Energy in Simple Harmonic Motion

AP Physics 1· AP Physics 1 CED — Oscillations· 14 min read

1. Core Principles of Energy in SHM★★☆☆☆⏱ 3 min

Energy analysis of SHM connects two core units of AP Physics 1, accounting for ~2% of total exam score, and appears in both multiple-choice and free-response sections. For undamped SHM, where no friction or non-conservative forces do work, total mechanical energy is always conserved.

Energy continuously converts between kinetic energy (associated with motion of the oscillating mass) and potential energy (elastic for mass-spring systems, gravitational for pendulums) associated with displacement from equilibrium.

📘 Definition

Energy Distribution in SHM

For any undamped SHM system, total mechanical energy is constant: all energy is potential at maximum displacement (amplitude $A$ , velocity = 0), and all energy is kinetic at equilibrium (displacement = 0, speed is maximum).

Example:

A mass-spring system has all elastic potential energy at $x = \pm A$ , and all kinetic energy at $x = 0$ .

2. Conservation of Energy for SHM Calculations★★☆☆☆⏱ 4 min

The core relationship for all undamped SHM is:

E_{total} = KE + PE = \text{constant}

We find total energy by evaluating it at maximum displacement, where $K E = 0$ , so $E_{t o t a l} = P E_{ma x}$ . For a horizontal mass-spring system, elastic potential energy is $P E_{s} = \frac{1}{2} k x^{2}$ , so at $x = A$ :

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

For any displacement $x$ , the full energy balance is:

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2

Cancel the $\frac{1}{2}$ factor and rearrange to solve for speed $v$ at any $x$ :

v = \sqrt{\frac{k}{m}(A^2 - x^2)}

For small-amplitude simple pendulums, set $P E = 0$ at the equilibrium (lowest) point. Total energy at maximum angle $θ_{ma x}$ is $E_{t o t a l} = m g L (1 - cos θ_{ma x})$ , following the same energy conservation logic.

📐 Worked Example

A 0.2 kg mass attached to a spring with $k = 10$ N/m oscillates with amplitude $A = 0.3$ m on a frictionless horizontal surface. What is the speed of the mass when its displacement is $x = 0.2$ m from equilibrium?

1
Start with conservation of energy: total energy at amplitude equals sum of KE and PE at displacement $x$ :
$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
2
Cancel the common factor of $\frac{1}{2}$ from all terms to simplify:
$kA^2 = mv^2 + kx^2$
3
Rearrange to isolate $v^{2}$ :
$v^2 = \frac{k(A^2 - x^2)}{m}$
4
Substitute the given values:
$v^2 = \frac{10\left[(0.3)^2 - (0.2)^2\right]}{0.2} = \frac{10(0.05)}{0.2} = 2.5 \implies v = \sqrt{2.5} \approx 1.6 \text{ m/s}$

Exam tip:

Always set $P E = 0$ at the equilibrium position for SHM energy problems to simplify math and avoid sign errors from extra offset terms.

3. Energy Graphs for SHM★★★☆☆⏱ 3 min

AP Physics 1 regularly tests interpretation of energy graphs for SHM. For a mass-spring system, $P E (x) = \frac{1}{2} k x^{2}$ and $K E (x) = \frac{1}{2} k (A^{2} - x^{2})$ . Both are quadratic functions of position, so their energy vs position graphs are parabolas, not trigonometric curves.

Potential energy $P E (x)$ is an upward-opening parabola with minimum $P E = 0$ at $x = 0$ (equilibrium) and maximum PE at $x = \pm A$ . Kinetic energy $K E (x)$ is a downward-opening parabola with maximum KE at $x = 0$ and $K E = 0$ at $x = \pm A$ . Total energy is a horizontal straight line because it is constant for undamped motion.

📐 Worked Example

For an undamped horizontal mass-spring system oscillating between $x = - A$ and $x = + A$ , find the displacement $x$ where the $K E (x)$ and $P E (x)$ curves intersect.

1
At the intersection point, kinetic energy equals potential energy: $KE = PE
2
Substitute the energy expressions:
$\frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kx^2$
3
Cancel common terms ( $\frac{1}{2}$ and $k$ ) from both sides:
$A^2 - x^2 = x^2$
4
Rearrange to solve for $x$ :
$A^2 = 2x^2 \implies x = \pm \frac{A}{\sqrt{2}} \approx \pm 0.71A$

The curves intersect at two points, ~71% of the amplitude from equilibrium on either side.

Exam tip:

Do not mix up graph shapes: KE and PE are parabolas for energy vs position, and squared sine/cosine curves for energy vs time, which oscillate twice per full period.

4. Parameter Dependence of SHM Energy★★★☆☆⏱ 4 min

A common AP exam question asks how changing system parameters (amplitude, mass, spring constant, pendulum length) changes total energy or maximum speed.

For mass-spring systems, total energy $E_{t o t a l} = \frac{1}{2} k A^{2}$ , so $E_{t o t a l}$ depends only on $k$ and $A$ , not on mass $m$ . Changing mass at fixed amplitude does not change total energy, but it does change maximum speed: since $E_{t o t a l} = \frac{1}{2} m v_{ma x}^{2}$ , we get $v_{ma x} = A \frac{k}{m}$ , so increasing mass decreases $v_{ma x}$ even when total energy is constant.

For small-amplitude pendulums, total energy $E_{t o t a l} = m g L (1 - cos θ_{ma x})$ , so increasing mass, amplitude angle, or pendulum length all increase total energy. For lightly damped SHM, the only damped case covered in AP Physics 1, friction does non-conservative work so total energy and amplitude decrease over time.

📐 Worked Example

A horizontal mass-spring system oscillates with amplitude $A$ . A student replaces the 0.5 kg mass with a 2.0 kg mass, and stretches the spring to the same amplitude $A$ . By what factor does the maximum speed of the mass change, compared to the original system?

