Energy of a simple harmonic oscillator

What you'll learn

How kinetic energy and potential energy interchange during simple harmonic motion.
Why the total mechanical energy of an ideal oscillator stays constant.
How to use the SHM energy equations involving $m$ , $ω\omega$ , $A$ and $x$ .
How to interpret and sketch energy-displacement graphs for a simple harmonic oscillator.

The SHM ideas you need first

A simple harmonic oscillator is a system that undergoes simple harmonic motion: its acceleration is directly proportional to its displacement from equilibrium and acts in the opposite direction.

The defining equation is:

-\omega^2 x

where $a$ is acceleration in metres per second squared, $x$ is displacement from equilibrium in metres, and $ω\omega$ is angular frequency in radians per second.

Definition

Key positions in SHM

The equilibrium position is where the resultant force is zero. The amplitude $A$ is the maximum displacement from equilibrium. The turning points are at $x = + A$ and $x = - A$ , where the oscillator momentarily has zero speed.

For a particle of mass $m$ , Newton’s second law gives:

-m\omega^2x

So the restoring force is proportional to displacement. This is why SHM energy graphs have smooth, symmetric curves.

Energy forms in an oscillator

In SHM, energy moves between two stores:

Kinetic energy $E_k$ : energy due to motion, given by $Ek=12mv2E_k = \frac{1}{2}mv^2$ .
Potential energy $E_p$ : energy stored because the oscillator is displaced from equilibrium.

For a horizontal mass-spring oscillator, the potential energy is elastic strain energy. For a pendulum, it is gravitational potential energy. In both cases, for small oscillations that obey SHM, the potential energy is lowest at equilibrium and greatest at the turning points.

Mass-spring oscillator showing energy transfer at the turning points and equilibrium

Key Idea

Energy interchange

In an ideal undamped oscillator, total mechanical energy is constant. At the turning points, all the energy is potential. At equilibrium, the potential energy is minimum and the kinetic energy is maximum.

The word ideal matters: it means no energy is transferred away to the surroundings by friction, air resistance or internal heating.

Potential energy as a function of displacement

Because the restoring force is proportional to displacement, the work done to move the oscillator slowly from equilibrium to displacement $x$ is the area under a straight-line force-displacement graph.

That area is triangular, so:

Ep=12mω2x2E_p = \frac{1}{2}m\omega^2x^2

This assumes we choose the potential energy at equilibrium to be zero.

For a spring oscillator, since $mω2=km\omega^2 = k$ , you may also write:

Ep=12kx2E_p = \frac{1}{2}kx^2

where $k$ is the spring constant in newtons per metre.

Tip

Why the square matters

Potential energy depends on $x^2$ , so the potential energy is the same at $+ x$ and $- x$ . Being equally far left or right of equilibrium stores the same energy.

Total energy of the oscillator

The maximum potential energy occurs at the turning points, where $\pm A$ and the speed is zero.

So the total mechanical energy is:

ET=12mω2A2E_T = \frac{1}{2}m\omega^2A^2

For a spring oscillator:

ET=12kA2E_T = \frac{1}{2}kA^2

This tells you an important scaling rule: if the amplitude doubles, the total energy becomes four times larger.

Kinetic energy at any displacement

At any displacement, the total energy is split between potential and kinetic energy:

E_T = E_k + E_p

So:

Ek=ET−Ep=12mω2A2−12mω2x2=12mω2(A2−x2)\begin{aligned} E_k &= E_T - E_p \\ &= \frac{1}{2}m\omega^2A^2 - \frac{1}{2}m\omega^2x^2 \\ &= \frac{1}{2}m\omega^2(A^2 - x^2) \end{aligned}

This is largest at $x = 0$ and zero at $\pm A$ .

Common Mistake

Using x instead of A for total energy

The total energy depends on the amplitude $A$ , not the current displacement $x$ . Use $ET=12mω2A2E_T = \frac{1}{2}m\omega^2A^2$ for the total, then use $Ep=12mω2x2E_p = \frac{1}{2}m\omega^2x^2$ for the energy stored at a particular position.

Example

Calculating energy at a displacement

A mass of 0.150 kg oscillates with frequency 2.00 Hz and amplitude 5.0 cm. Calculate the total energy, the potential energy and the kinetic energy when the displacement is 3.0 cm.

