Representing and Analyzing SHM
Why this matters
Imagine you're at a playground, pushing your younger cousin, Leo, on a swing. You pull the swing back as far as it can go and let go. He swings forward, reaches the highest point on the other side, and swings back. You watch this repeating motion: back and forth, back and forth.
That predictable, rhythmic motion is a perfect example of what physicists call Simple Harmonic Motion (SHM). But how could we describe it with precision? Where is Leo moving fastest? Where does he momentarily stop? Where does he feel the biggest push?
Answering these questions isn't just about swings. It's the key to understanding everything from the vibrations of a guitar string to the design of earthquake-proof buildings. In this lesson, we'll learn how to represent this motion with graphs and equations, turning that intuitive playground scene into the language of physics.
Diagram
Concept map
flowchart TD
A[Position at +A] --> B{v = 0, a = max negative};
B --> C[Position at 0];
C --> D{v = max negative, a = 0};
D --> E[Position at -A];
E --> F{v = 0, a = max positive};
F --> G[Position at 0];
G --> H{v = max positive, a = 0};
H --> A;
Core explanation
When we talk about Simple Harmonic Motion (SHM), we're describing a very specific kind of oscillation. It's the motion that happens when an object has a stable equilibrium position, and if you move it away, a restoring force pulls it back. The farther you pull it, the stronger the force. A mass on a spring or a pendulum swinging through a small angle are the classic examples.
Let's build our understanding from the ground up, using a horizontal mass on a frictionless spring.
Describing Position with an Equation
Imagine our mass starts at its maximum displacement from the center. We'll call this center point x = 0 (the equilibrium position). The maximum distance it's pulled to is the amplitude, which we label A.
If we release the mass from x = +A at time t = 0, it will oscillate back and forth between +A and -A. Its position at any given time t can be described by a cosine function:
x(t) = A cos(2πft)
Let's break that down:
x(t)is the position at timet.Ais the amplitude: the maximum displacement from equilibrium.fis the frequency: how many full cycles (back-and-forth motions) happen per second, measured in Hertz (Hz).tis the time in seconds.
Sometimes you'll see this written with period T instead of frequency f. Since f = 1/T, the equation is the same: x(t) = A cos(2πt / T).
What if we started timing when the mass was passing through equilibrium? In that case, we'd use a sine function instead: x(t) = A sin(2πft). For the AP exam, you need to be comfortable with both, but they describe the same physical motion—just with a different starting point for your stopwatch.
The Relationship Between Position, Velocity, and Acceleration
This is the heart of the topic, and where we need to be very clear. Let's think about our mass on the spring.
- At the endpoints (
x = +Aorx = -A): The mass momentarily stops to change direction. This means its velocity is zero. The spring is stretched or compressed the most, so the restoring force is at its maximum. SinceF = ma, the acceleration is also at its maximum, but it points in the opposite direction of the displacement (back toward equilibrium). - At the equilibrium position (
x = 0): The spring is not stretched or compressed, so the restoring force is zero. This means the acceleration is zero. The mass isn't being pushed or pulled at this instant. However, it's not stopping! This is where it's moving the fastest. The velocity is at its maximum.
This is where many students get confused. They think that where the object is moving fastest, the acceleration must also be large. But that's not true! Acceleration is about the change in velocity, driven by force. At equilibrium, the force is zero, so the acceleration is zero, even though the speed is at its peak.
Visualizing SHM with Graphs
The best way to see these relationships is with graphs of position, velocity, and acceleration versus time.
(Imagine the three stacked graphs from the visual plan here)
-
Position vs. Time (
xvs.t): If we start at maximum displacement, this is a cosine curve. It starts at+A, goes to0, then-A, then0, and back to+Ain one full period,T. -
Velocity vs. Time (
vvs.t): Velocity is the slope of the position graph.- At
t=0, the position graph is flat at its peak, so the slope (velocity) is zero. - As the object moves toward equilibrium (
t=0tot=T/4), the position slope is negative and gets steeper. So, velocity is negative, reaching its most negative value (-v_max) when the object passes through equilibrium (x=0). - This pattern continues, creating a negative sine curve.
