Mass-spring systems and simple pendulum — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Angular frequency and period calculation for mass-spring systems, small-angle approximation for simple pendulums, energy conservation in undamped oscillations, and the effect of constant forces on equilibrium position and period.

You should already know: Hooke's law and Newton's second law, definition of simple harmonic motion, small-angle approximation for trigonometric functions.

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 C: Mechanics 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 Mass-spring systems and simple pendulum?

Mass-spring systems and simple pendulums are the two core physical examples of systems that undergo simple harmonic motion (SHM), the focus of Unit 6 Oscillations in the AP Physics C: Mechanics CED. This subtopic accounts for approximately 6-8% of the total AP exam score, and it appears regularly on both multiple choice (MCQ) and free response (FRQ) sections, often combined with energy, force, or differential equation questions. A mass-spring system consists of a mass attached to an ideal massless spring, with a restoring force proportional to displacement from equilibrium that obeys Hooke's law. A simple pendulum consists of a point mass attached to a massless, inextensible string fixed at one end, with a gravitational restoring force that approximates Hooke's law for small displacements. Standard notation for this topic uses $m$ for oscillating mass, $k$ for spring constant, $L$ for pendulum length, $T$ for period, $\omega$ for angular frequency, $A$ for amplitude, and $\theta$ for angular displacement of the pendulum. This topic is the foundation for all further work on oscillations, so mastery of its core relationships is required for exam success.

2. Angular Frequency and Period for Mass-Spring Oscillators

To find the period of a mass-spring oscillator, we start with Newton's second law for displacements from equilibrium. For any mass-spring system, the restoring force follows Hooke's law: $F=-kx$, where $x$ is displacement from equilibrium. Substituting into Newton's second law $F=ma$ gives: $$-kx=ma⟹a=-\left(\frac{k}{m}\right)x$$ By definition, SHM has acceleration of the form $a=-{\omega }^{2}x$, where $\omega$ is angular frequency. Matching terms gives $\omega =\sqrt{\frac{k}{m}}$, so period $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$. A common point of confusion is the effect of gravity for vertical or incline mass-spring systems. Gravity adds a constant force to the mass, which only shifts the equilibrium position of the system. When we measure displacement from this new equilibrium, gravity cancels out of the restoring force, so the same formula for $\omega$ and $T$ holds regardless of orientation.

Worked Example

A 200 g block is attached to a vertical spring with spring constant $k=5.0\text{N/m}$. When the block is hung at rest, the spring stretches 39.2 cm from its unstretched length to reach equilibrium. What is the period of small amplitude oscillations of the block?

1. Step 1: Recognize that gravity only shifts the equilibrium position of a vertical spring, it does not change the restoring force constant or period. The equilibrium stretch given is a distractor for this calculation.
2. Step 2: Convert mass to SI units: $m=200\text{g}=0.200\text{kg}$.
3. Step 3: Use the standard period formula for mass-spring oscillations: $T=2\pi \sqrt{\frac{m}{k}}$.
4. Step 4: Substitute values: $T=2\pi \sqrt{\frac{0.200}{5.0}}=2\pi \sqrt{0.04}=0.4\pi \approx 1.3\text{s}$.

Exam tip: If a problem gives you the equilibrium stretch of a vertical spring alongside mass and spring constant, the stretch is almost always a distractor for period calculation. You only need $m$ and $k$ to find $T$.

3. The Simple Pendulum and Small-Angle Approximation

A simple pendulum consists of a point mass $m$ on a massless, inextensible string of length $L$ fixed at one end. The restoring force acts tangential to the arc of the pendulum's motion, and is given by $F=-mg\sin \theta$, where $\theta$ is the angular displacement from equilibrium. For SHM, we need a restoring force proportional to displacement. For small angles (typically less than ~10°), the approximation $\sin \theta \approx \theta$ (for $\theta$ in radians) holds, so the restoring force simplifies to: $$F\approx -mg\theta$$ The arc length displacement from equilibrium is $s=L\theta$, so $\theta =\frac{s}{L}$. Substituting gives $F=-\left(\frac{mg}{L}\right)s$, which matches Hooke's law with effective spring constant $k_{\text{eff}}=\frac{mg}{L}$. Using the SHM angular frequency formula $\omega =\sqrt{\frac{k_{\text{eff}}}{m}}$ gives $\omega =\sqrt{\frac{g}{L}}$, so period: $$T=2\pi \sqrt{\frac{L}{g}}$$ Note that the mass cancels out, so the period of a simple pendulum is independent of the mass of the bob and (for small angles) independent of amplitude.

