Kinematics of Simple Harmonic Motion

AP Physics 1· AP Physics 1 CED — Simple Harmonic Motion and Waves· 14 min read

1. What is SHM Kinematics?★★☆☆☆⏱ 3 min

Kinematics of simple harmonic motion (SHM) is the study of how position, velocity, and acceleration of an oscillating object change over time, without analyzing the forces that cause the motion (force analysis is covered in SHM dynamics).

📘 Definition

Simple Harmonic Motion (SHM)

A specific type of periodic motion defined by a fixed proportional relationship between acceleration and displacement from equilibrium.

Example:

Ideal mass-spring system, small-angle pendulum swings

We use standard notation: $A$ for amplitude (maximum displacement from equilibrium), $ω$ for angular frequency, $f$ for frequency (cycles per second), $T$ for period (seconds per cycle), $ϕ$ for phase constant, and $x (t), v (t), a (t)$ for position, velocity, and acceleration at time $t$ , respectively. SHM is the foundational prerequisite for all other SHM and wave topics in Unit 7 of AP Physics 1.

2. Defining Condition and Angular Frequency Conversions★★☆☆☆⏱ 3 min

While all periodic motion repeats over a fixed period, only SHM follows the specific proportional relationship between acceleration and displacement that defines it. This relationship separates SHM from other types of periodic motion, and is the starting point for all kinematic analysis.

a(t) = -\omega^2 x(t)

The negative sign is critical: it indicates acceleration always points back toward equilibrium, opposite the direction of displacement from equilibrium. This matches intuition: if you pull a mass on a spring to the right of equilibrium, acceleration pulls left back to center, and vice versa.

Angular frequency is directly related to the more familiar measurable quantities period ( $T$ , time per cycle) and frequency ( $f$ , cycles per second) by two core relationships:

\omega = 2\pi f = \frac{2\pi}{T}

📐 Worked Example

A block on a spring undergoes SHM. At an instant when the block's displacement from equilibrium is $x = + 0.15 m$ , its acceleration is measured as $a = - 13.5 m/s^{2}$ . What is the period of the block's SHM?

1
Start with the defining SHM relationship:
$a = -\omega^2 x$
2
Cancel negative signs and solve for $ω^{2}$ :
$\omega^2 = \frac{-a}{x} = \frac{-(-13.5)}{0.15} = 90 \text{ rad}^2/\text{s}^2$
3
Take the positive square root (angular frequency is always positive):
$\omega = \sqrt{90} \approx 9.49 \text{ rad/s}$
4
Relate $ω$ to period:
$T = \frac{2\pi}{\omega} \approx \frac{6.28}{9.49} \approx 0.66 \text{ s}$

Exam tip:

Always cancel the negative signs first when solving for $ω$ from the defining relationship. $ω$ is always positive, so you will never end up with a negative angular frequency, even if displacement or acceleration are negative.

3. Kinematic Equations for Position, Velocity, and Acceleration★★★☆☆⏱ 4 min

From the defining acceleration-displacement relationship, we can derive equations for position, velocity, and acceleration as functions of time. The general form of the position function (written with cosine for consistency with AP conventions) is:

x(t) = A \cos(\omega t + \phi)

$A$ is the amplitude (maximum displacement from equilibrium, always positive), and $ϕ$ is the phase constant, which adjusts the equation to match the initial position and velocity of the oscillator at $t = 0$ . Velocity is the first derivative of position, acceleration is the second derivative:

v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi)

a(t) = \frac{dv}{dt} = -A \omega^2 \cos(\omega t + \phi) = -\omega^2 x(t)

To find $ϕ$ , use both initial position and initial velocity at $t = 0$ . Common cases: if the oscillator is at maximum positive displacement at $t = 0$ , $ϕ = 0$ . If the oscillator is at equilibrium moving positive at $t = 0$ , $ϕ = - π /2$ , simplifying to $x (t) = A sin (ω t)$ .

📐 Worked Example

A 0.2 kg block undergoes SHM with amplitude 0.4 m and period 2.0 s. At $t = 0$ , the block is at position $x = 0$ m and moving in the positive x-direction. Write the equation for the position of the block as a function of time.

1
Calculate angular frequency first:
$\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \text{ rad/s}. We know A = 0.4 \text{ m}, so only find $\phi$.$
2
Apply initial position condition at $t = 0$ :
$x(0) = A \cos(\phi) = 0 \implies \cos(\phi) = 0, so \phi = +\pi/2 or -\pi/2$
3
Use initial velocity (positive at $t = 0$ ) to find the correct sign:
$v(0) = -A \omega \sin(\phi) > 0 \implies \sin(\phi) < 0 \implies \phi = -\pi/2$
4
Substitute back and simplify with trig identity:
$x(t) = 0.4 \cos\left(\pi t - \frac{\pi}{2}\right) = 0.4 \sin(\pi t) \text{ meters}$

Exam tip:

Always check both position and velocity at $t = 0$ to find the correct sign of the phase constant. Many students stop after only matching the initial position and end up with the wrong phase.

4. Graphical Analysis of SHM Kinematics★★★☆☆⏱ 4 min

AP Physics 1 frequently tests graphical reasoning for SHM, asking you to relate graphs of $x (t)$ , $v (t)$ , and $a (t)$ or extract properties like amplitude and period from a given graph. The key is understanding the fixed phase differences between the three quantities:

Acceleration is 180° ( $π$ radians) out of phase with position: when position is maximum positive, acceleration is maximum negative, and vice versa.
Velocity is 90° ( $π /2$ radians) out of phase with position: when position is zero (at equilibrium), velocity is maximum magnitude, and when position is maximum, velocity is zero.