- At
-
Acceleration vs. Time (
avs.t): Acceleration is the slope of the velocity graph. It's also directly proportional to displacement but in the opposite direction (ais proportional to-x).- At
t=0, the positionxis at its maximum positive value (+A). Therefore, accelerationamust be at its maximum negative value (-a_max). - When
x=0,a=0. - When
x=-A,a=+a_max. - This creates a negative cosine curve—it looks like the position graph, but flipped upside down.
- At
An Important Independence: Amplitude and Period
Here's a critical concept that often appears on the exam. Let's say you have a mass on a spring. You pull it 5 cm and let it go. It takes 2 seconds to complete a cycle.
Now, what happens if you pull it 10 cm and let it go? How long will its period be?
The surprising answer is: it's still 2 seconds.
For a simple harmonic oscillator (like a mass-spring system or a simple pendulum with small angles), the period does not depend on the amplitude.
Think about it this way: if you pull it farther, the restoring force is stronger (F = -kx). This stronger force causes a greater acceleration, making the mass move faster over the longer distance. These two effects—a longer distance to travel and a higher average speed—perfectly cancel each other out, resulting in the same period. This is a defining feature of SHM.
Worked examples
Interpreting the SHM Equation
A 0.5 kg block attached to a spring is undergoing SHM. Its position is described by the equation x(t) = 0.10 cos(4πt), where x is in meters and t is in seconds.
(a) What are the amplitude and period of the motion?
(b) What is the block's position at t = 0.25 s?
(c) Where is the block when its velocity is maximum? Where is it when its acceleration is maximum?
Solution:
(a) Amplitude and Period
- Step 1: Identify the general formThe general equation is
x(t) = A cos(2πft). - Step 2: Compare the given equation to the general form
- Our equation is
x(t) = 0.10 cos(4πt). - By direct comparison, the amplitude
Ais the number out front. So,A = 0.10m.
- Our equation is
- Step 3: Find the frequency, then the period
- The term inside the cosine is
2πft. In our equation, that term is4πt. - So,
2πft = 4πt. We can cancelπtfrom both sides to get2f = 4, which means the frequencyf = 2Hz. - The period
Tis the inverse of the frequency:T = 1/f. - Therefore,
T = 1/2 = 0.5s.
- The term inside the cosine is
(b) Position at a specific time
- Step 1: Plug the time into the equationWe want to find
xwhent = 0.25s.x(0.25) = 0.10 cos(4π * 0.25)x(0.25) = 0.10 cos(π)
- Step 2: Evaluate the cosineRemember your unit circle!
cos(π)is -1.x(0.25) = 0.10 * (-1) = -0.10m.
- Why this makes senseThe period is 0.5 s. The time
t = 0.25s is exactly half a period (T/2). The object starts at+Aatt=0, so after half a cycle, it should be at the opposite extreme,-A. Our calculation confirms this.
(c) Locations of max velocity and max acceleration
- This is a conceptual question, not a calculation
- Maximum VelocityVelocity is greatest when the object passes through the equilibrium position, where the net force is zero. So, max velocity occurs at
x = 0. - Maximum AccelerationAcceleration is greatest where the net force is greatest (
F=ma). The spring force is greatest at maximum displacement. So, max acceleration occurs at the endpoints,x = +0.10m andx = -0.10m.
Try it yourself
Practice Problem
A child's bouncy seat acts like a spring. A 10 kg baby, Liam, sits in the seat, and it oscillates with a period of 1.5 seconds and an amplitude of 20 cm.
- If you were to graph Liam's position versus time, starting the clock when he is at the lowest point of his bounce, would the graph be best modeled by a sine, cosine, negative sine, or negative cosine function? Why?
- At what point(s) in the motion is Liam's acceleration zero? At what point(s) is his speed at its maximum?
- If Liam's older sister, Priya, pushes down on him to increase the amplitude of the bounce to 30 cm, what will the new period of the motion be?
Hints:
- The "lowest point" is the maximum negative displacement. What does the cosine graph look like when it's flipped upside down?
- Remember the relationship between force, acceleration, and position. Where is the "spring" of the seat not stretched or compressed from its new equilibrium with the baby in it?
- Think about the key rule connecting amplitude and period for SHM.
Practice — 8 questions
In simple terms, this lesson is about how to describe and graph the back-and-forth motion of oscillating objects, like a pendulum or a mass on a spring, using position, velocity, and acceleration.
- 7.3.A: Describe the displacement, velocity, and acceleration of an object exhibiting SHM.
