# Simple Harmonic Motion

> AP Physics C: Mechanics · Unit 6: Oscillations
> Source: https://www.owlsprep.com/study/ap-physics-c-mech-u6-simple-harmonic-motion/

This sub-topic covers core definitions, kinematics, and energy of undamped simple harmonic motion (SHM) for AP Physics C: Mechanics. Includes worked examples and exam-specific tips for common problem types.

**Prerequisites:** Newton's second law for translational motion; Solving second-order linear homogeneous differential equations; [Conservation of mechanical energy](https://www.owlsprep.com/study/ap-physics-c-mech-u4-conservation-of-mechanical-energy/)

## Learning objectives

- Define simple harmonic motion (SHM) via force and differential equation definitions
- Solve for kinematic quantities, amplitude, and phase constant from initial conditions
- Apply energy conservation to SHM systems
- Calculate angular frequency for common SHM systems

## Definition and Differential Equation of SHM

Simple harmonic motion (SHM) is periodic motion where the restoring force is directly proportional to displacement from stable equilibrium, and opposite in direction to displacement. SHM is the simplest undamped oscillatory motion, accounting for 12-18% of your total AP Physics C: Mechanics exam score. It appears in both multiple-choice and free-response questions, often paired with forces, energy, or rotational motion.

**Undamped Simple Harmonic Motion** — Periodic motion with a linear restoring force $F_{\text{net}} = -kx$, resulting in constant amplitude and no energy loss

*Notation:* $x(t)$ = displacement, $k$ = spring constant, $m$ = mass, $\omega$ = angular frequency

*Example:* Horizontal/vertical mass-spring systems, small-angle simple pendulums

Applying Newton's second law $F_{\text{net}} = ma = m \frac{d^2x}{dt^2}$ rearranges the force definition to the standard SHM differential equation:

$$\frac{d^2x}{dt^2} + \omega^2 x = 0$$

Where $\omega^2 = \frac{k}{m}$ for a mass-spring system. Any motion satisfying this ODE is SHM, regardless of the physical system. The general solution is $x(t) = A \cos(\omega t + \phi)$, where $A$ = amplitude (maximum displacement), $\phi$ = phase constant (adjusts for initial conditions), and $\omega = \frac{2\pi}{T} = 2\pi f$.

**Worked example:** A 2.0 kg block on a frictionless horizontal track is attached to a spring with $k = 50$ N/m. Write the differential equation for the block's motion, then calculate $\omega$ and $T$.

1. Start with net restoring force and Newton's second law:

   $$F_{\text{net}} = -kx = m \frac{d^2x}{dt^2}$$
2. Rearrange to standard SHM form, substitute values:

   $$\frac{d^2x}{dt^2} + \frac{k}{m}x = 0 \implies \frac{d^2x}{dt^2} + 25x = 0$$
3. By definition, $\omega^2 = \frac{k}{m}$, so:

   $$\omega = \sqrt{25} = 5 \text{ rad/s}$$
4. Calculate period from $\omega$:

   $$T = \frac{2\pi}{\omega} = \frac{2\pi}{5} \approx 1.26 \text{ s}$$

> **Exam tip:** If a problem gives you a non-standard system (e.g., mass between two springs, floating object), always derive the ODE from Newton's second law to find $\omega$, do not guess from memorized standard formulas.

## Kinematics of SHM and Initial Conditions

Once you have the position function for SHM, velocity and acceleration are found via differentiation. For the standard position function $x(t) = A \cos(\omega t + \phi)$:

1. Velocity (first derivative): $v(t) = -A\omega \sin(\omega t + \phi)$
2. Acceleration (second derivative): $a(t) = -A\omega^2 \cos(\omega t + \phi) = -\omega^2 x(t)$

Key relationships: Maximum speed $v_{\text{max}} = A\omega$, which occurs at equilibrium ($x=0$) when all energy is kinetic. Maximum acceleration $a_{\text{max}} = A\omega^2$, which occurs at maximum displacement ($x = \pm A$) when the restoring force is largest.

To find amplitude $A$ and phase constant $\phi$, use initial conditions ($x_0 = x(0), v_0 = v(0)$). Amplitude follows $A^2 = x_0^2 + \left(\frac{v_0}{\omega}\right)^2$, and phase constant from $\tan\phi = -\frac{v_0}{\omega x_0}$. Always check the quadrant of $\phi$ using the signs of $x_0$ and $v_0$.

**Worked example:** For the 2.0 kg block from the previous example, initial conditions at $t=0$ are $x_0 = 0.10$ m and $v_0 = -1.0$ m/s, and $\omega = 5$ rad/s. Find $A$ and $\phi$, then write the full position function.

