Unit 7.3 | Representing and Analyzing SHM
- 13 minutes
- 0 formulas
- 20 questions
- 2 videos
Position is a cosine, velocity is its slope, acceleration is it flipped. This lesson shows how to read and match every SHM graph and equation on the AP exam.
Unit 7 Breakdown
In this lesson:
- Write position as a function of time: [katex] x = A\cos(2\pi f t) [/katex] and [katex] x = A\sin(2\pi f t) [/katex]
- Read amplitude, period and frequency off a graph
- Sketch velocity and acceleration graphs from a position graph, and the other way around
- The 4 problem types: reading a position graph, using the equation, matching x, v and a, and force vs. position graphs
Introduction
This lesson covers how to represent simple harmonic motion: with an equation, with SHM graphs of position, velocity and acceleration against time, and with a force vs. position graph.
You know what SHM is (Unit 7.1) and how long a cycle takes (Unit 7.2). The graphs here use the same slope rules as the motion graphs from Unit 1, so nothing new to memorize, just a new shape.
The Equation of SHM
Position in SHM traces a cosine or a sine wave. Which one depends on where the motion starts.
| Starts at | Equation |
| A turning point, [katex] x = +A [/katex], released from rest | [katex] x = A\cos(2\pi f t) [/katex] |
| Equilibrium, [katex] x = 0 [/katex], moving in the + direction | [katex] x = A\sin(2\pi f t) [/katex] |
Both are on the AP equation sheet. [katex] A [/katex] is the amplitude, [katex] f [/katex] is the frequency, and [katex] 2\pi f = \frac{2\pi}{T} [/katex]. The angle [katex] 2\pi f t [/katex] is in radians, so put your calculator in radian mode. One full cycle is [katex] 2\pi [/katex] radians, so the motion repeats every [katex] T [/katex] seconds.
A pendulum or a block released from rest at [katex] x = -A [/katex] fits [katex] x = -A\cos(2\pi f t) [/katex]. Check any equation at [katex] t = 0 [/katex] before you use it.
▶ Watch: the SHM lines of the AP equation sheet, and how [katex] T = \frac{1}{f} [/katex] and [katex] \omega = 2\pi f [/katex] connect (starts at 18:42).
Position, Velocity and Acceleration Graphs
Here is a block released from rest at [katex] x = +0.20 \text{ m} [/katex], with a period of [katex] 2.0\,\text{s} [/katex]:
Position of the block
The other two SHM graphs follow from rules you already know:
- Velocity is the slope of the position graph. The slope is zero at every peak and trough (the turning points) and steepest where the graph crosses zero (equilibrium).
- Acceleration is [katex] a = -\frac{k}{m}x [/katex], so its graph is the position graph flipped upside down. When [katex] x [/katex] is at its most positive, [katex] a [/katex] is at its most negative.
x, v and a over one and a half cycles
| Where | x | v | a |
| Turning point [katex] +A [/katex] | [katex] +A [/katex] (max) | 0 | [katex] -a_{max} [/katex] |
| Equilibrium | 0 | [katex] \pm v_{max} [/katex] | 0 |
| Turning point [katex] -A [/katex] | [katex] -A [/katex] (min) | 0 | [katex] +a_{max} [/katex] |
Every SHM representation question on the AP exam is one of the four types below.
| Problem type | The clue | The setup |
| 1. Reading a position graph | an [katex] x [/katex] vs. [katex] t [/katex] graph: “amplitude”, “period”, “when is the speed greatest” | Peak height = [katex] A [/katex], peak to peak = [katex] T [/katex], zero crossings = fastest |
| 2. Using the equation | an equation like [katex] x = (0.12)\cos(4\pi t) [/katex], or “write an equation” | Match it to [katex] x = A\cos(2\pi f t) [/katex], check [katex] t = 0 [/katex] |
| 3. Matching x, v and a | pick the velocity or acceleration graph, or the signs at a given time | v = slope of x; a is x flipped |
| 4. Force vs. position graphs | a graph of the restoring force against [katex] x [/katex] | Slope [katex] = -k [/katex]; area = stored energy |
Problem Type 1: Reading a position graph
How to spot it: A graph of position against time for a block, glider or pendulum, with questions about amplitude, period, frequency, or when the speed or acceleration is largest.
Setup: Amplitude is the height of a peak above the middle line, not peak to trough. Period is the time from one peak to the next. The object is fastest where the graph crosses the middle line and stopped at every peak and trough.
Use the position graph above. Find the amplitude, period and frequency. At which times in the first [katex] 4\,\text{s} [/katex] is the block moving fastest? When does it have its largest positive acceleration?
Amplitude [katex] 0.20\,\text{m} [/katex]. Period [katex] 2.0\,\text{s} [/katex] (peaks at 0, 2.0 and [katex] 4.0\,\text{s} [/katex]), so [katex] f = \frac{1}{2.0} = 0.50 \text{ Hz} [/katex]. Fastest where the graph crosses zero: 0.5, 1.5, 2.5 and [katex] 3.5\,\text{s} [/katex]. Largest positive acceleration at [katex] x = -A [/katex], the troughs: 1.0 and [katex] 3.0\,\text{s} [/katex].
