# Mass-spring systems and simple pendulum

> AP Physics C: Mechanics · CED Unit 6: Oscillations
> Source: https://www.owlsprep.com/study/ap-physics-c-mech-u6-mass-spring-systems-and-simple/

This guide 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.

**Prerequisites:** Hooke's law and Newton's second law; Definition of simple harmonic motion; Small-angle approximation for trigonometric functions

## Learning objectives

- Calculate angular frequency and period for mass-spring systems in any orientation
- Apply the small-angle approximation to find the period of a simple pendulum
- Use conservation of mechanical energy to solve problems in undamped SHM
- Explain the effect of constant forces like gravity on equilibrium position and period

## Core Overview of Oscillator Systems

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, appearing regularly on both multiple choice (MCQ) and free response (FRQ) sections, often combined with energy, force, or differential equation questions.

**Mass-spring system** — A system consisting of an oscillating mass attached to an ideal massless spring, with a restoring force proportional to displacement from equilibrium that obeys Hooke's law.

**Simple pendulum** — A system consisting 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.

Mastery of core relationships here is required for all further work on oscillations in AP Physics C: Mechanics.

## 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:

$$-kx = ma \implies a = -\left(\frac{k}{m}\right)x$$

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. 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. Convert mass to SI units:

   $$m = 200 \ \text{g} = 0.200 \ \text{kg}$$
3. Use the standard period formula for mass-spring oscillations:

   $$T = 2\pi \sqrt{\frac{m}{k}}$$
4. Substitute values and solve:

   $$T = 2\pi \sqrt{\frac{0.200}{5.0}} = 2\pi \sqrt{0.04} = 0.4\pi \approx 1.3 \ \text{s}$$

> **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 $T$.

## 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:

$$F = -mg\sin\theta$$

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:

$$F = -\left(\frac{mg}{L}\right)s$$

This 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. Write the period formula for small-angle simple pendulum for both locations:

   $$T_E = 2\pi \sqrt{\frac{L}{g_E}} \quad \text{and} \quad T_M = 2\pi \sqrt{\frac{L}{g_M}}$$
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. Substitute $T_E = 2.0 \ \text{s}$ to get the final result:

   $$T_M = 1.62 \times 2.0 \approx 3.2 \ \text{s}$$

> **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.

## Energy Conservation in Undamped 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:

$$E = KE + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}$$

At the maximum displacement (turning point), $x = A$ (amplitude) and $v=0$, so all energy is potential: $E = \frac{1}{2}kA^2$. This gives the energy conservation relation:

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

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 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. Use conservation of mechanical energy for undamped SHM: total energy at any displacement equals maximum potential energy at the turning point:

   $$\frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$$
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. Substitute values and solve:

   $$A^2 - x^2 = (0.10)^2 - (0.05)^2 = 0.0075 \ \text{m}^2 \\ v = \sqrt{\frac{200 \times 0.0075}{0.50}} = \sqrt{3} \approx 1.7 \ \text{m/s}$$

> **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.

## Common pitfalls

- **Wrong:** Including $g$ in the angular frequency calculation for a vertical mass-spring system, incorrectly adjusting for gravity
  - Why it fails: You may incorrectly assume gravity changes the restoring force, when it only shifts the equilibrium position of the oscillator.
  - Correct: Measure all displacements from the new equilibrium position; constant forces like gravity cancel out of the restoring force, so use $\omega = \sqrt{k/m}$ regardless of orientation.
- **Wrong:** Scaling the period of a simple pendulum proportional to $\sqrt{m}$, copying the mass dependence from mass-spring systems
  - Why it fails: You transfer the mass dependence from mass-spring systems to pendulums without re-deriving the relationship.
  - Correct: Recall mass cancels out of the simple pendulum period formula, so $T$ is independent of bob mass for a given pendulum length.
- **Wrong:** Applying the small-angle approximation $\sin\theta \approx \theta$ with $\theta$ measured in degrees
  - Why it fails: The small-angle approximation comes from the Taylor series of $\sin\theta$, which is only valid when $\theta$ is in radians.
  - Correct: Convert all angular displacements to radians before applying the small-angle approximation to restoring force.
- **Wrong:** Writing energy conservation as $\frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2}kA$, omitting the square on amplitude
  - Why it fails: You confuse Hooke's force $F=kx$ with elastic potential energy $U=\frac{1}{2}kx^2$.
  - Correct: Check units after writing energy terms: $kA$ has units of force (N), not energy (J), so the square on displacement is always required.
- **Wrong:** Assuming increasing a simple pendulum's amplitude from 2° to 8° doubles its period, because period is amplitude-independent
  - Why it fails: You forget that amplitude independence is only an approximation that holds for small angles where $\sin\theta \approx \theta$.
  - Correct: Recognize that amplitude independence holds only for small angles; small increases within the small-angle regime produce negligible, not proportional, changes to period.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| All Mass-Spring Angular Frequency | $\omega = \sqrt{\frac{k}{m}}$ | Constant forces (gravity) only shift equilibrium, do not change $\omega$ or period |
| Mass-Spring Period | $T = 2\pi \sqrt{\frac{m}{k}}$ | Applies to all mass-spring SHM, regardless of orientation |
| Simple Pendulum (Small Angles) Angular Frequency | $\omega = \sqrt{\frac{g}{L}}$ | $\theta$ must be in radians for the small-angle approximation |
| Simple Pendulum (Small Angles) Period | $T = 2\pi \sqrt{\frac{L}{g}}$ | Independent of bob mass and amplitude (for small $\theta$ only) |
| Total Energy (Mass-Spring SHM) | $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2$ | Total energy is constant for undamped (frictionless) SHM |
| Maximum Speed (Mass-Spring SHM) | $v_{\text{max}} = A\omega = A\sqrt{\frac{k}{m}}$ | Occurs at equilibrium position $x=0$ |
| Total Energy (Simple Pendulum SHM) | $E = \frac{1}{2}mv^2 + mgL(1-\cos\theta) = mgL(1-\cos\theta_{\text{max}})$ | Zero potential energy defined at equilibrium |
| Maximum Kinetic Energy (Undamped SHM) | $KE_{\text{max}} = U_{\text{max}}$ | Always true, regardless of oscillator type |

## What's next

This subtopic establishes the core relationships and problem-solving techniques for simple undamped oscillators that form 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 Physics C: Mechanics 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.

- [Gravitation Overview](https://www.owlsprep.com/study/ap-physics-c-mech-u7-overview/)
- [Gravitational Forces](https://www.owlsprep.com/study/ap-physics-c-mech-u7-gravitational-forces/)
- [Orbits of planets and satellites](https://www.owlsprep.com/study/ap-physics-c-mech-u7-orbits-of-planets-and-satellites/)

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