1. Start with initial conditions: $x_0 = A\cos\phi = 0.10$, and $v_0 = -A\omega \sin\phi = -1.0$, so rearrange to $A\sin\phi = 0.20$.
2. Calculate amplitude using the Pythagorean identity:

   $$A^2 = (A\cos\phi)^2 + (A\sin\phi)^2 = 0.10^2 + 0.20^2 = 0.05 \implies A \approx 0.22 \text{ m}$$
3. Calculate phase constant: $\tan\phi = 2$, both terms are positive so $\phi$ is first quadrant:

   $$\phi = \arctan(2) \approx 1.11 \text{ rad}$$
4. Final position function:

   $$x(t) = 0.22 \cos(5t + 1.11) \text{ m}$$

> **Exam tip:** Calculators only return arctangent values in the first/fourth quadrants. If $\phi$ is in the second/third quadrant, you must adjust the result manually to get the correct value.

## Energy in Undamped SHM

For undamped SHM, total mechanical energy is conserved, converting between kinetic energy of the mass and potential energy of the restoring force. For a mass-spring system, $K = \frac{1}{2}mv^2$ and $U = \frac{1}{2}kx^2$.

Substituting SHM position and velocity functions, and using $\omega^2 = \frac{k}{m}$, total energy simplifies to:

$$E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$$

Total energy is constant, proportional to the square of the amplitude. A very useful result from energy conservation is the speed at any displacement: $v = \omega\sqrt{A^2 - x^2}$, which avoids working through phase constants entirely.

**Worked example:** A 0.5 kg mass-spring system oscillates with SHM of amplitude 0.4 m and period 1.0 s. What is the total mechanical energy, and what is the speed when displacement is 0.2 m?

1. Calculate angular frequency from period:

   $$\omega = \frac{2\pi}{T} = 2\pi \text{ rad/s}$$
2. Calculate total energy:

   $$E = \frac{1}{2}m\omega^2 A^2 = 0.5(0.5)(2\pi)^2(0.4)^2 \approx 1.58 \text{ J}$$
3. Use the energy-derived speed formula:

   $$v = \omega\sqrt{A^2 - x^2} = 2\pi\sqrt{0.4^2 - 0.2^2} \approx 2.18 \text{ m/s}$$
4. Sanity check: Maximum speed $v_{\text{max}} = A\omega \approx 2.51$ m/s, which is larger than our result as expected away from equilibrium.

> **Exam tip:** If asked for speed at a given displacement, using energy conservation is almost always faster and less error-prone than differentiating position and calculating phase terms.

## AP-Style Worked Problems

**Worked example:** A particle undergoes SHM described by $x(t) = 3 \cos(2t - \pi/4)$ (all SI units). What is the magnitude of the maximum acceleration of the particle?

(A) 3 m/s²
(B) 6 m/s²
(C) 12 m/s²
(D) 24 m/s²

1. For any SHM, $a(t) = -\omega^2 x(t)$, so maximum acceleration magnitude is $a_{\text{max}} = \omega^2 A$.
2. From the position function, identify $A = 3$ m, $\omega = 2$ rad/s.
3. Substitute values:

   $$a_{\text{max}} = (2)^2(3) = 12 \text{ m/s}^2$$
4. The most common mistake is forgetting to square $\omega$, leading to the incorrect answer 6 m/s². The correct answer is (C).

**Worked example:** A 1.5 kg block is attached to two identical springs each with $k = 30$ N/m. The block sits between two fixed walls, with one spring attached to each end, and both springs are at natural length when the block is at $x=0$.

(a) Show that the block undergoes SHM when displaced from $x=0$, and derive the angular frequency.
(b) At $t=0$, the block is pulled to $x = 0.2$ m and released from rest. Write the complete position function.
(c) Calculate total energy and kinetic energy at $x = 0.1$ m.

1. Part (a): When displaced $x$ right, left spring pulls left with $-kx$, right spring pushes left with $-kx$. Net force:

   $$F_{\text{net}} = -2kx = m \frac{d^2x}{dt^2}$$
2. Rearrange to standard SHM form, confirming SHM:

   $$\frac{d^2x}{dt^2} + \left(\frac{2k}{m}\right)x = 0$$
3. Calculate $\omega$:

   $$\omega^2 = \frac{2(30)}{1.5} = 40 \implies \omega \approx 6.32 \text{ rad/s}$$
4. Part (b): Initial conditions $x(0) = 0.2$ m, $v(0) = 0$. $v(0) = -A\omega \sin\phi = 0$, so $\phi = 0$, $A = 0.2$ m. Final position:

   $$x(t) = 0.2 \cos\left(\sqrt{40} t\right) \text{ m}$$
5. Part (c): Equivalent spring constant $k_{\text{eq}} = 60$ N/m. Total energy:

   $$E = \frac{1}{2}k_{\text{eq}} A^2 = 1.2 \text{ J}$$
6. Kinetic energy from conservation of energy:

   $$K = E - \frac{1}{2}k_{\text{eq}}x^2 = 1.2 - 0.3 = 0.9 \text{ J}$$

**Worked example:** A car's suspension is modeled as a mass-spring SHM system. The empty 1200 kg car oscillates with $\omega = 2.0$ rad/s after a bump. What is the new angular frequency when carrying 500 kg of load, assuming $k$ is constant? Explain the result for the ride.