AP trap: Reading the amplitude as the distance from peak to trough. That is [katex] 2A [/katex]. And the period is peak to peak, not peak to trough.
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Problem Type 2: Using the equation
How to spot it: The question gives an equation for the position, asks you to write one, or asks where the object is at a given time.
Setup: Line it up with [katex] x = A\cos(2\pi f t) [/katex]: the number in front is [katex] A [/katex], the number multiplying [katex] t [/katex] is [katex] 2\pi f [/katex]. Check the starting point at [katex] t = 0 [/katex]. Use radians.
A block’s position is [katex] x = (0.12 \text{ m})\cos(4\pi t) [/katex]. Find its amplitude, frequency and period. A second block is released from rest at [katex] 5.0\,\text{cm} [/katex] from equilibrium and has a period of [katex] 0.80\,\text{s} [/katex]. Write its equation and find its position at [katex] 0.10\,\text{s} [/katex] and at [katex] 0.20\,\text{s} [/katex].
First block: [katex] A = 0.12 \text{ m} [/katex], [katex] 2\pi f = 4\pi [/katex] so [katex] f = 2.0 \text{ Hz} [/katex] and [katex] T = 0.50 \text{ s} [/katex].
Second block: released from rest, so it is a cosine: [katex] x = (5.0 \text{ cm})\cos\left(\frac{2\pi t}{0.80}\right) [/katex]. At [katex] 0.10\,\text{s} [/katex]: [katex] 5.0\cos\left(\frac{\pi}{4}\right) = 3.5 \text{ cm} [/katex]. At [katex] 0.20\,\text{s} [/katex], a quarter period: [katex] 5.0\cos\left(\frac{\pi}{2}\right) = 0 [/katex], it is passing through equilibrium.
Notice that at one eighth of a period the block is at [katex] 3.5\,\text{cm} [/katex], not halfway ([katex] 2.5\,\text{cm} [/katex]). It starts slow at the turning point, so it covers less than half the distance in the first half of the quarter cycle.
AP trap: Leaving the calculator in degree mode. [katex] \cos(4\pi \cdot 0.10) [/katex] means the cosine of 1.26 radians, not 1.26 degrees.
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Problem Type 3: Matching position, velocity and acceleration
How to spot it: Pick the velocity or acceleration graph that goes with a position graph, or give the signs of [katex] x [/katex], [katex] v [/katex] and [katex] a [/katex] at a certain time.
Setup: Velocity: read the slope of the position graph. Acceleration: flip the position graph. At each turning point [katex] v = 0 [/katex] and [katex] a [/katex] is largest, pointing toward equilibrium.
A pendulum is released from rest at [katex] x = +A [/katex] at [katex] t = 0 [/katex] and has period [katex] T [/katex]. Describe its position, velocity and acceleration at [katex] t = \frac{T}{4} [/katex] and at [katex] t = \frac{T}{2} [/katex].
At [katex] \frac{T}{4} [/katex] it is passing through equilibrium on its way to [katex] -A [/katex]: [katex] x = 0 [/katex], speed at its maximum and velocity negative, [katex] a = 0 [/katex]. At [katex] \frac{T}{2} [/katex] it has reached the other side: [katex] x = -A [/katex], [katex] v = 0 [/katex], and the acceleration is at its largest and positive, pointing back toward equilibrium.
AP trap: Drawing the acceleration graph in step with the position graph. Acceleration always points toward equilibrium, so where [katex] x [/katex] is positive, [katex] a [/katex] is negative.
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Problem Type 4: Force vs. position graphs
How to spot it: A graph of the force on the object (or the spring force) against its position, often from a lab with a force sensor.
Setup: A straight line through the origin with a negative slope means SHM, and the slope is [katex] -k [/katex]. With [katex] k [/katex] you have the period. The area between the line and the axis is the energy stored in the spring, which is the next lesson.
Force on a 0.40 kg block
The graph shows the force on a [katex] 0.40\,\text{kg} [/katex] block on a spring. Find the spring constant and the period of the block.
Read one point on the line: [katex] F = -6.0 \text{ N} [/katex] at [katex] x = 15 \text{ cm} = 0.15 \text{ m} [/katex]. Slope: [katex] k = \frac{6.0}{0.15} = 40 \text{ N/m} [/katex]. Period: [katex] T = 2\pi\sqrt{\frac{0.40}{40}} = 2\pi(0.10) \approx 0.63 \text{ s} [/katex]. If the block oscillates with amplitude [katex] 0.15\,\text{m} [/katex], the triangle between the line and the axis from 0 to [katex] 0.15\,\text{m} [/katex] has area [katex] \tfrac{1}{2}(0.15)(6.0) = 0.45 \text{ J} [/katex]: the energy stored in the spring at the turning point.
AP trap: Reading [katex] k [/katex] as the force at one point. The spring constant is the slope, force divided by displacement, in N/m.
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Lesson 7.4 Preview
The area under the force graph was energy. In the last lesson of the unit we follow that energy around the cycle: kinetic at the middle, potential at the ends, and the same total the whole way.
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This article was written by Jason