Energy of Simple Harmonic Oscillators

Learning Goals

After this lesson, you should be able to:

• choose a system that contains the relevant potential-energy interaction;
• calculate spring potential, kinetic, and total mechanical energy;
• derive \(E_{\mathrm{total}}=\frac12kA^2\);
• identify energy extrema at equilibrium and turning points;
• calculate speed at any displacement using energy conservation;
• interpret \(U(x)\), \(K(x)\), and total-energy graphs;
• compare the periods of position and energy graphs;
• solve energy-fraction questions without confusing displacement and amplitude;
• predict how amplitude, stiffness, external work, and damping change energy; and
• test mechanical-energy conservation using motion data.

1. Select the Energy System

For a horizontal mass–spring oscillator, choose the block and spring as the system. The spring interaction then contributes elastic potential energy internally:

For a pendulum, include the bob and Earth so gravitational potential energy belongs to the system. A clear system boundary prevents treating the same interaction as both internal potential energy and external work.

2. Spring Potential and Kinetic Energy

Measure signed displacement \(x\) from equilibrium. For an ideal linear spring,

The moving object has

Both energies are nonnegative. Spring potential energy is the same at \(+x\) and \(-x\), while the direction of motion does not affect kinetic energy.

4. Total Energy Is Set by Amplitude

At a turning point \(x=\pm A\), the speed is zero. Therefore,

For ideal undamped motion, this value remains constant everywhere in the cycle.

5. Energy Conservation at Any Position

At displacement \(x\),

The spring potential energy lost while moving toward equilibrium becomes kinetic energy. Moving away from equilibrium reverses the transfer.

6. Energy at Equilibrium and Turning Points

Minimum kinetic energy is zero. Maximum potential energy equals the total mechanical energy in the ideal model.

7. Deriving Speed as a Function of Position

Rearrange the conservation equation:

Energy determines speed magnitude, not direction. At most positions the object passes once in each direction during a cycle.

8. Maximum Speed

At equilibrium \(x=0\), all energy is kinetic:

A larger amplitude or stiffer spring raises the maximum speed; a larger oscillating mass lowers it when \(k\) and \(A\) are fixed.

9. Worked Example: Energy and Speed

A \(0.50\,\mathrm{kg}\) block oscillates on an ideal spring with \(k=80\,\mathrm{N/m}\) and amplitude \(A=0.120\,\mathrm m\). Find the energies and speed at \(x=0.072\,\mathrm m\).

Answer: \(E=0.576\,\mathrm J\), \(U=0.207\,\mathrm J\), \(K=0.369\,\mathrm J\), and the speed is \(1.21\,\mathrm{m/s}\).

10. Energy Fractions Depend on \(x/A\)

Divide the potential energy by total energy:

If \(x=qA\), then \(U/E=q^2\) and \(K/E=1-q^2\). These fractions do not depend on \(m\) or \(k\).

11. Where Are Kinetic and Potential Energies Equal?

Set \(K=U=E/2\):

The energies are not equal at \(x=A/2\). There, \(U/E=1/4\) and \(K/E=3/4\) because energy depends on displacement squared.

12. Energy–Position Graphs

For a spring oscillator:

• \(U(x)=\frac12kx^2\) is an upward-opening parabola;
• \(E_{\mathrm{total}}=\frac12kA^2\) is a horizontal line;
• \(K(x)=E_{\mathrm{total}}-U(x)\) is a downward-opening parabola over the allowed interval;
• turning points occur where \(K=0\), at \(x=\pm A\); and
• positions outside \([-A,A]\) are inaccessible for that total energy.

Raising the total-energy line moves the turning points farther from equilibrium.

13. Energy–Time Graphs Repeat Twice per Cycle

For an oscillator released at \(+A\),

Because squaring removes the sign, both \(U\) and \(K\) repeat after \(T/2\), while position repeats after \(T\). During one full position cycle, each energy reaches its maximum twice.

14. Amplitude Controls Energy Quadratically

At fixed spring constant,

For two amplitudes,

• Double amplitude \(\rightarrow\) four times the energy.
• Triple amplitude \(\rightarrow\) nine times the energy.
• Half amplitude \(\rightarrow\) one fourth the energy.