1
Total energy of the system is unchanged, because $E_{t o t a l} = \frac{1}{2} k A^{2}$ , and both $k$ and $A$ are held constant: $E_{n e w} = E_{o l d}$
2
Maximum speed occurs at equilibrium, where all energy is kinetic:
$E = \frac{1}{2}mv_{max}^2 \implies v_{max} = \sqrt{\frac{2E}{m}}$
3
Take the ratio of new to old maximum speed:
$\frac{v_{new}}{v_{old}} = \sqrt{\frac{2E/m_{new}}{2E/m_{old}}} = \sqrt{\frac{m_{old}}{m_{new}}} = \sqrt{\frac{0.5}{2.0}} = \frac{1}{2}$

The maximum speed is reduced by a factor of $\frac{1}{2}$ .

✓ Quick check

Test your understanding of SHM energy relationships with this AP-style multiple choice question:

A simple pendulum oscillates with small amplitude between $θ = - θ_{ma x}$ and $θ = + θ_{ma x}$ . Gravitational potential energy is set to zero at the equilibrium position ( $θ = 0$ ). Which of the following correctly describes the kinetic energy of the pendulum as a function of $θ$ for small amplitudes?
- KE is proportional to $θ^{2}$ , with maximum KE at $θ = \pm θ_{ma x}$
- KE is proportional to $(θ_{ma x}^{2} - θ^{2})$ , with maximum KE at $θ = 0$
- KE is proportional to $(θ_{ma x} - θ)$ , with maximum KE at $θ = 0$
- KE is proportional to $cos θ$ , with maximum KE at $θ = \pm θ_{ma x}$
Reveal answer
1 —
For small angles, $1 - cos θ \approx \frac{θ ^{2}}{2}$ , so $P E \approx \frac{m g L}{2} θ^{2}$ , so $K E = E_{t o t a l} - P E = \frac{m g L}{2} (θ_{ma x}^{2} - θ^{2})$ , which is maximum when $θ = 0$ .

Exam tip:

Always separate total energy and maximum speed when answering parameter change questions: total energy does not depend on mass for mass-spring SHM at fixed amplitude, but maximum speed does.

5. Common Pitfalls

Wrong move:

Claims increasing mass of a mass-spring system (fixed amplitude) increases total oscillation energy.

Why:

Students confuse total energy with period: period increases with mass, so they incorrectly assume energy also increases.

Correct move:

Remember $E_{t o t a l} = \frac{1}{2} k A^{2}$ , which depends only on $k$ and $A$ , so energy stays the same regardless of mass at fixed $A$ and $k$ .

Wrong move:

Gets a negative potential energy for negative displacement $x$ .

Why:

Students forget PE depends on displacement squared, so negative positions have the same PE as equal-magnitude positive positions.

Correct move:

Always square displacement when calculating PE for SHM, so PE is always non-negative for any $x$ .

Wrong move:

Draws sine/cosine curves for energy vs position graphs.

Why:

Students remember position and velocity are trigonometric functions of time, so incorrectly assume energy is also trigonometric in position.

Correct move:

For energy vs position, PE and KE are parabolas; for energy vs time, they are squared sine/cosine curves that oscillate twice per period.

Wrong move:

Adds an extra gravitational PE term for vertical mass-spring systems, leading to incorrect total energy.

Why:

Students think gravity changes the energy equation, but forget that shifting the equilibrium position absorbs the gravitational term.

Correct move:

For vertical mass-spring SHM, measure $x$ from the new equilibrium position, and the energy equation is identical to horizontal SHM.

Wrong move:

Memorizes $v_{ma x} = A ω$ and uses it incorrectly after a system change like a collision that changes mass.

Why:

Students rely on memorized formulas instead of deriving from energy, leading to wrong amplitude calculations.

Correct move:

Always re-derive total energy and $v_{ma x}$ from conservation rules when the system changes, rather than relying on pre-derived relationships.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Conservation of Energy (Undamped SHM)	$E_{t o t a l} = K E + P E = constant$	Applies only when no non-conservative work done
Total Energy (Mass-Spring SHM)	$E_{t o t a l} = \frac{1}{2} k A^{2}$	x measured from equilibrium; mass does not affect $E_{t o t a l}$ at fixed A
Energy Balance (Any Displacement)	$\frac{1}{2} k A^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}$	Cancel $\frac{1}{2}$ to simplify, solve for v at any x
Maximum Speed (Mass-Spring SHM)	$v_{ma x} = A \frac{k}{m}$	Occurs at x=0 (equilibrium), derived from energy conservation
Total Energy (Small-Angle Pendulum)	$E_{t o t a l} = m g L (1 - cos θ_{ma x})$	PE=0 set at equilibrium; all energy is PE at max angle
Maximum Speed (Pendulum SHM)	$v_{ma x} = 2 g L (1 - cos θ_{ma x})$	Mass cancels out, so maximum speed is independent of mass
PE vs Position (Mass-Spring)	$P E (x) = \frac{1}{2} k x^{2}$	Upward-opening parabola, maximum at $x = \pm A$
KE vs Position (Mass-Spring)	$K E (x) = \frac{1}{2} k (A^{2} - x^{2})$	Downward-opening parabola, maximum at x=0
Total Energy vs Position	$E_{t o t a l} (x) = \frac{1}{2} k A^{2}$	Horizontal straight line, constant for all x

What's Next

This topic connects the earlier work-and-energy unit to simple harmonic motion, and completes Unit 7 (Oscillations). The energy-conservation reasoning you practiced here reinforces skills tested across the course, from work-energy to rotational motion. Next you move on to the final unit, Unit 8: Fluids.