Convert the distances into metres and calculate angular frequency:
$s−1\begin{aligned} A &= 5.0\text{ cm} = 0.050\text{ m} \\ x &= 3.0\text{ cm} = 0.030\text{ m} \\ \omega &= 2\pi f = 2\pi \times 2.00\text{ Hz} = 12.6\text{ rad s}^{-1} \end{aligned}$
Use the amplitude to calculate the total energy:
$J\begin{aligned} E_T &= \frac{1}{2}m\omega^2A^2 \\ &= \frac{1}{2}(0.150\text{ kg})(12.6\text{ rad s}^{-1})^2(0.050\text{ m})^2 \\ &= 2.96 \times 10^{-2}\text{ J} \end{aligned}$
Use the current displacement to calculate the potential energy:
$J\begin{aligned} E_p &= \frac{1}{2}m\omega^2x^2 \\ &= \frac{1}{2}(0.150\text{ kg})(12.6\text{ rad s}^{-1})^2(0.030\text{ m})^2 \\ &= 1.07 \times 10^{-2}\text{ J} \end{aligned}$
Subtract to find the kinetic energy:
$J\begin{aligned} E_k &= E_T - E_p \\ &= 2.96 \times 10^{-2}\text{ J} - 1.07 \times 10^{-2}\text{ J} \\ &= 1.89 \times 10^{-2}\text{ J} \end{aligned}$

Energy-displacement graphs

An energy-displacement graph shows how energy depends on position $x$ , not on time.

For a simple harmonic oscillator:

Total energy $E_T$ is a horizontal line.
Potential energy $E_p$ is an upward-opening parabola.
Kinetic energy $E_k$ is a downward-opening parabola.
Both $E_p$ and $E_k$ are symmetric about $x = 0$ .
The allowed motion is only between $x = - A$ and $x = + A$ .

Energy-displacement graph for a simple harmonic oscillator showing total, kinetic and potential energy

At the turning points:

\pm A,\quad E_k = 0,\quad E_p = E_T

At equilibrium:

0,\quad E_p = 0,\quad E_k = E_T

Common Mistake

This is not a time graph

An energy-displacement graph does not tell you whether the oscillator is moving left or right. The oscillator passes most displacements twice per cycle, once in each direction, with the same kinetic and potential energies.

Reading graph values using fractions

You can often solve energy graph questions without knowing $m$ or $ω\omega$ .

Since:

Ep=12mω2x2E_p = \frac{1}{2}m\omega^2x^2

and:

ET=12mω2A2E_T = \frac{1}{2}m\omega^2A^2

dividing gives:

EpET=x2A2\frac{E_p}{E_T} = \frac{x^2}{A^2}

So:

Ep=ET(xA)2E_p = E_T\left(\frac{x}{A}\right)^2

Then:

E_k = E_T - E_p

Example

Reading an energy-displacement graph

An oscillator has amplitude 8.0 cm and total energy 0.32 J. Find the kinetic and potential energies when the displacement is 4.0 cm.

Compare the displacement with the amplitude:
$cm=0.50\frac{x}{A} = \frac{4.0\text{ cm}}{8.0\text{ cm}} = 0.50$
Square the fraction to find the fraction of total energy stored as potential energy:
$EpET=(0.50)2=0.25\frac{E_p}{E_T} = (0.50)^2 = 0.25$
Calculate the potential energy:
$JE_p = 0.25 \times 0.32\text{ J} = 0.080\text{ J}$
Use conservation of energy to calculate the kinetic energy:
$JE_k = 0.32\text{ J} - 0.080\text{ J} = 0.24\text{ J}$

A quick method for exam calculations

When you see an SHM energy question, use this order:

Identify the amplitude $A$ and the displacement $x$ separately.
Convert all distances into metres.
Find $ω\omega$ if needed, usually from $ω=2πf\omega = 2\pi f$ .
Calculate total energy using $A$ .
Calculate potential energy using $x$ .
Calculate kinetic energy by subtraction.

Exam technique

In the exam

Label the turning points $x = - A$ and $x = + A$ before interpreting an energy graph.
Check whether the question gives amplitude, displacement, frequency, period or spring constant; choose the matching energy equation.
Sanity-check your answer: $E_k$ cannot be negative, and $E_p$ cannot exceed $E_T$ for a valid displacement.

Self review

Check yourself

Why is kinetic energy zero at the turning points but maximum at equilibrium?
If the amplitude of an ideal oscillator is tripled, what happens to its total energy?
On an energy-displacement graph, why do the curves have the same values at $+ x$ and $- x$ ?