Changing amplitude does not change the ideal spring period, even though total energy changes.

15. Changing Stiffness: State What Is Held Fixed

If amplitude is fixed,

If total energy is fixed instead,

Thus, doubling \(k\) doubles energy at fixed \(A\), but reduces amplitude by \(1/\sqrt2\) at fixed \(E\). A comparison is incomplete unless the controlled quantity is stated.

16. External Work Changes the Amplitude

Work crossing the system boundary changes mechanical energy:

For a spring oscillator,

Positive external work increases amplitude; negative external work decreases it, provided the system remains within the SHM model.

17. Vertical Spring Energy About Equilibrium

For a vertical spring, gravity shifts the equilibrium position. If \(y\) is measured from that equilibrium, the combined spring–Earth potential energy can be written, apart from an irrelevant additive constant, as

The energy equation about equilibrium has the same form as the horizontal case:

Do not measure \(y\) from the spring's natural length and then omit gravitational potential energy.

18. Small-Angle Pendulum Energy

Taking the lowest point as zero gravitational potential, an ideal pendulum has

For small angles in radians, \(1-\cos\theta\approx\theta^2/2\), so

At \(\theta=\pm\theta_{\max}\), kinetic energy is zero; at \(\theta=0\), gravitational potential is minimum and kinetic energy is maximum. This mirrors the spring oscillator.

19. Damping and Mechanical-Energy Loss

Real drag or friction transforms mechanical energy into thermal or other internal energy. Then

On a graph, damping appears as a decreasing amplitude envelope and a declining mechanical-energy curve. The total energy of a sufficiently inclusive system is still conserved, but the oscillator's mechanical energy is not constant.

Ideal SHM energy equations apply over intervals where dissipative effects are negligible.

20. Investigation: Test Energy Conservation

Use a cart, spring, level track, motion sensor, balance, and independently measured spring constant.

1. Define the cart–spring system and mark equilibrium.
2. Measure \(m\), \(k\), and initial amplitude.
3. Record position and velocity over several cycles.
4. Calculate \(U(t)=\frac12kx(t)^2\) and \(K(t)=\frac12mv(t)^2\).
5. Plot \(U\), \(K\), and \(E=U+K\) on common time axes.
6. Check whether \(U\) and \(K\) repeat every \(T/2\).
7. Compare measured \(E\) with \(\frac12kA^2\).
8. Repeat at several amplitudes and graph \(E\) versus \(A^2\); the predicted slope is \(k/2\).
9. Discuss sensor differentiation noise, friction, spring mass, track tilt, equilibrium error, and uncertainty in \(k\).

21. A Reliable Energy Workflow

1. Select the system and identify energy transfers across its boundary.
2. Measure displacement from equilibrium and identify amplitude.
3. Write \(E_i=E_f\) only when nonconservative transfer is negligible.
4. For a spring, use \(U=\frac12kx^2\), \(K=\frac12mv^2\), and \(E=\frac12kA^2\).
5. Solve symbolically before substituting numbers.
6. Use energy to find speed magnitude; use motion information for velocity sign.
7. Check \(0\le U\le E\), \(0\le K\le E\), and \(K+U=E\).
8. For comparisons, distinguish what is fixed: amplitude, energy, stiffness, or mass.
9. Check whether damping or external work makes mechanical energy nonconstant.

22. Common Traps and Key Takeaways

• Total mechanical energy is \(K+U\), not either form alone.
• At turning points, \(K=0\) and \(U=E\).
• At equilibrium, \(U\) is minimum and \(K=E\).
• Spring energy depends on \(x^2\), so it is identical at \(+x\) and \(-x\).
• Use \(E=\frac12kA^2\), not \(\frac12mA^2\).
• Kinetic and potential energy are equal at \(|x|=A/\sqrt2\), not \(A/2\).
• Energy gives speed, not velocity direction.
• Doubling amplitude quadruples energy.
• Energy graphs repeat every \(T/2\).
• With damping, oscillator mechanical energy decreases even though total energy for an enlarged system is conserved.