7.4 Energy of Simple Harmonic Oscillators

Simple harmonic oscillators, like springs and pendulums, showcase energy conservation in action. As these systems move back and forth, energy constantly shifts between kinetic and potential forms, with the total energy remaining constant throughout the motion.

Understanding the energy dynamics in simple harmonic motion is crucial for grasping broader concepts in physics. This topic connects ideas of energy conservation, periodic motion, and the relationship between force and displacement in oscillating systems.

Mechanical energy in SHM

more resources to help you study

practice multiple choiceFRQ practice & scoringcheatsheetsscore calculator

Total energy components

In a system undergoing simple harmonic motion (SHM), the total energy consists of the sum of the system's kinetic energy (K) and potential energy (U) 🔋

$E_{total} = U + K$

• Kinetic energy comes from the object's motion and depends on its mass and velocity
• Potential energy relates to the object's position and the restoring force acting on it
• For springs, potential energy is stored as elastic potential energy in the deformed spring
• For pendulums, potential energy is stored as gravitational potential energy

Conservation of total energy

The fundamental principle of energy conservation states that energy cannot be created or destroyed in an isolated system—it only changes form.

$E_{total} = U + K = \text{constant}$

• During oscillation, energy continuously converts between kinetic and potential forms
• At any instant, the sum of kinetic and potential energy equals the total energy
• In ideal SHM, the total mechanical energy remains constant while energy shifts between kinetic and potential forms
• In real systems, friction gradually reduces total energy, damping the oscillation

Maximum kinetic energy

A system undergoing SHM reaches its maximum kinetic energy when passing through the equilibrium position.

$K_{max} = \frac{1}{2}mv_{max}^2$

• At equilibrium, displacement is zero and velocity reaches its maximum
• All energy exists as kinetic energy at this point
• The object experiences zero net force at equilibrium
• This moment represents the transition between acceleration directions

Maximum potential energy

Potential energy in SHM reaches its maximum value at the extreme positions of oscillation 📈

At the extreme positions, the system's potential energy is at its maximum and the kinetic energy is zero, so $E_{total} = U_{max}$. For a spring-object system specifically, $E_{total} = U_{max} = \frac{1}{2}kA^2$.

• At maximum displacement (amplitude), velocity becomes zero
• All energy exists as potential energy at these turning points
• The restoring force reaches its maximum magnitude here
• After reaching maximum displacement, the object reverses direction

Minimum Kinetic Energy

The kinetic energy of a system exhibiting SHM reaches its minimum value of zero at the points of maximum displacement.

• At the turning points, the oscillating object momentarily stops before changing direction
• The velocity at these points is zero, resulting in zero kinetic energy
• These points represent the conversion of all energy to potential form
• The minimum kinetic energy can be expressed as: $K_{min} = 0$

Effects of Changing Amplitude

For a spring-object system, changing the amplitude changes the maximum potential energy and therefore the total energy. Using $E_{total} = \frac{1}{2}kA^2$, increasing amplitude increases total energy, and doubling the amplitude quadruples the total energy.

$E_{total} = \frac{1}{2}kA^2$

• Increasing the amplitude increases the maximum potential energy
• Since total energy equals maximum potential energy, amplitude changes alter total energy
• Doubling the amplitude quadruples the total energy (energy ∝ amplitude²)

Practice Problem 1: Spring Energy Conservation

Solution:

(a) Total energy of the system: $E_{total} = \frac{1}{2}kA^2 = \frac{1}{2} \times 20 \text{ N/m} \times (0.10 \text{ m})^2 = 0.10 \text{ J}$

(b) Maximum speed occurs at equilibrium when all energy is kinetic: $K_{max} = E_{total} = \frac{1}{2}mv_{max}^2$

Solving for $v_{max}$: $v_{max} = \sqrt{\frac{2E_{total}}{m}} = \sqrt{\frac{2 \times 0.10 \text{ J}}{0.5 \text{ kg}}} = 0.63 \text{ m/s}$

(c) Speed at 5 cm from equilibrium: At this position, we need to find how the energy is divided: $U = \frac{1}{2}kx^2 = \frac{1}{2} \times 20 \text{ N/m} \times (0.05 \text{ m})^2 = 0.025 \text{ J}$

Since $E_{total} = K + U$: $K = E_{total} - U = 0.10 \text{ J} - 0.025 \text{ J} = 0.075 \text{ J}$

Speed at this position: $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 0.075 \text{ J}}{0.5 \text{ kg}}} = 0.55 \text{ m/s}$

Practice Problem 2: Amplitude and Energy

Solution:

(a) Total energy with 8 cm amplitude: $E_{total} = \frac{1}{2}kA^2 = \frac{1}{2} \times 15 \text{ N/m} \times (0.08 \text{ m})^2 = 0.048 \text{ J}$

(b) Total energy with 16 cm amplitude: $E_{total} = \frac{1}{2}kA^2 = \frac{1}{2} \times 15 \text{ N/m} \times (0.16 \text{ m})^2 = 0.192 \text{ J}$

(c) Relationship between amplitude and energy: The new total energy (0.192 J) is exactly 4 times the original energy (0.048 J). This demonstrates that energy in a spring-object system is proportional to the square of the amplitude. When amplitude doubles, energy quadruples. This relationship is required for spring-object systems in SHM.