# General Mechanics/Energy Analysis

## Energy Analysis

Figure 11.2: Potential, kinetic, and total energy of a one-dimensional harmonic oscillator plotted as a function of spring displacement.

For a spring, Hooke's law says the total force is proportional to the displacement, and in the opposite direction.

$F=-kx$

Since this is independent of velocity, it is a conservative force. We can integrate to find the potential energy of the mass-spring system,

$V(x)=\frac{1}{2}kx^2$

Since a potential energy exists, the total energy

$E=\frac{1}{2}kx^2+\frac{1}{2}m\dot{x}^2$

is conserved, i. e., is constant in time.

We can now use energy conservation to determine the velocity in terms of the position

$\dot{x}=\pm \sqrt{\frac{2E}{m}-\frac{k}{m}x^2}$

We could integrate this to determine the position as a function of time, but we can deduce quite a bit from this equation as it is.

It is fairly evident how the mass moves. From Hooke's law, the mass is always accelerating toward the equilibrium position, so we know which sign of the square root to take.

The velocity is zero when

$x=\pm \sqrt{\frac{2E}{k}}$

If x were larger than this the velocity would have to be imaginary, clearly impossible, so the mass must be confined between these values. We can call them the turning points.

If the mass is moving to the left, it slows down as it approaches the left turning point. It stops when it reaches this point and begins to move to the right. It accelerates until it passes the equilibrium position and then begins to decelerate, stopping at the right turning point, accelerating toward the left, etc. The mass thus oscillates between the left and right turning points.

How does the period of the oscillation depend on the total energy of the system? We can get a general idea without needing to solve the differential equation.

There are only two parameters the period, T, could depend on; the mass, m, and the spring constant, k.

We know T is measured in seconds, and m in kilograms. For the units in Hooke's law to match, k must be measured in N·m-1, or equivalently, in kg·s-2.

We immediately see that the only way to combine m and k to get something measured in seconds is to divide, cancelling out the kg's.

Therefore T$\sqrt{m/k}$

We have established the general way the period depends on the parameters of the problem, without needing to use calculus.

This technique is called dimensional analysis, and has wide application. E.g., if we couldn't calculate the proportionality constant exactly, using calculus, we'd be able to deduce by doing one experiment. Without dimensional analysis, if calculus failed us we'd have to do scores of experiments, each for different combinations of m and k.

Fortunately, the proportionality constants are typically small numbers, like $3\sqrt{2}$ or $\sqrt{\pi}/2$.