- 7.3.A.1
- For an object exhibiting SHM, the displacement of that object measured from its equilibrium position can be represented by the equations x = A cos(2πft) or x = A sin(2πft).
- 7.3.A.1.i
- Minima, maxima, and zeros of displacement, velocity, and acceleration are features of harmonic motion.
- 7.3.A.1.ii
- Recognizing the positions or times at which the displacement, velocity, and acceleration for SHM have extrema or zeros can help in qualitatively describing the behavior of the motion.
- 7.3.A.2
- Changing the amplitude of a system exhibiting SHM will not change the period of that system.
- 7.3.A.3
- Properties of SHM can be determined and analyzed using graphical representations.
flowchart TD
A[Position at +A] --> B{v = 0, a = max negative};
B --> C[Position at 0];
C --> D{v = max negative, a = 0};
D --> E[Position at -A];
E --> F{v = 0, a = max positive};
F --> G[Position at 0];
G --> H{v = max positive, a = 0};
H --> A;
Read what Saavi narrates
Hi everyone, it's Saavi from Shrutam.
Have you ever pushed a little kid on a swing? You pull the swing back, let go, and watch it go back and forth. That simple, repeating motion is what we're talking about today: Simple Harmonic Motion. We're going to learn how to describe it precisely, using the language of physics. Where is the swing moving fastest? Where does it stop? We'll find out.
Essentially, we're translating that physical motion into math. We'll see how the swing's position, its velocity, and its acceleration all change over time in a beautiful, predictable rhythm that we can capture with graphs and equations.
Let's walk through an example. Imagine a block on a spring, and its position is given by the equation x equals zero point one zero times the cosine of four pi t.
First, what's the amplitude? That's the number in front, so the amplitude is zero point one zero meters.
Next, the period. The term inside the cosine is two pi f t. In our equation, it's four pi t. So, two f must equal four, which means the frequency, f, is two Hertz. The period is just one over the frequency, so the period T is one half, or zero point five seconds.
Now, let's think conceptually. Where is the block when its velocity is maximum? This is a classic point of confusion. The velocity is greatest when the block zips through the middle, the equilibrium position, where x equals zero. And where is acceleration maximum? That happens at the endpoints, where the spring is pulling or pushing the hardest.
A very common mistake I see every year is thinking that maximum velocity happens at the endpoints. But think about the swing again. At the highest point of its arc, the swing stops for a split second. The velocity is zero. It's moving fastest at the very bottom. So, always remember: maximum speed happens at the equilibrium point, x equals zero. Zero speed happens at the endpoints.
Keep practicing these relationships. Look at the graphs, think about the physical motion, and you'll build that intuition. You've got this.
At maximum displacement, the object momentarily stops to change direction. Its velocity is zero. Maximum velocity occurs as it passes through the equilibrium point (`x=0`), where the net force is zero and it's no longer accelerating (in that instant).
Always associate maximum speed with the equilibrium position (`x=0`) and zero speed with the endpoints (`x = ±A`).
Acceleration is caused by force (`F=ma`). The restoring force from the spring (`F=-kx`) is strongest at the endpoints where the stretch/compression is greatest. Therefore, acceleration is also maximum at the endpoints.
Remember that acceleration is maximum where force is maximum: at the endpoints (`x = ±A`). Acceleration is zero where force is zero: at equilibrium (`x=0`).
For a standard mass-spring system or a simple pendulum, the period is independent of the amplitude. A larger amplitude means the object travels a greater distance, but it also has a greater average speed. These two effects cancel out perfectly.
On the AP exam, if a question asks how the period of a mass-spring system changes when you double the amplitude, the answer is "it doesn't change."
The three graphs have a precise mathematical relationship (based on calculus, though you don't need calculus for this course). The velocity graph is the slope of the position graph, and the acceleration graph is the slope of the velocity graph. They are phase-shifted from one another.
Memorize the key relationships: When `x` is max/min, `v` is zero. When `v` is max/min, `a` is zero. When `x` is max positive, `a` is max negative. Practice sketching the `v` and `a` graphs given an `x` graph.
`cos(0) = 1` and `sin(0) = 0`. Using the wrong function means your description of the motion is incorrect at `t=0`.
If the object starts at its maximum displacement (`x=A`), use the cosine function: `x = A cos(...)`. If the object starts at the equilibrium position and is moving in the positive direction, use the sine function: `x = A sin(...)`.