What you'll learn

How kinetic energy and potential energy interchange during simple harmonic motion.
Why the total mechanical energy of an ideal oscillator stays constant.
How to use the SHM energy equations involving $m$ , $ω\omega$ , $A$ and $x$ .
How to interpret and sketch energy-displacement graphs for a simple harmonic oscillator.

The SHM ideas you need first

The defining equation is:

-\omega^2 x

where $a$ is acceleration in metres per second squared, $x$ is displacement from equilibrium in metres, and $ω\omega$ is angular frequency in radians per second.

Definition

Key positions in SHM

For a particle of mass $m$ , Newton’s second law gives:

-m\omega^2x

So the restoring force is proportional to displacement. This is why SHM energy graphs have smooth, symmetric curves.

Energy forms in an oscillator

In SHM, energy moves between two stores:

Kinetic energy $E_k$ : energy due to motion, given by $Ek=12mv2E_k = \frac{1}{2}mv^2$ .
Potential energy $E_p$ : energy stored because the oscillator is displaced from equilibrium.

Mass-spring oscillator showing energy transfer at the turning points and equilibrium

Key Idea

Energy interchange

The word ideal matters: it means no energy is transferred away to the surroundings by friction, air resistance or internal heating.

Potential energy as a function of displacement

That area is triangular, so:

Ep=12mω2x2E_p = \frac{1}{2}m\omega^2x^2

This assumes we choose the potential energy at equilibrium to be zero.

For a spring oscillator, since $mω2=km\omega^2 = k$ , you may also write:

Ep=12kx2E_p = \frac{1}{2}kx^2

where $k$ is the spring constant in newtons per metre.

Tip

Why the square matters

Potential energy depends on $x^2$ , so the potential energy is the same at $+ x$ and $- x$ . Being equally far left or right of equilibrium stores the same energy.

Total energy of the oscillator

The maximum potential energy occurs at the turning points, where $\pm A$ and the speed is zero.

So the total mechanical energy is:

ET=12mω2A2E_T = \frac{1}{2}m\omega^2A^2

For a spring oscillator:

ET=12kA2E_T = \frac{1}{2}kA^2

This tells you an important scaling rule: if the amplitude doubles, the total energy becomes four times larger.

Kinetic energy at any displacement

At any displacement, the total energy is split between potential and kinetic energy:

E_T = E_k + E_p

So:

Ek=ET−Ep=12mω2A2−12mω2x2=12mω2(A2−x2)\begin{aligned} E_k &= E_T - E_p \\ &= \frac{1}{2}m\omega^2A^2 - \frac{1}{2}m\omega^2x^2 \\ &= \frac{1}{2}m\omega^2(A^2 - x^2) \end{aligned}

This is largest at $x = 0$ and zero at $\pm A$ .

Common Mistake

Using x instead of A for total energy

Example

Calculating energy at a displacement

A mass of 0.150 kg oscillates with frequency 2.00 Hz and amplitude 5.0 cm. Calculate the total energy, the potential energy and the kinetic energy when the displacement is 3.0 cm.

Convert the distances into metres and calculate angular frequency:
$s−1\begin{aligned} A &= 5.0\text{ cm} = 0.050\text{ m} \\ x &= 3.0\text{ cm} = 0.030\text{ m} \\ \omega &= 2\pi f = 2\pi \times 2.00\text{ Hz} = 12.6\text{ rad s}^{-1} \end{aligned}$
Use the amplitude to calculate the total energy:
$J\begin{aligned} E_T &= \frac{1}{2}m\omega^2A^2 \\ &= \frac{1}{2}(0.150\text{ kg})(12.6\text{ rad s}^{-1})^2(0.050\text{ m})^2 \\ &= 2.96 \times 10^{-2}\text{ J} \end{aligned}$
Use the current displacement to calculate the potential energy:
$J\begin{aligned} E_p &= \frac{1}{2}m\omega^2x^2 \\ &= \frac{1}{2}(0.150\text{ kg})(12.6\text{ rad s}^{-1})^2(0.030\text{ m})^2 \\ &= 1.07 \times 10^{-2}\text{ J} \end{aligned}$
Subtract to find the kinetic energy:
$J\begin{aligned} E_k &= E_T - E_p \\ &= 2.96 \times 10^{-2}\text{ J} - 1.07 \times 10^{-2}\text{ J} \\ &= 1.89 \times 10^{-2}\text{ J} \end{aligned}$

Energy-displacement graphs

An energy-displacement graph shows how energy depends on position $x$ , not on time.

