Energy of
Simple Harmonic Oscillators

Discover how kinetic energy and potential energy transform back and forth in oscillating systems, and why the total mechanical energy stays constant.

AP Physics 1

What is Simple Harmonic Motion?

Simple harmonic motion (SHM) occurs when a restoring force is directly proportional to displacement from equilibrium and directed toward it. The hallmark equation is:

Examples include a mass on a spring and a simple pendulum (for small angles). In SHM, energy continuously converts between kinetic and potential forms.

The Three Energies in SHM

Kinetic Energy

$ K = \tfrac{1}{2}mv^2 $: energy of motion, maximum at equilibrium.

Potential Energy

$ U = \tfrac{1}{2}kx^2 $: stored energy, maximum at maximum displacement.

Total Energy

$ E = K + U = \tfrac{1}{2}kA^2 $: stays constant throughout the motion.

Why Energy Matters in SHM

Energy methods let you solve SHM problems without tracking forces or accelerations at every instant. Since total energy is conserved (no friction), you can relate speed at one position to speed at another.

1

Find speed at any position: Use $ \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2 $.

2

Find amplitude from initial conditions: If you know $ v $ and $ x $, solve for $ A $.

3

Compare systems: Doubling the amplitude quadruples the total energy.

Core Energy Equations

Energy vs. Position Diagram

Energy vs. Time

A rotating peg whose shadow becomes an oscillator

Bench with lamp, peg wheel, blank screen and a spring connecting a fixed support to a sliding cart.

In the undamped spring model, kinetic and elastic potential energies exchange while their sum remains constant. Both energies oscillate at twice the displacement frequency.

Energy versus displacement

$Energy versus displacement. For an undamped spring oscillator, U/E = (x/A)² and K/E = 1 − (x/A)². The same energy fractions occur at positive and negative displacement of equal magnitude.$ For an undamped spring oscillator, U/E = (x/A)² and K/E = 1 − (x/A)². The same energy fractions occur at positive and negative displacement of equal magnitude.

Spring vs. Pendulum Energy

	Mass-Spring	Simple Pendulum
Potential Energy	$ \tfrac{1}{2}kx^2 $ (elastic)	$ mgh $ (gravitational)
At equilibrium	$ x = 0 $, spring relaxed	Lowest point, $ h = 0 $
At max displacement	$ x = \pm A $, $ v = 0 $	$ \theta = \theta_{\max} $, $ v = 0 $
Total Energy	$ \tfrac{1}{2}kA^2 $	$ mgL(1 - \cos\theta_{\max}) $
Period depends on	$ m $ and $ k $	$ L $ and $ g $

Speed at Any Position

From energy conservation, solve for the speed $ v $ at displacement $ x $:

$ x = 0 $

$ v = \omega A = v_{\max} $: maximum speed at equilibrium.

$ x = \pm A $

$ v = 0 $: momentarily at rest at the turning points.

How to Solve Energy Problems

1

Identify the system: What oscillates? Is it a spring ($ U = \tfrac{1}{2}kx^2 $) or pendulum ($ U = mgh $)?

2

Write total energy: Use $ E = \tfrac{1}{2}kA^2 $ or given initial conditions to find $ E $.

3

Set up conservation: $ K_i + U_i = K_f + U_f $. Plug in known values at two positions.

4

Solve for the unknown: Typically $ v $ at some position, or $ A $ from given speed and position.

5

Check limiting cases: At $ x = 0 $, all energy should be kinetic. At $ x = \pm A $, all potential.

Example 1: Finding Speed

A $0.5\;\text{kg}$ mass on a spring ($ k = 200\;\text{N/m} $) oscillates with amplitude $ A = 0.10\;\text{m} $. Find the speed when $ x = 0.06\;\text{m} $.

Example 2: Finding Amplitude

A mass of $2.0\;\text{kg}$ on a spring ($ k = 50\;\text{N/m} $) passes through equilibrium at $ v = 3.0\;\text{m/s} $. What is the amplitude?

Example 3: Energy Ratios

At what fraction of the amplitude is the kinetic energy equal to the potential energy?

What Happens When You Double $ A $?

Total Energy

$ E = \tfrac{1}{2}kA^2 $ so doubling $ A $ makes $ E $ go up by a factor of 4.

Max Speed

$ v_{\max} = A\omega $ so doubling $ A $ makes $ v_{\max} $ go up by a factor of 2.

Period

$ T = 2\pi\sqrt{m/k} $: period does not depend on amplitude. Unchanged!

