Energy in Simple Harmonic Motion — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Conservation of mechanical energy for undamped simple harmonic motion, kinetic and potential energy functions, energy-position and energy-time graphs, energy-based calculations for amplitude and maximum speed, and parameter dependence for mass-spring and pendulum systems.

You should already know: Conservation of mechanical energy for closed systems. Hooke’s law and elastic potential energy. Basic dynamics of simple harmonic motion.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Energy in Simple Harmonic Motion?

Energy analysis of simple harmonic motion (SHM) is a core connecting topic in AP Physics 1, accounting for ~2% of the total exam score weight per the AP Physics 1 CED, and appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. For undamped SHM—no friction or other non-conservative forces doing work—total mechanical energy of the oscillating system is always conserved. Energy continuously converts between two forms: kinetic energy (KE) associated with the motion of the oscillating mass, and potential energy (PE, elastic for horizontal/vertical mass-spring systems, gravitational for simple pendulums) associated with displacement from equilibrium. At maximum displacement (equal to the amplitude $A$ of the motion), velocity is zero, so all energy is potential; at equilibrium position, displacement is zero, so all energy is kinetic, and speed is maximum. Energy methods avoid solving for acceleration or force to find speed at a given position, making them a quick, reliable tool for many AP problems that connect energy conservation to SHM properties.

2. Conservation of Energy for SHM

The core relationship for all undamped SHM comes from conservation of mechanical energy: total energy $E_{total}=KE+PE=\text{constant}$. We can find the total energy by evaluating energy at the amplitude, where $KE=0$, so $E_{total}=P{E}_{max}$. For a horizontal mass-spring system, elastic potential energy is $P{E}_s=\frac{1}{2}k{x}^2$, where $x$ is displacement from equilibrium and $k$ is the spring constant. At $x=A$, $P{E}_{max}=\frac{1}{2}k{A}^2$, so total energy becomes: $$E_{total}=\frac{1}{2}k{A}^2$$ For any position $x$, we can write the full energy balance: $$\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$$ We can cancel the $\frac{1}{2}$ factor from all terms and rearrange to solve for speed $v$ at any $x$: $$v=\sqrt{\frac{k}{m}(A^2-x^2)}$$ This formula can be derived directly from energy conservation, so you do not need to memorize it if you remember the core energy balance. For a simple pendulum, potential energy is gravitational: if we set $PE=0$ at the lowest (equilibrium) point, for small amplitude ${\theta }_{max}$, the height of the bob at maximum displacement is $h=L(1-\cos {\theta }_{max})$, so total energy $E_{total}=mgL(1-\cos {\theta }_{max})$, following the same logic: all energy is PE at maximum displacement, all KE at equilibrium.

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 the sum of KE and PE at displacement $x$: $\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$
2. Cancel the common factor of $\frac{1}{2}$ from all terms to simplify: $k{A}^2=m{v}^2+k{x}^2$
3. Rearrange to isolate $v^2$: $v^2=\frac{k(A^2-x^2)}{m}$
4. Substitute values: $v^2=\frac{10\left[(0.3{)}^2-(0.2{)}^2\right]}{0.2}=\frac{10(0.05)}{0.2}=2.5$, so $v=\sqrt{2.5}\approx 1.6$ m/s.

Exam tip: Always set $PE=0$ at the equilibrium position for SHM energy problems—this simplifies the math and avoids sign errors from extra offset terms.

3. Energy Graphs for SHM

AP Physics 1 regularly tests graphs of KE and PE for SHM, so understanding their shape and key features is critical for exam success. From the energy balance above, $KE(x)=E_{total}-PE(x)=\frac{1}{2}k(A^2-x^2)$ for a mass-spring system, while $PE(x)=\frac{1}{2}k{x}^2$. Both KE and PE are quadratic functions of position $x$, meaning their graphs are parabolas, not trigonometric curves. $PE(x)$ is an upward-opening parabola with minimum at $x=0$ ($PE=0$ at equilibrium, maximum $PE=\frac{1}{2}k{A}^2$ at $x=\pm A$). $KE(x)$ is a downward-opening parabola with maximum $KE=\frac{1}{2}k{A}^2$ at $x=0$, and $KE=0$ at $x=\pm A$. Total energy is a horizontal straight line at $E=\frac{1}{2}k{A}^2$, since it is constant for undamped motion. For small-amplitude pendulums, the shape of the graphs is nearly identical, because gravitational PE is proportional to $x^2$ for small angles.

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: $\frac{1}{2}k(A^2-x^2)=\frac{1}{2}k{x}^2$
3. Cancel common terms ($\frac{1}{2}$ and $k$) from both sides: $A^2-x^2=x^2$
4. Rearrange to solve for $x$: $A^2=2x^2⟹x=\pm \frac{A}{\sqrt{2}}\approx \pm 0.71A$

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

Exam tip: When asked to identify or draw energy graphs, remember: KE and PE are parabolas for energy vs position, while they are squared sine/cosine curves for energy vs time—do not mix these up.

4. Parameter Dependence of Total Energy

A common AP exam question asks how changing a system parameter (amplitude, mass, spring constant, pendulum length) changes the total energy or maximum speed of an oscillating system. For mass-spring systems, total energy is $E_{total}=\frac{1}{2}k{A}^2$, so $E_{total}$ depends only on $k$ and $A$, not on mass $m$. This is a frequent point of confusion: changing the mass of the oscillating object, while keeping amplitude the same, does not change total energy. It does change maximum speed, however: since $v_{max}$ comes from $E_{total}=\frac{1}{2}m{v}_{max}^2$, $v_{max}=\sqrt{\frac{2E_{total}}{m}}=A\sqrt{\frac{k}{m}}$, so increasing mass decreases $v_{max}$ even though total energy stays the same. For simple pendulums, total energy is $E_{total}=mgL(1-\cos {\theta }_{max})$, 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 $E_{total}=\frac{1}{2}k{A}^2$, and both $k$ and $A$ are held constant: $E_{new}=E_{old}$
2. Maximum speed occurs at equilibrium, where all energy is kinetic: $E=\frac{1}{2}m{v}_{max}^2⟹v_{max}=\sqrt{\frac{2E}{m}}$
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 $\frac{1}{2}$.