Worked Example

A simple pendulum on Earth ($g=9.81{\text{m/s}}^2$) has a measured period of $2.0\text{s}$ for small oscillations. What is the approximate period of the same pendulum for small oscillations on the surface of Mars, where gravitational acceleration is 38% of Earth's $g$?

1. Step 1: Write the period formula for small-angle simple pendulum for both locations: $T_E=2\pi \sqrt{\frac{L}{g_E}}$ and $T_M=2\pi \sqrt{\frac{L}{g_M}}$.
2. Step 2: Divide $T_M$ by $T_E$ to eliminate constants: $\frac{T_M}{T_E}=\sqrt{\frac{g_E}{g_M}}=\frac{1}{\sqrt{g_M/g_E}}=\frac{1}{\sqrt{0.38}}\approx 1.62$.
3. Step 3: Substitute $T_E=2.0\text{s}$: $T_M=1.62\times 2.0\approx 3.2\text{s}$.

Exam tip: Always confirm the problem asks for the period of small oscillations before using $T=2\pi \sqrt{L/g}$. For large angular displacements, the actual period is always longer than the small-angle prediction.

4. Energy Conservation in Oscillating Systems

For undamped (frictionless) SHM, total mechanical energy is conserved. This relationship is frequently used to solve for speed at a given displacement, faster than integrating the SHM differential equation. For mass-spring systems, total energy is the sum of kinetic energy and elastic potential energy: $$E=KE+U=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\text{constant}$$ At the maximum displacement (turning point), $x=A$ (amplitude) and $v=0$, so all energy is potential: $E=\frac{1}{2}k{A}^2$. This gives the energy conservation relation: $$\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2$$ For simple pendulums, total energy is the sum of kinetic energy and gravitational potential energy. Taking equilibrium as the zero potential reference, at maximum angular displacement ${\theta }_{\text{max}}$, the bob rises by $h=L(1-\cos {\theta }_{\text{max}})$, so maximum potential energy is $mgh=mgL(1-\cos {\theta }_{\text{max}})$, which equals total energy.

Worked Example

A horizontal mass-spring system on a frictionless surface has $m=0.50\text{kg}$, $k=200\text{N/m}$, and amplitude of oscillation $A=0.10\text{m}$. What is the speed of the mass when it is $0.05\text{m}$ from equilibrium?

1. Step 1: Use conservation of mechanical energy for undamped SHM: total energy at any displacement equals maximum potential energy at the turning point, giving $\frac{1}{2}k{A}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$.
2. Step 2: Cancel the common factor of $\frac{1}{2}$ from all terms, rearrange to solve for $v$: $v=\sqrt{\frac{k(A^2-x^2)}{m}}$.
3. Step 3: Substitute values: $A^2-x^2=(0.10{)}^2-(0.05{)}^2=0.0075{\text{m}}^2$, so $v=\sqrt{\frac{200\times 0.0075}{0.50}}=\sqrt{3}\approx 1.7\text{m/s}$.

Exam tip: When asked for speed at a non-equilibrium, non-turning point position, energy conservation is almost always faster and less error-prone than using the explicit SHM position-velocity formula.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Incorrectly adjusting the angular frequency of a vertical or incline mass-spring to account for gravity, getting a result that includes $g$ in the final $\omega$. Why: Students incorrectly assume gravity adds to the restoring force, rather than just shifting the equilibrium position. Correct move: Always measure displacement from the equilibrium position; constant forces like gravity cancel out of the restoring force, so $\omega =\sqrt{k/m}$ holds for all mass-spring systems regardless of orientation.
• Wrong move: Scaling the period of a simple pendulum proportional to $\sqrt{m}$ after seeing period scales with $\sqrt{m}$ for mass-spring systems. Why: Students transfer the mass dependence from mass-spring systems to pendulums without deriving the relationship. Correct move: Recall mass cancels out of the simple pendulum period formula, so $T$ does not depend on bob mass for a given length.
• Wrong move: Using $\sin \theta \approx \theta$ with $\theta$ measured in degrees when deriving the simple pendulum restoring force. Why: The small-angle approximation only holds for radians, as it comes from the Taylor series of $\sin \theta$ around 0, which uses radians by definition. Correct move: Convert all angular displacements to radians before applying the approximation.
• Wrong move: Writing the energy conservation relation as $\frac{1}{2}m{v}_{\text{max}}^2=\frac{1}{2}kA$, omitting the square on amplitude. Why: Students confuse Hooke's force ($F=kx$) with elastic potential energy ($U=\frac{1}{2}k{x}^2$). Correct move: Check units after writing energy terms: $kA$ has units of force (N), not energy (J), so the square is required.
• Wrong move: Assuming increasing the amplitude of a simple pendulum from 2 degrees to 8 degrees doubles the period. Why: Students memorize that period is independent of amplitude without remembering this is only an approximation for small angles. Correct move: Recognize that amplitude independence holds only for small angles where $\sin \theta \approx \theta$; small increases in amplitude within the small-angle regime produce negligible changes to period.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A simple pendulum of length $L$ has period $T$ when its maximum angular displacement is 2 degrees. What is the approximate new period if the maximum angular displacement is increased to 8 degrees, with all other variables unchanged? A) $T$ B) $2T$ C) $1.02T$ D) $4T$

