Kinematics of Simple Harmonic Motion — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: The defining acceleration-displacement relationship for SHM, angular frequency/period/frequency conversions, kinematic equations for position/velocity/acceleration, phase constants, and graphical analysis of SHM kinematics.

You should already know: Basic kinematics for non-constant acceleration. Derivatives of trigonometric functions. Definitions of period and frequency for periodic 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 Kinematics of Simple Harmonic Motion?

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 (that analysis is left for SHM dynamics). According to the AP Physics 1 Course and Exam Description (CED), this topic makes up approximately 1-2% of the total exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with SHM dynamics or energy questions. We use standard notation: $A$ for amplitude (maximum displacement from equilibrium), $\omega$ for angular frequency, $f$ for frequency (cycles per second), $T$ for period (seconds per cycle), $\varphi$ for phase constant, and $x(t),v(t),a(t)$ for position, velocity, and acceleration at time $t$, respectively. SHM is a specific type of periodic motion, defined by a unique relationship between acceleration and displacement. AP Physics 1 primarily tests your ability to relate graphs of the three kinematic quantities, write correct kinematic equations for given initial conditions, and connect angular frequency to measurable period and frequency. This topic is the foundational prerequisite for all other SHM and wave topics in Unit 7.

2. Defining Condition for Simple Harmonic Motion

While all periodic motion repeats over a fixed period, only SHM follows a specific proportional relationship between acceleration and displacement from equilibrium. This is the defining feature of SHM that separates it from other types of periodic motion, and it is the starting point for all kinematic analysis of SHM. The relationship is written as: $$a(t)=-{\omega }^{2}x(t)$$ Here, $a(t)$ is acceleration at time $t$, $x(t)$ is displacement from equilibrium at time $t$, and $\omega$ is the constant angular frequency of the oscillation. 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: $$\omega =2\pi f=\frac{2\pi }{T}$$ This relationship is one of the most frequently tested on the AP exam, as it connects the abstract angular frequency to real measurable quantities you can get from experiments or graphs.

Worked Example

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

1. Start with the defining SHM relationship: $a=-{\omega }^{2}x$.
2. Cancel the negative signs on both sides (both $a$ and $x$ have matching signs here, so the negatives cancel) to solve for ${\omega }^2$: ${\omega }^2=\frac{-a}{x}=\frac{-(-13.5)}{0.15}=90{\text{ rad}}^2/{\text{s}}^2$.
3. Take the positive square root to get angular frequency (angular frequency is always positive): $\omega =\sqrt{90}\approx 9.49\text{ rad/s}$.
4. Relate $\omega$ 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 $\omega$ from the defining relationship. $\omega$ 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

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: $$x(t)=A\cos (\omega t+\varphi )$$ $A$ is the amplitude (maximum displacement from equilibrium, always positive), and $\varphi$ 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 with respect to time, and acceleration is the second derivative: $$v(t)=\frac{dx}{dt}=-A\omega \sin (\omega t+\varphi )$$ $$a(t)=\frac{dv}{dt}=-A{\omega }^2\cos (\omega t+\varphi )=-{\omega }^{2}x(t)$$ Notice that the second derivative gives us back the original defining relationship for SHM, which confirms the equations are correct.

To find $\varphi$, use both initial position and initial velocity at $t=0$. If the oscillator is at maximum positive displacement at $t=0$, $\varphi =0$, and the equation simplifies to $x(t)=A\cos (\omega t)$. If the oscillator is at equilibrium moving in the positive direction at $t=0$, $\varphi =-\pi /2$, which simplifies to $x(t)=A\sin (\omega t)$ using the trig identity $\cos (\theta -\pi /2)=\sin \theta$.

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. First calculate angular frequency: $\omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi \text{ rad/s}$. Amplitude $A=0.4\text{ m}$, so we only need to find the phase constant $\varphi$.
2. Use the initial position condition: $x(0)=A\cos (\varphi )=0$, so $\cos (\varphi )=0$, meaning $\varphi =+\pi /2$ or $-\pi /2$.
3. Use the initial velocity condition to find the correct sign: $v(0)=-A\omega \sin (\varphi )>0$ (velocity is positive at $t=0$), so $-\sin (\varphi )>0⟹\sin (\varphi )<0$, which means $\varphi =-\pi /2$.
4. Substitute back into the general equation: $x(t)=0.4\cos \left(\pi t-\frac{\pi }{2}\right)=0.4\sin (\pi t)$ 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, leading to incorrect velocity directions.

4. Graphical Analysis of SHM Kinematics

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 to these questions is understanding the fixed phase differences between the three quantities:

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

These phase differences come directly from the kinematic equations, and they are consistent for all SHM regardless of the system.

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: Amplitude $A=2.0\text{ cm}=0.02\text{ m}$, period $T=4.0\text{ s}$ (time between identical points on the graph).
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 $v_{max}=A\omega$, since the sine term in the velocity equation reaches a maximum magnitude of 1: $v_{max}=(0.02\text{ m})(1.57\text{ rad/s})\approx 0.031\text{ m/s}$.
4. Maximum acceleration is $a_{max}=A{\omega }^2$, since the cosine term in the acceleration equation reaches a maximum magnitude of 1: $a_{max}=(0.02\text{ m})(1.57^2{\text{ rad}}^2/{\text{s}}^2)\approx 0.049{\text{ m/s}}^2$.