Common Mistakes to Avoid

✕ Energy proportional to A

Total energy is proportional to $ A^2 $, not $ A $. Doubling amplitude quadruples energy.

✕ K = U at half amplitude

$ K = U $ at $ x = A/\sqrt{2} \approx 0.71A $, not at $ x = A/2 $.

✕ Period changes with A

For ideal SHM, the period $ T = 2\pi\sqrt{m/k} $ is completely independent of amplitude.

Summary of Key Relationships

Position	K	U	Speed
$ x = 0 $	$ \tfrac{1}{2}kA^2 $ (max)	0	$ v_{\max} = A\omega $
$ x = \pm\tfrac{A}{\sqrt{2}} $	$ \tfrac{1}{4}kA^2 $	$ \tfrac{1}{4}kA^2 $	$ \tfrac{v_{\max}}{\sqrt{2}} $
$ x = \pm A $	0	$ \tfrac{1}{2}kA^2 $ (max)	0

Question 1 of 5

A block on a spring oscillates with amplitude $ A $. If the amplitude is tripled, by what factor does the total mechanical energy change?

Since $ E = \tfrac{1}{2}kA^2 $, energy is proportional to $ A^2 $. Tripling $ A $ gives $ (3)^2 = 9 $ times the energy.

Question 2 of 5

In SHM, at which position is the kinetic energy at its maximum?

At equilibrium ($ x = 0 $), all potential energy has been converted to kinetic energy, so $ K $ is maximum and equals $ \tfrac{1}{2}kA^2 $.

Question 3 of 5

A $0.4\;\text{kg}$ block on a spring ($ k = 100\;\text{N/m} $) has amplitude $ 0.20\;\text{m} $. What is the maximum speed of the block?

$ v_{\max} = A\omega = A\sqrt{k/m} = 0.20\sqrt{100/0.4} = 0.20\sqrt{250} = 0.20 \times 15.81 \approx 3.16\;\text{m/s} $.

Question 4 of 5

At what displacement (as a fraction of $ A $) is the potential energy equal to three times the kinetic energy?

If $ U = 3K $ and $ U + K = E $, then $ U = \tfrac{3}{4}E = \tfrac{3}{4}\cdot\tfrac{1}{2}kA^2 $. So $ \tfrac{1}{2}kx^2 = \tfrac{3}{4}\cdot\tfrac{1}{2}kA^2 \implies x^2 = \tfrac{3}{4}A^2 \implies x = \tfrac{A\sqrt{3}}{2} $.

Question 5 of 5

A simple pendulum swings with small amplitude. If the mass is doubled but the length and amplitude remain the same, what happens to the total energy?

For a pendulum, $ E = mgL(1 - \cos\theta_{\max}) $. With the same $ L $, $ g $, and $ \theta_{\max} $, energy is directly proportional to $ m $. Doubling $ m $ doubles $ E $.

	Mass-Spring	Simple Pendulum
Potential Energy	\( \tfrac{1}{2}kx^2 \) (elastic)	\( mgh \) (gravitational)
At equilibrium	\( x = 0 \), spring relaxed	Lowest point, \( h = 0 \)
At max displacement	\( x = \pm A \), \( v = 0 \)	\( \theta = \theta_{\max} \), \( v = 0 \)
Total Energy	\( \tfrac{1}{2}kA^2 \)	\( mgL(1 - \cos\theta_{\max}) \)
Period depends on	\( m \) and \( k \)	\( L \) and \( g \)

Position	K	U	Speed
\( x = 0 \)	\( \tfrac{1}{2}kA^2 \) (max)	0	\( v_{\max} = A\omega \)
\( x = \pm\tfrac{A}{\sqrt{2}} \)	\( \tfrac{1}{4}kA^2 \)	\( \tfrac{1}{4}kA^2 \)	\( \tfrac{v_{\max}}{\sqrt{2}} \)
\( x = \pm A \)	0	\( \tfrac{1}{2}kA^2 \) (max)	0

Energy ofSimple Harmonic Oscillators

What is Simple Harmonic Motion?

The Three Energies in SHM

Why Energy Matters in SHM

Core Energy Equations

Energy vs. Position Diagram

Energy vs. Time

A rotating peg whose shadow becomes an oscillator

Energy versus displacement

Spring vs. Pendulum Energy

Speed at Any Position

How to Solve Energy Problems

Example 1: Finding Speed

Example 2: Finding Amplitude

Example 3: Energy Ratios

What Happens When You Double \( A \)?

Common Mistakes to Avoid

Summary of Key Relationships

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Energy of
Simple Harmonic Oscillators