Worked Solution: The approximation that period is independent of amplitude only holds for small angles where $\sin \theta \approx \theta$. Even at 8 degrees, $\sin \theta$ differs from $\theta$ by less than 1%, so the period increases by only a tiny fraction from its small-angle value $T$. It cannot double or quadruple, as that would require a much larger change in restoring force. Options B and D are incorrect, and option A ignores the small deviation from the approximation. The correct answer is C.

Question 2 (Free Response)

A 0.4 kg block is attached to a horizontal spring with $k=10\text{N/m}$ on a frictionless surface. The block is pulled 0.3 m from equilibrium and released from rest at $t=0$. (a) Derive the period of oscillation of the block, and calculate its numerical value. (b) Calculate the maximum kinetic energy of the block, and state the position where this maximum occurs. (c) The same block and spring are now attached to the top of a frictionless 30° incline. The block is pulled 0.3 m from the new equilibrium position and released. What is the new period of oscillation? Justify your answer.

Worked Solution: (a) Starting from Newton's second law, the restoring force on the block is $F=-kx=ma$, so acceleration is $a=-\frac{k}{m}x$. For SHM, acceleration follows $a=-{\omega }^{2}x$, so $\omega =\sqrt{\frac{k}{m}}$. Period is $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$. Substituting values: $T=2\pi \sqrt{\frac{0.4}{10}}=0.4\pi \approx 1.26\text{s}$. (b) Maximum kinetic energy equals maximum elastic potential energy at the turning point ($x=A$, $v=0$), so $K{E}_{\text{max}}=\frac{1}{2}k{A}^2=0.5(10)(0.3{)}^2=0.45\text{J}$. Maximum kinetic energy occurs at the equilibrium position $x=0$, where all energy is kinetic. (c) The period is unchanged at $\approx 1.26\text{s}$. The component of gravity along the incline adds a constant force to the block, which only shifts the equilibrium position down the incline. It does not change the restoring force for displacements from the new equilibrium, so $\omega$ and $T$ remain the same as in the horizontal case.

Question 3 (Application / Real-World Style)

Geophysicists studying a potential mineral deposit use a simple pendulum to measure local gravitational acceleration. They use a 1.000 m long massless string with a 0.500 kg lead bob, and measure a period of 2.008 s for small oscillations. What is the local gravitational acceleration at this site? Is it higher or lower than the standard gravitational acceleration of $9.81{\text{m/s}}^2$?

Worked Solution: For small oscillations of a simple pendulum, the period formula is $T=2\pi \sqrt{\frac{L}{g}}$. Rearrange to solve for $g$: $g=\frac{4{\pi }^{2}L}{T^2}$. Substitute the given values: $L=1.000\text{m}$, $T=2.008\text{s}$, so $g=\frac{4{\pi }^2(1.000)}{(2.008{)}^2}\approx \frac{39.48}{4.032}\approx 9.79{\text{m/s}}^2$. This local gravitational acceleration is slightly lower than the standard $9.81{\text{m/s}}^2$, consistent with a lower-density ore body that reduces the local mass density.

7. Quick Reference Cheatsheet

8. What's Next

This chapter establishes the core relationships and problem-solving techniques for simple undamped oscillators that are the foundation of all Unit 6 Oscillations content. Next you will study damped and driven oscillations, which extend these basic relationships to include friction and external driving forces, commonly tested in AP C:M FRQs. Without mastering the period, angular frequency, and energy relationships for mass-spring and pendulum systems covered here, you cannot correctly set up or solve the differential equations for more complex oscillating systems. This topic also connects force and energy concepts from earlier units, and prepares you for wave motion, where individual elements of a wave undergo SHM.

• Damped and driven oscillations
• Simple harmonic motion differential equations
• Energy in simple harmonic motion
• Restoring forces in SHM