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 (and how to avoid them)

• Wrong move: Claims that increasing the mass of a mass-spring system (at fixed amplitude) increases the total energy of the oscillation. Why: Students confuse total energy with period—period increases with mass, so they incorrectly assume energy also increases. Correct move: Remember total energy is $E_{total}=\frac{1}{2}k{A}^2$, which depends only on $k$ and $A$, so energy stays the same for fixed $A$ and $k$ regardless of mass.
• 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 positive positions of equal magnitude. 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 they 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 $v_{max}=A\omega$ and uses it incorrectly after 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 $v_{max}$ from conservation rules when the system changes, rather than relying on pre-derived relationships.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

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? A) KE is proportional to ${\theta }^2$, with maximum KE at $\theta =\pm {\theta }_{max}$ B) KE is proportional to $({\theta }_{max}^2-{\theta }^2)$, with maximum KE at $\theta =0$ C) KE is proportional to $({\theta }_{max}-\theta )$, with maximum KE at $\theta =0$ D) KE is proportional to $\cos \theta$, with maximum KE at $\theta =\pm {\theta }_{max}$

Worked Solution: Start with conservation of energy: $E_{total}=KE+PE$, so $KE=E_{total}-PE$. For small angles, $1-\cos \theta \approx \frac{{\theta }^2}{2}$, so $PE=mgL(1-\cos \theta )\approx \frac{mgL}{2}{\theta }^2$, and $E_{total}=\frac{mgL}{2}{\theta }_{max}^2$. Substituting gives $KE=\frac{mgL}{2}({\theta }_{max}^2-{\theta }^2)$, so KE is proportional to $({\theta }_{max}^2-{\theta }^2)$, with maximum KE at $\theta =0$ where PE is minimum. Correct answer: B.

Question 2 (Free Response)

A 0.4 kg block attached to a vertical spring with $k=20$ N/m is pulled down 0.15 m from its equilibrium position and released from rest, resulting in undamped SHM. (a) Show that the total mechanical energy of the oscillation, measured relative to the equilibrium position, is 0.225 J. (b) Calculate the maximum kinetic energy of the block during the oscillation. (c) A small piece of putty with mass 0.1 kg is dropped vertically onto the block when the block is at its equilibrium position, and sticks to it. The collision happens very quickly, so the position of the block does not change during the collision. Does the amplitude of the oscillation increase, decrease, or stay the same after the collision? Justify your answer.

Worked Solution: (a) When released from rest at displacement $A=0.15$ m, kinetic energy is zero, so total energy equals potential energy relative to equilibrium. For vertical SHM, with $x$ measured from equilibrium: $$E_{total}=\frac{1}{2}k{A}^2=\frac{1}{2}(20\text{N/m})(0.15\text{m}{)}^2=10(0.0225)=0.225\text{J}$$ This matches the required value. (b) Maximum kinetic energy occurs at equilibrium, where potential energy relative to equilibrium is zero. By conservation of energy for the undamped system, all energy is kinetic at this point, so $K{E}_{max}=E_{total}=0.225\text{J}$. (c) The amplitude will decrease. Justification: The collision is perfectly inelastic, so kinetic energy is lost during the collision. At equilibrium, potential energy is zero, so the new total energy of the oscillating system equals the kinetic energy immediately after collision. Momentum is conserved during the collision, so the new speed is lower than the original maximum speed. Even after accounting for the increased mass, the new kinetic energy is less than the original total energy. Since amplitude is $A=\sqrt{\frac{2E_{total}}{k}}$, a lower total energy gives a smaller amplitude.

Question 3 (Application / Real-World Style)

A playground swing can be modeled as a 2.5 m long simple pendulum. A student pulls the swing back to an angle of 15 degrees from vertical before releasing it from rest. Assuming no air resistance, calculate the maximum speed of the swing at the lowest point, if the total mass of the student and swing is 55 kg. Interpret your result in context.

Worked Solution: Use conservation of energy: gravitational potential energy at maximum angle converts entirely to kinetic energy at the lowest point: $mgh=\frac{1}{2}m{v}_{max}^2$. The height difference is $h=L(1-\cos \theta )$, so mass cancels from both sides: $$v_{max}=\sqrt{2gL(1-\cos \theta )}$$ Substitute values: $g=9.8{\text{m/s}}^2$, $L=2.5\text{m}$, $\cos 15^{\circ }\approx 0.9659$, so: $$v_{max}=\sqrt{2(9.8)(2.5)(1-0.9659)}=\sqrt{1.67}\approx 1.3\text{m/s}$$ This maximum speed of ~1.3 m/s (≈4.7 km/h) is a slow, safe speed for a playground swing, consistent with real-world experience.

7. Quick Reference Cheatsheet

8. 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 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 mastered here. Without mastering energy analysis for SHM, you will struggle to solve problems involving wave intensity and energy transfer by waves, and may miss key connections between oscillation energy and wave behavior. This topic also reinforces energy conservation skills that are tested across all units of AP Physics 1, from rotational motion to DC circuits.

Next topics to study:

• Simple Harmonic Motion Period and Frequency
• Mechanical Wave Properties
• Wave Interference and Standing Waves
• Conservation of Mechanical Energy