved.

The energy diagram looks like this:

The potential energy function $V(y)$ is a straight line, with slope $mg$. Given a fixed total energy $E$, the system cannot occupy positions $y>y_{\text{max}}.$ The turning point (and thus the maximum height an object can reach) is $y_{\text{max}}$.

Example: A projectile is launched vertically upward from ground level, with initial speed $y=98$m/s. What is the maximum height the projectile will reach?

The potential energy corresponding to the spring force $F=-kx$ can be calculated as

Setting $c=0,$ the system is characterised by the energy equation

Since $\displaystyle x(t)=A\cos(\omega t-\phi),$ $\ds \frac{dx}{dt}=-A\omega\sin(\omega t-\phi).$ Equation $\clubsuit$ becomes

Setting $c=0,$ the system is characterised by the energy equation

Since $\displaystyle x(t)=A\cos(\omega t-\phi),$ $\ds \frac{dx}{dt}=-A\omega\sin(\omega t-\phi).$ Equation $\clubsuit$ becomes

The energy of this system is proportional both to square of the angular frequency, and to the square of the amplitude.

The amplitude $A$ is the maximum extent of the object's oscillation.

The amplitude $A$ is the maximum extent of the object's oscillation.

Hence, it represents a turning point, where $V(y)=E,$ and kinetic energy is $0$.

Hence we can calculate:

Example: Recall the undamped, oscillating spring system from page 78. It had a mass of $9$kg, and a spring constant of $4$N/m. It was pulled down $1$m and given an initial upward kick of $-0.5$m/s.

How fast is the mass moving when it passes through its equilibrium position?