📐 Worked Example

The position vs time graph of an object undergoing SHM has a maximum displacement of 2.0 cm, and consecutive peaks (maximum positive displacements) occur at $t = 0$ s and $t = 4.0$ s. Calculate the maximum speed and maximum acceleration of the object.

1
Extract amplitude and period from the graph:
$A = 2.0 \text{ cm} = 0.02 \text{ m}, \quad T = 4.0 \text{ s}$
2
Calculate angular frequency:
$\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \approx 1.57 \text{ rad/s}$
3
Maximum speed is $v_{ma x} = A ω$ (sine term max magnitude = 1):
$v_{max} = (0.02 \text{ m})(1.57 \text{ rad/s}) \approx 0.031 \text{ m/s}$
4
Maximum acceleration is $a_{ma x} = A ω^{2}$ (cosine term max magnitude = 1):
$a_{max} = (0.02 \text{ m})(1.57^2 \text{ rad}^2/\text{s}^2) \approx 0.049 \text{ m/s}^2$

✓ Quick check

Test your understanding with this AP-style multiple choice question:

An object undergoes simple harmonic motion described by the equation $x (t) = 3 cos (\frac{π}{2} t + \frac{π}{2})$ where $x$ is in meters and $t$ is in seconds. What is the speed of the object at $t = 0$ ?
- 0 m/s
- $\frac{3 π}{2}$ m/s
- $3 π$ m/s
- $\frac{9 π}{4}$ m/s
Reveal answer
1 —
Correct. Speed is the magnitude of velocity. Using $v (t) = - A ω sin (ω t + ϕ)$ , substituting values gives $v (0) = - \frac{3 π}{2}$ m/s, so speed is $\frac{3 π}{2}$ m/s. The other options correspond to common student mistakes.

Exam tip:

When reading period from a position vs time graph, always measure between two identical points (two consecutive peaks, or two consecutive zero crossings with the same slope direction), not just any two zero crossings.

5. Common Pitfalls

Wrong move:

Using $ω = T / (2 π)$ instead of $ω = 2 π / T$ when relating angular frequency to period.

Why:

Students mix up the inverse relationship between period and frequency, confusing which quantity goes in the numerator.

Correct move:

Always start from $f = 1/ T$ and $ω = 2 π f$ , substitute to get $ω = 2 π / T$ , and write this relationship at the top of your page before starting calculations.

Wrong move:

Forgetting the negative sign in $v (t) = - A ω sin (ω t + ϕ)$ and getting the wrong direction of velocity at $t = 0$ .

Why:

Students remember to take the derivative of cosine, but forget that the derivative of cosine is negative sine.

Correct move:

After taking the derivative of $x (t)$ to get $v (t)$ , always verify the negative sign is in place before solving for phase constant or initial direction.

Wrong move:

Calculating period as the distance between a positive-slope zero crossing and negative-slope zero crossing on an $x (t)$ graph.

Why:

Students see two zero crossings and assume they are one period apart.

Correct move:

On any SHM position graph, mark two consecutive peaks (or two consecutive zero crossings with the same slope sign) to measure the full period.

Wrong move:

Claiming that maximum acceleration occurs when velocity is maximum.

Why:

Students confuse SHM kinematics with constant acceleration kinematics, where maximum acceleration is independent of velocity.

Correct move:

Memorize the rule: maximum position = zero velocity = maximum acceleration; zero position = maximum velocity = zero acceleration, and reference this for every graph or calculation question.

Wrong move:

Using $v_{ma x} = ω r$ (from circular motion) for SHM, but substituting the wrong value for $r$ .

Why:

Students recall the relationship between tangential speed and angular speed from uniform circular motion, but misapply it to SHM.

Correct move:

Remember that for SHM, $v_{ma x} = ω A$ , where $A$ is the amplitude of the SHM; derive it quickly from the derivative of $x (t)$ if you forget.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Defining SHM condition	$a = - ω^{2} x$	True for all SHM, regardless of system
Angular frequency → period	$ω = \frac{2 π}{T}$	$ω$ in rad/s, $T$ in s, always positive
Angular frequency → frequency	$ω = 2 π f$	$f$ in Hz (cycles per second)
General position function	$x (t) = A cos (ω t + ϕ)$	$A$ = amplitude (always positive), $ϕ$ = phase constant
Velocity function	$v (t) = - A ω sin (ω t + ϕ)$	Derived from derivative; negative sign critical
Acceleration function	$a (t) = - A ω^{2} cos (ω t + ϕ)$	Reduces to defining $a = - ω^{2} x$
Maximum speed	$v_{ma x} = A ω$	Occurs when $x = 0$ (at equilibrium)
Maximum acceleration	$a_{ma x} = A ω^{2}$	Occurs when $x = \pm A$ (max displacement)
Phase relationship 1	$a$ 180° out of phase with $x$	When $x$ is max positive, $a$ is max negative
Phase relationship 2	$v$ 90° out of phase with $x$	When $x = 0$ , $v$ has maximum magnitude

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Find period from a-x SHM measurement
2022 · FRQ
Write SHM position equation from initial conditions

What's Next

This topic lays the kinematic foundation for all SHM topics in Unit 7 of AP Physics 1. Immediately next, you will connect the kinematic definition of SHM to dynamics, relating angular frequency to system properties like mass and spring constant for a mass-spring system, or length and gravitational acceleration for a pendulum. Without mastering the kinematic relationships between $ω$ , $T$ , and the $a - x$ condition, you will not be able to derive or predict the period of SHM for different systems, a high-weight FRQ topic on the AP exam. This topic also feeds into the study of energy in SHM, wave kinematics, and interference later in Unit 7, since all wave motion relies on SHM principles for individual oscillating particles.

→Energy in Simple Harmonic Motion