AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/7/7-3)
Unit 7 · Topic 7.3
7.3 Representing and Analyzing SHM
An object in SHM has a position that follows a sine or cosine curve in time, x = A cos(2πft) or x = A sin(2πft), so its velocity and acceleration graphs are sinusoidal too. Speed is greatest at equilibrium, where acceleration is zero; at the turning points the speed is zero and the acceleration is greatest, pointing back toward equilibrium.
Key terms
- amplitude
- sinusoidal graph
- turning point
- maximum speed
- maximum acceleration
The position equation
Measured from equilibrium, an SHM object's position is x = A cos(2πft) or x = A sin(2πft). The amplitude A is the largest displacement from equilibrium, and f is the frequency. The angle 2πft is in radians.
Which one you use depends on where the motion starts. Use cosine if the object is released from rest at x = +A when t = 0. Use sine if it passes through equilibrium moving in the + direction at t = 0.
The object travels between x = +A and x = −A, so the total distance from one turning point to the other is 2A, and it covers 4A in one full period.
Three key positions
Everything follows from a = −(k/m)x: acceleration is proportional to displacement and points the other way. Combine that with the fact that the object stops at each turning point.
| Position | Velocity | Acceleration and net force |
|---|---|---|
| x = +A (turning point) | zero | largest, pointing in the − direction |
| x = 0 (equilibrium) | largest speed | zero |
| x = −A (turning point) | zero | largest, pointing in the + direction |
Reading the graphs together
Picture a block released from x = +A. Its position–time graph is a cosine curve: it starts at its peak, crosses zero a quarter period later, reaches −A at half a period and returns to +A after one period.
Velocity is the slope of the position graph. The slope is zero at each peak and valley and steepest where the curve crosses zero. So the velocity graph starts at zero, dips to its most negative value at a quarter period (the block is moving fastest, toward −x), returns to zero at half a period, and so on. It's an upside-down sine curve.
Acceleration is always opposite to position, so the acceleration–time graph is the position graph flipped upside down: it starts at its most negative value, exactly when position is most positive.
All three graphs have the same period. The velocity graph is shifted a quarter cycle from the position graph, and the acceleration graph is shifted half a cycle from it, which is why it looks like the position graph upside down.
Maximum speed and maximum acceleration
The largest acceleration occurs at the turning points: a_max = kA/m for a spring, straight from Hooke's law. The largest speed occurs at equilibrium. You can get it from energy (7.4): v_max = A√(k/m).
Both are proportional to A. Double the amplitude and the object moves twice as fast and accelerates twice as hard, but the period doesn't change. That's why the period of SHM doesn't depend on amplitude: a longer trip is covered at proportionally higher speeds.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Using the position equation
A block on a spring has position x = (0.080 m) cos(2π · 2.5t), with t in seconds. Find (a) the amplitude, frequency and period, (b) the position at t = 0.050 s, t = 0.10 s and t = 0.20 s.
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- Step 1: (a) Compare with x = A cos(2πft): A = 0.080 m and f = 2.5 Hz, so T = 1/f = 0.40 s.
- Step 2: (b) At t = 0.050 s: 2π(2.5)(0.050) = π/4 rad, and cos(π/4) ≈ 0.707, so x ≈ 0.057 m. Set your calculator to radians.
- Step 3: At t = 0.10 s (a quarter period): the angle is π/2 and cos(π/2) = 0, so x = 0. The block is at equilibrium, moving fastest.
- Step 4: At t = 0.20 s (half a period): the angle is π and cos π = −1, so x = −0.080 m, the other turning point.
Answer: (a) A = 0.080 m, f = 2.5 Hz, T = 0.40 s (b) about 0.057 m, 0 and −0.080 m
- Example 2Calculator allowed
Reading a position–time graph
A graph of an oscillating cart's position against time starts at x = 0 at t = 0, rises to a peak of +0.15 m at t = 0.25 s, returns to 0 at t = 0.50 s, reaches −0.15 m at t = 0.75 s and is back at 0 at t = 1.00 s, repeating after that. Find A, T and f, write x(t), and state when the cart's speed is greatest and when its acceleration is greatest in the + direction.
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- Step 1: The peak is 0.15 m above equilibrium, so A = 0.15 m. One full cycle takes 1.00 s, so T = 1.0 s and f = 1.0 Hz.
- Step 2: It starts at equilibrium heading in the + direction, so use sine: x = (0.15 m) sin(2πt).
- Step 3: Speed is greatest where the graph is steepest, at the zero crossings: t = 0, 0.50 s and 1.00 s.
- Step 4: Acceleration points opposite the displacement, so it's largest in the + direction where x is most negative: t = 0.75 s.
Answer: A = 0.15 m, T = 1.0 s, f = 1.0 Hz, x = (0.15 m) sin(2πt); fastest at t = 0, 0.50 s, 1.00 s; largest + acceleration at t = 0.75 s
- Example 3Calculator allowed
Where is acceleration largest? (classic trap)
A 0.50 kg block on a spring with k = 50 N/m oscillates with amplitude 0.10 m. A student says the acceleration is greatest at equilibrium, because that's where the speed is greatest. Find the maximum acceleration and maximum speed, and say where each occurs.
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- Step 1: Acceleration comes from the net force, which is the spring force kx. That's zero at equilibrium (x = 0), so the acceleration there is zero, even though the speed is greatest.
- Step 2: The force is largest at the turning points, x = ±A: a_max = kA/m = (50)(0.10)/0.50 = 10 m/s², pointing toward equilibrium.
- Step 3: The speed is greatest at equilibrium: v_max = A√(k/m) = (0.10)√(50/0.50) = (0.10)(10) = 1.0 m/s.
- Step 4: The student mixed up velocity with acceleration. Large speed doesn't mean large acceleration; acceleration tracks how quickly the velocity is changing.
Answer: a_max = 10 m/s² at the turning points; v_max = 1.0 m/s at equilibrium
Common mistakes
- Saying acceleration is greatest at equilibrium. It's zero there; it's greatest at the turning points.
- Using degrees instead of radians in x = A cos(2πft).
- Reading the amplitude as the full peak-to-valley height. A is measured from equilibrium to one peak.
- Drawing the acceleration graph with the same shape as the position graph. It's the position graph flipped upside down.
On the exam
- Translation questions ask you to sketch velocity or acceleration graphs from a position graph, or the reverse. Line up the zeros and extremes carefully, and keep the same period.
- Expect questions asking where (or when) speed, acceleration, force or displacement is zero or greatest. Justify with the restoring force: it's proportional to displacement and points toward equilibrium.
Connected topics
Videos
Check yourself
4 questions on 7.3 Representing and Analyzing SHM. Pick an answer to see if you got it, and why.
An object's position is x = (0.050 m) cos(4πt), with t in seconds. What are the amplitude and period of the motion?
A block on an ideal spring oscillates between x = −A and x = +A. Which describes the block at x = +A?
A block on an ideal horizontal spring passes through its equilibrium position, moving in the +x direction. Which describes it at that instant?
An oscillator's position is x = A cos(2πt/T). Which describes the object at t = T/4?
0 of 4 answered