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. Two zero crossings (one positive slope, one negative slope) are only half a period apart.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $\omega =T/(2\pi )$ instead of $\omega =2\pi /T$ 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 $\omega =2\pi f$, substitute to get $\omega =2\pi /T$, and write this relationship at the top of your page before starting calculations.
• Wrong move: Forgetting the negative sign in $v(t)=-A\omega \sin (\omega t+\varphi )$ 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 $v_{max}=\omega r$ (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, $v_{max}=\omega A$, where $A$ is the amplitude of the SHM; derive it quickly from the derivative of $x(t)$ if you forget.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

An object undergoes simple harmonic motion described by the equation $x(t)=3\cos \left(\frac{\pi }{2}t+\frac{\pi }{2}\right)$ where $x$ is in meters and $t$ is in seconds. What is the speed of the object at $t=0$? A) 0 m/s B) $\frac{3\pi }{2}$ m/s C) $3\pi$ m/s D) $\frac{9\pi }{4}$ m/s

Worked Solution: First, we use the velocity derivative formula: $v(t)=-A\omega \sin (\omega t+\varphi )$. Here, amplitude $A=3$ m, angular frequency $\omega =\pi /2$ rad/s, and phase constant $\varphi =\pi /2$. Substitute $t=0$: $v(0)=-(3)(\pi /2)\sin (\pi /2)=-(3\pi /2)(1)=-3\pi /2$ m/s. Speed is the magnitude of velocity, so the speed is $3\pi /2$ m/s. The other options correspond to common student mistakes: option A is what you get if you incorrectly evaluate $x(0)$ as speed, option C comes from doubling $\omega$ by mistake, and option D comes from squaring $\omega$ incorrectly. Correct answer: B.

Question 2 (Free Response)

A student conducts an experiment with a mass on a spring that undergoes horizontal SHM. The student measures position as a function of time, and records that at $t=0$, the mass is at $x=+2.0$ cm, its speed is zero, and the time between consecutive maximum negative displacements is 1.2 seconds. (a) Calculate the angular frequency of the SHM. (b) Write the complete equation for the position of the mass as a function of time. (c) For what range of times between $t=0$ and $t=1.2$ s is the acceleration of the mass negative? Justify your answer.

Worked Solution: (a) The time between consecutive identical maximum displacements is the full period of SHM, so $T=1.2$ s. Angular frequency is calculated as: $$\omega =\frac{2\pi }{T}=\frac{2\pi }{1.2}=\frac{5\pi }{3}\approx 5.24\text{ rad/s}$$ (b) At $t=0$, $x(0)=+A=+2.0$ cm, and speed is zero, so phase constant $\varphi =0$. The equation (in units of centimeters) is: $$x(t)=2.0\cos \left(\frac{5\pi }{3}t\right)\text{ cm}$$ Or in meters: $x(t)=0.02\cos \left(\frac{5\pi }{3}t\right)\text{ m}$, which is also acceptable. (c) From the defining SHM relationship $a=-{\omega }^{2}x$, ${\omega }^2$ is always positive, so acceleration is negative when $x>0$. One full period is 1.2 s, so $x$ is positive between $0<t<0.3$ s (as the mass moves from +2 cm to 0) and between $0.9<t<1.2$ s (as the mass moves from 0 back to +2 cm). Acceleration is negative for these two time intervals.

Question 3 (Application / Real-World Style)

A seismograph measures the horizontal motion of the ground during a small earthquake. The seismograph's internal mass undergoes simple harmonic motion with amplitude 1.5 cm and frequency 2.5 Hz. Assuming the mass starts at maximum positive displacement at $t=0$, calculate the maximum acceleration of the mass during the SHM, then find the maximum force the spring exerts on the 0.1 kg mass.

Worked Solution: First, convert frequency to angular frequency: $\omega =2\pi f=2\pi (2.5)=5\pi \approx 15.7$ rad/s. Convert amplitude to meters: $A=1.5$ cm $=0.015$ m. Maximum acceleration for SHM is: $$a_{max}=A{\omega }^2=(0.015\text{ m})(15.7\text{ rad/s}{)}^2\approx 3.7{\text{ m/s}}^2$$ Use Newton's second law to find maximum force: $F_{max}=m{a}_{max}=(0.1\text{ kg})(3.7{\text{ m/s}}^2)=0.37\text{ N}$. In context, this means the spring exerts a maximum force of ~0.4 N on the seismograph mass when the mass is at its maximum displacement from equilibrium, consistent with the motion from a small earthquake.

7. Quick Reference Cheatsheet

8. What's Next

This chapter lays the kinematic foundation for all SHM topics in Unit 7. 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 $\omega$, $T$, 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.

• Dynamics of Simple Harmonic Motion
• Energy in Simple Harmonic Motion
• Properties of Traveling Waves