For a simple harmonic oscillator:

Total energy $E_T$ is a horizontal line.
Potential energy $E_p$ is an upward-opening parabola.
Kinetic energy $E_k$ is a downward-opening parabola.
Both $E_p$ and $E_k$ are symmetric about $x = 0$ .
The allowed motion is only between $x = - A$ and $x = + A$ .

Energy-displacement graph for a simple harmonic oscillator showing total, kinetic and potential energy

At the turning points:

\pm A,\quad E_k = 0,\quad E_p = E_T

At equilibrium:

0,\quad E_p = 0,\quad E_k = E_T

Common Mistake

This is not a time graph

Reading graph values using fractions

You can often solve energy graph questions without knowing $m$ or $ω\omega$ .

Since:

Ep=12mω2x2E_p = \frac{1}{2}m\omega^2x^2

and:

ET=12mω2A2E_T = \frac{1}{2}m\omega^2A^2

dividing gives:

EpET=x2A2\frac{E_p}{E_T} = \frac{x^2}{A^2}

So:

Ep=ET(xA)2E_p = E_T\left(\frac{x}{A}\right)^2

Then:

E_k = E_T - E_p

Example

Reading an energy-displacement graph

An oscillator has amplitude 8.0 cm and total energy 0.32 J. Find the kinetic and potential energies when the displacement is 4.0 cm.

Compare the displacement with the amplitude:
$cm=0.50\frac{x}{A} = \frac{4.0\text{ cm}}{8.0\text{ cm}} = 0.50$
Square the fraction to find the fraction of total energy stored as potential energy:
$EpET=(0.50)2=0.25\frac{E_p}{E_T} = (0.50)^2 = 0.25$
Calculate the potential energy:
$JE_p = 0.25 \times 0.32\text{ J} = 0.080\text{ J}$
Use conservation of energy to calculate the kinetic energy:
$JE_k = 0.32\text{ J} - 0.080\text{ J} = 0.24\text{ J}$

A quick method for exam calculations

When you see an SHM energy question, use this order:

Identify the amplitude $A$ and the displacement $x$ separately.
Convert all distances into metres.
Find $ω\omega$ if needed, usually from $ω=2πf\omega = 2\pi f$ .
Calculate total energy using $A$ .
Calculate potential energy using $x$ .
Calculate kinetic energy by subtraction.

Exam technique

In the exam

Label the turning points $x = - A$ and $x = + A$ before interpreting an energy graph.
Check whether the question gives amplitude, displacement, frequency, period or spring constant; choose the matching energy equation.
Sanity-check your answer: $E_k$ cannot be negative, and $E_p$ cannot exceed $E_T$ for a valid displacement.

Self review

Check yourself

Why is kinetic energy zero at the turning points but maximum at equilibrium?
If the amplitude of an ideal oscillator is tripled, what happens to its total energy?
On an energy-displacement graph, why do the curves have the same values at $+ x$ and $- x$ ?

What you'll learn

The SHM ideas you need first

Energy forms in an oscillator

Potential energy as a function of displacement

Total energy of the oscillator

Kinetic energy at any displacement

Energy-displacement graphs

Reading graph values using fractions

A quick method for exam calculations

Review Energy of a simple harmonic oscillator by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Energy of a simple harmonic oscillator

What you'll learn

The SHM ideas you need first

Energy forms in an oscillator

Potential energy as a function of displacement

Total energy of the oscillator

Kinetic energy at any displacement

Energy-displacement graphs

Reading graph values using fractions

A quick method for exam calculations

Review Energy of a simple harmonic oscillator by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Energy of a simple harmonic oscillator

What you'll learn

The SHM ideas you need first

Energy forms in an oscillator

Potential energy as a function of displacement

Total energy of the oscillator

Kinetic energy at any displacement

Energy-displacement graphs

Reading graph values using fractions

A quick method for exam calculations

Review Energy of a simple harmonic oscillator by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Energy of a simple harmonic oscillator Revision Guide

Energy of a simple harmonic oscillator

What you'll learn

The SHM ideas you need first

Energy forms in an oscillator

Potential energy as a function of displacement

Total energy of the oscillator

Kinetic energy at any displacement

Energy-displacement graphs

Reading graph values using fractions

A quick method for exam calculations

Review Energy of a simple harmonic oscillator by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Energy of a simple harmonic oscillator Revision Guide