# Energy in Simple Harmonic Motion

> AP Physics 1 · AP Physics 1 CED Unit 7
> Source: https://www.owlsprep.com/study/ap-physics-1-u7-energy-in-simple-harmonic-motion/

This module covers energy conservation in undamped simple harmonic motion (SHM), energy graphs, parameter dependence, and energy-based calculations for mass-spring and pendulum systems, regularly tested on AP Physics 1.

**Prerequisites:** [Conservation of mechanical energy for closed systems](https://www.owlsprep.com/study/ap-physics-1-u5-conservation-of-mechanical-energy/); Hooke's law and elastic potential energy; Basic dynamics of simple harmonic motion

## Learning objectives

- Apply conservation of mechanical energy to undamped simple harmonic motion systems
- Interpret energy-position graphs for SHM
- Calculate speed and amplitude for mass-spring and pendulum SHM systems
- Analyze how changing system parameters affects total energy and maximum speed

## Core Principles of Energy in SHM

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.

**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 $x = 0$.

## Conservation of Energy for SHM Calculations

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 $KE = 0$, so $E_{total} = PE_{max}$. For a horizontal mass-spring system, elastic potential energy is $PE_s = \frac{1}{2}kx^2$, 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 $PE = 0$ at the equilibrium (lowest) point. Total energy at maximum angle $\theta_{max}$ is $E_{total} = mgL(1-\cos\theta_{max})$, 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 $PE = 0$ at the equilibrium position for SHM energy problems to simplify math and avoid sign errors from extra offset terms.

## Energy Graphs for SHM

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

Potential energy $PE(x)$ is an upward-opening parabola with minimum $PE = 0$ at $x=0$ (equilibrium) and maximum PE at $x = \pm A$. Kinetic energy $KE(x)$ is a downward-opening parabola with maximum KE at $x=0$ and $KE=0$ 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 $KE(x)$ and $PE(x)$ 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.

## Parameter Dependence of SHM Energy

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 $E_{total} = \frac{1}{2}kA^2$, so $E_{total}$ 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 $E_{total} = \frac{1}{2}mv_{max}^2$, we get $v_{max} = A\sqrt{\frac{k}{m}}$, so increasing mass decreases $v_{max}$ even when total energy is constant.

For small-amplitude pendulums, total energy $E_{total} = mgL(1-\cos\theta_{max})$, 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 $E_{total} = \frac{1}{2}kA^2$, and both $k$ and $A$ are held constant: $E_{new} = E_{old}$
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 $\frac{1}{2}$.

**Check your understanding**

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

1. A simple pendulum oscillates with small amplitude between $\theta = -\theta_{max}$ and $\theta = +\theta_{max}$. Gravitational potential energy is set to zero at the equilibrium position ($\theta = 0$). Which of the following correctly describes the kinetic energy of the pendulum as a function of $\theta$ for small amplitudes?

   - KE is proportional to $\theta^2$, with maximum KE at $\theta = \pm\theta_{max}$
   - KE is proportional to $(\theta_{max}^2 - \theta^2)$, with maximum KE at $\theta = 0$
   - KE is proportional to $(\theta_{max} - \theta)$, with maximum KE at $\theta = 0$
   - KE is proportional to $\cos\theta$, with maximum KE at $\theta = \pm\theta_{max}$

   *Answer:* KE is proportional to $(\theta_{max}^2 - \theta^2)$, with maximum KE at $\theta = 0$

   *Why:* For small angles, $1-\cos\theta \approx \frac{\theta^2}{2}$, so $PE \approx \frac{mgL}{2}\theta^2$, so $KE = E_{total} - PE = \frac{mgL}{2}(\theta_{max}^2 - \theta^2)$, which is maximum when $\theta=0$.

> **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.

## Common pitfalls

- **Wrong:** Claims increasing mass of a mass-spring system (fixed amplitude) increases total oscillation energy.
  - Why it fails: Students confuse total energy with period: period increases with mass, so they incorrectly assume energy also increases.
  - Correct: Remember $E_{total} = \frac{1}{2}kA^2$, which depends only on $k$ and $A$, so energy stays the same regardless of mass at fixed $A$ and $k$.
- **Wrong:** Gets a negative potential energy for negative displacement $x$.
  - Why it fails: Students forget PE depends on displacement squared, so negative positions have the same PE as equal-magnitude positive positions.
  - Correct: Always square displacement when calculating PE for SHM, so PE is always non-negative for any $x$.
- **Wrong:** Draws sine/cosine curves for energy vs position graphs.
  - Why it fails: Students remember position and velocity are trigonometric functions of time, so incorrectly assume energy is also trigonometric in position.
  - Correct: 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:** Adds an extra gravitational PE term for vertical mass-spring systems, leading to incorrect total energy.
  - Why it fails: Students think gravity changes the energy equation, but forget that shifting the equilibrium position absorbs the gravitational term.
  - Correct: For vertical mass-spring SHM, measure $x$ from the new equilibrium position, and the energy equation is identical to horizontal SHM.
- **Wrong:** Memorizes $v_{max} = A\omega$ and uses it incorrectly after a system change like a collision that changes mass.
  - Why it fails: Students rely on memorized formulas instead of deriving from energy, leading to wrong amplitude calculations.
  - Correct: Always re-derive total energy and $v_{max}$ from conservation rules when the system changes, rather than relying on pre-derived relationships.

## Cheatsheet

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

## What's next

This topic connects the earlier work and energy unit to the simple harmonic motion unit, creating a critical foundation for all remaining topics in Unit 7. Immediately after mastering energy in SHM, you will study mechanical wave properties, including energy transport by traveling waves, which relies directly on the same energy conservation principles you learned here. Without mastering energy analysis for SHM, you will struggle to solve problems involving wave intensity and energy transfer by waves, and miss key connections between oscillation energy and wave behavior. This topic also reinforces energy conservation skills tested across all units of AP Physics 1, from rotational motion to DC circuits.

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