KAHIBARO
Table of Contents
Forms of Energy in a Mass-Spring Oscillator
A mass attached to a spring and moving in simple harmonic motion continuously exchanges energy between motion and spring deformation. The total mechanical energy stays constant if there is no friction or other energy loss. What changes with time is how that energy is shared between kinetic energy and elastic potential energy.
When the mass moves fastest, its kinetic energy is greatest. When the spring is stretched or compressed the most, its elastic potential energy is greatest. This back and forth exchange is one of the clearest features of simple harmonic motion.
Elastic Potential Energy
A spring stores energy when it is stretched or compressed from its equilibrium position. If the displacement from equilibrium is $x$, the elastic potential energy is
$$
U = \frac{1}{2}kx^2
$$
where $k$ is the spring constant.
This formula shows that the stored energy depends on the square of the displacement. If the displacement doubles, the elastic potential energy becomes four times larger.
At the equilibrium position, $x = 0$, so
$$
U = 0
$$
This means the spring has its minimum elastic potential energy at equilibrium.
For a spring in simple harmonic motion, the elastic potential energy is
$$
U = \frac{1}{2}kx^2
$$
It is always zero or positive, and it is maximum at the turning points.
Kinetic Energy
The moving mass has kinetic energy
$$
K = \frac{1}{2}mv^2
$$
where $m$ is the mass and $v$ is its speed.
The kinetic energy is greatest when the mass passes through equilibrium, because that is where the speed is greatest. At the turning points, the mass stops for an instant, so $v = 0$, and the kinetic energy becomes zero.
The kinetic energy of the oscillating mass is
$$
K = \frac{1}{2}mv^2
$$
It is maximum at equilibrium and zero at the turning points.
Total Mechanical Energy
In an ideal mass-spring system, the total mechanical energy is the sum of kinetic and elastic potential energy:
$$
E = K + U
$$
So,
$$
E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2
$$
Since energy is conserved,
$$
E = \text{constant}
$$
The maximum displacement is the amplitude $A$. At the turning points, the speed is zero, so all the energy is elastic potential energy:
$$
E = \frac{1}{2}kA^2
$$
This gives an important expression for the total energy of simple harmonic motion.
For ideal simple harmonic motion in a spring-mass system,
$$
E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2
$$
The total mechanical energy remains constant.
Energy at Special Positions
The energy changes form depending on position in the motion.
At the equilibrium position, $x = 0$, so
$$
U = 0
$$
and all the energy is kinetic:
$$
K = E = \frac{1}{2}kA^2
$$
At the turning points, $x = \pm A$, so
$$
U = \frac{1}{2}kA^2
$$
and
$$
K = 0
$$
At positions between these extremes, the total energy is shared between kinetic and potential energy.
| Position | Displacement $x$ | Speed $v$ | Kinetic Energy $K$ | Potential Energy $U$ |
|---|---|---|---|---|
| Equilibrium | $0$ | Maximum | Maximum | $0$ |
| Turning point | $\pm A$ | $0$ | $0$ | Maximum |
| Intermediate point | between $0$ and $\pm A$ | intermediate | intermediate | intermediate |
Energy as a Function of Position
Because the total energy is constant,
$$
K = E - U
$$
Using $E = \frac{1}{2}kA^2$ and $U = \frac{1}{2}kx^2$, we get
$$
K = \frac{1}{2}kA^2 - \frac{1}{2}kx^2
$$
or
$$
K = \frac{1}{2}k\left(A^2 - x^2\right)
$$
This shows how the kinetic energy depends on the position of the mass. As $x$ increases in magnitude, the kinetic energy decreases.
Since $K = \frac{1}{2}mv^2$, we can also write
$$
\frac{1}{2}mv^2 = \frac{1}{2}k\left(A^2 - x^2\right)
$$
which gives
$$
v^2 = \frac{k}{m}\left(A^2 - x^2\right)
$$
and therefore
$$
v = \pm \sqrt{\frac{k}{m}\left(A^2 - x^2\right)}
$$
This relation connects speed directly to position.
A very useful energy relation in SHM is
$$
K = \frac{1}{2}k\left(A^2 - x^2\right)
$$
and therefore
$$
v^2 = \frac{k}{m}\left(A^2 - x^2\right)
$$
This lets you find the speed from the displacement without using time.
Energy and Amplitude
The total energy depends on the amplitude:
$$
E = \frac{1}{2}kA^2
$$
So a larger amplitude means more total energy. Because of the square, if the amplitude doubles, the total energy becomes four times as large.
This is important physically. A larger oscillation is not just a little more energetic, it can be much more energetic.
Visual Picture of Energy Exchange
The motion can be pictured as a smooth transfer of energy. Near the center, most energy is kinetic. Near the ends, most energy is stored in the spring.
Energy Graphs
The elastic potential energy depends on $x^2$, so its graph against position is a parabola opening upward. The total energy is a horizontal line. The kinetic energy is the difference between the total energy line and the potential energy curve.
A Simple Example
Suppose a mass-spring system has spring constant $k = 200\ \text{N/m}$ and amplitude $A = 0.10\ \text{m}$. The total energy is
$$
E = \frac{1}{2}kA^2
$$
$$
E = \frac{1}{2}(200)(0.10)^2 = 1.0\ \text{J}
$$
So the oscillator has a total mechanical energy of $1.0\ \text{J}$.
If the displacement is $x = 0.06\ \text{m}$, then the potential energy is
$$
U = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.06)^2 = 0.36\ \text{J}
$$
The kinetic energy is
$$
K = E - U = 1.0 - 0.36 = 0.64\ \text{J}
$$
So at that instant, $0.36\ \text{J}$ is stored in the spring and $0.64\ \text{J}$ is in the motion of the mass.
What Conservation of Energy Tells Us
Energy methods are powerful because they let us describe the motion without following every moment in time. If we know the amplitude and position, we can find the kinetic energy and speed immediately.
In simple harmonic motion, the essential idea is that the oscillator does not lose energy in the ideal case. It only changes the form of that energy, from kinetic to elastic potential and back again, over and over.
In ideal SHM, energy is not lost. It is continuously exchanged between
$$
K = \frac{1}{2}mv^2
$$
and
$$
U = \frac{1}{2}kx^2
$$
while the total energy remains
$$
E = \frac{1}{2}kA^2
$$