1. For mass-spring SHM, $\omega = \sqrt{\frac{k}{m}}$. Solve for $k$ from the empty car:

   $$k = \omega^2 m = (2.0)^2(1200) = 4800 \text{ N/m}$$
2. New total mass $m_{\text{new}} = 1700$ kg. Calculate new $\omega$:

   $$\omega_{\text{new}} = \sqrt{\frac{4800}{1700}} \approx 1.7 \text{ rad/s}$$
3. Lower angular frequency means a longer period of oscillation, so the loaded car oscillates more slowly, resulting in a softer ride than the empty car.

## Common pitfalls

- **Wrong:** Writing $x(t) = A \sin(\omega t)$ by default, even when initial position at $t=0$ is non-zero
  - Why it fails: Students memorize the sine form for the specific case of starting at equilibrium, and forget to adjust for other starting positions
  - Correct: Always start with the general form $x(t) = A \cos(\omega t + \phi)$ and solve for $\phi$ from initial conditions, regardless of the starting position
- **Wrong:** Using $\omega = \sqrt{g/L}$ for a pendulum displaced by 45 degrees from equilibrium
  - Why it fails: The SHM result for simple pendulums only applies for small angles where $\sin\theta \approx \theta$, which is required to produce the SHM differential equation
  - Correct: For large-angle pendulum motion, explicitly derive the differential equation from Newton's second law and confirm it fits the SHM form before using SHM formulas
- **Wrong:** Forgetting the chain rule when differentiating $x(t)$, leading to $v(t) = -A \sin(\omega t + \phi)$ (missing the $\omega$ factor)
  - Why it fails: Students focus on the derivative of cosine and forget to differentiate the inner linear term in the argument
  - Correct: Always write the chain rule step explicitly: $\frac{d}{dt} f(\omega t + \phi) = \omega f'(\omega t + \phi)$
- **Wrong:** Changing $\omega$ to account for gravity in a vertical mass-spring system
  - Why it fails: Students think gravity changes the restoring force because it shifts the equilibrium position
  - Correct: Recognize gravity only shifts the equilibrium position of a vertical mass-spring; the restoring force is still proportional to displacement from the new equilibrium, so $\omega = \sqrt{k/m}$ remains unchanged
- **Wrong:** Using $1/k_{\text{eq}} = 1/k_1 + 1/k_2$ for two springs in parallel
  - Why it fails: Students confuse spring combinations with resistor combinations
  - Correct: For parallel springs (same displacement for both springs), $k_{\text{eq}} = k_1 + k_2$; for series springs (same force for both springs), $1/k_{\text{eq}} = 1/k_1 + 1/k_2$

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Defining SHM ODE | \frac{d^2x}{dt^2} + \omega^2 x = 0 | Any motion satisfying this is SHM, regardless of system |
| General position solution | x(t) = A \cos(\omega t + \phi) | $A$ = amplitude, $\phi$ = phase constant, $\omega$ = angular frequency |
| SHM velocity | v(t) = -A\omega \sin(\omega t + \phi) | Maximum speed $v_{\text{max}} = A\omega$, occurs at equilibrium $x=0$ |
| SHM acceleration | a(t) = -\omega^2 x(t) | Maximum acceleration $a_{\text{max}} = A\omega^2$, occurs at $x=\pm A$ |
| Period-frequency relation | T = \frac{2\pi}{\omega} = \frac{1}{f} | $T$ = period, $f$ = frequency in cycles per second |
| Mass-spring angular frequency | \omega = \sqrt{\frac{k_{\text{eq}}}{m}} | Same for horizontal and vertical systems; gravity only shifts equilibrium |
| Small-angle simple pendulum | \omega = \sqrt{\frac{g}{L}} | Only valid for $\theta < 10^\circ$ where $\sin\theta \approx \theta$ |
| Total energy undamped SHM | E = \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 | Total mechanical energy is constant, proportional to $A^2$ |

## What's next

This sub-topic gives you the foundational model for all oscillatory motion in AP Physics C: Mechanics, and the differential equation approach used here is applied to many other dynamic systems across the course. Next in Unit 6 Oscillations, you will extend the SHM model to damped and driven harmonic motion, which include energy loss and external driving forces. Mastering the differential equation, kinematics, and energy relations of SHM is required to solve more complex ODEs for damped motion and understand the commonly tested topic of resonance. SHM also connects directly to uniform circular motion and energy conservation, core concepts that appear across the entire AP exam.

- [Mass-spring systems and simple pendulum](https://www.owlsprep.com/study/ap-physics-c-mech-u6-mass-spring-systems-and-simple/)
- [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/)

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