7.3 Representing and analysing SHM

Four little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. A swing draws a wave

2. Reading the wave

Lena's swing wave is at its highest point. Where is Lena?

3. The seismograph

Read the story as text

An old seismograph has a heavy mass hung on a spring with a pen attached. Paper rolls past underneath at a steady speed. When the mass bobs up and down, the pen draws a smooth wave on the moving paper. The wave is not the path of the mass (the mass only moves up and down). It is a graph of position against time.

Scientists read these waves: the height tells them the amplitude, the spacing tells them the period.

The question: how do we write and read the graphs and the equation of simple harmonic motion?

The tops of a seismograph wave are 2 s apart. What is the period?

4. Check yourself

Think of your answer first, then tap to see it.

a) The swing's wave is at a top. Where is the kid, and how fast is she moving?

Show answer

At the far end of the swing (as far out as she goes), and stopped for an instant: the graph is flat there, so v = 0.

b) The wave crosses the middle line, going down. Where is she now?

Show answer

At the middle (the lowest point of the swing), going fastest, moving in the negative direction. The graph is steepest there.

c) The tops of a wave are 4 s apart and 0.5 m above the middle line. What are T and A?

Show answer

T = 4 s (top to next top) and A = 0.5 m (middle line to top, not top to bottom). So f = 1/T = 0.25 Hz.

Already know this?

Get all three right on the first try and you can skip ahead.

An x-t graph goes from +0.40 m at its tops to −0.40 m at its bottoms. What is the amplitude?

On an x-t graph of SHM, where is the object moving fastest?

An oscillator is pulled to +A and let go at t = 0. Which equation fits?

One equation and three wavy graphs tell you everything about a simple harmonic oscillator: where it is, how fast it is going and which way it is being pushed, at any moment.

The idea: a cosine wave

In pictures

  • A pen tied to the swing (or the spring) draws a wave on paper that slides by: a graph of position against time.
  • Top of the wave: at the far end, stopped for an instant. Crossing the middle line: in the middle, going fastest.
  • Top to next top is one period T. Middle line to top is the amplitude A.
  • Pulled out and let go, the wave starts at its top: that is a cosine.

Below is the same idea in words, and then with numbers.

If the object starts at x = +A and is let go at t = 0, its position is

x = A cos(2πft)

A is the amplitude, f is the frequency, t is time. When t goes up by one period T = 1/f, the angle 2πft goes up by 2π, and the motion repeats. Set your calculator to radians.

If instead it starts at x = 0 moving in the positive direction, use x = A sin(2πft). Both are on the AP equation sheet. Where the object is at t = 0 (and which way it is moving) is called the starting phase.

Reading a position-time graph

t x +A −A T T/2 P M Q R S
PointWherexv (slope of x-t)a = −(k/m)x
Ptop: x = +A+ (max)0− (most negative)
Mbetween top and zero, going down+−−
Qcrossing zero, going down0− (fastest)0
Rbottom: x = −A− (min)0+ (most positive)
Scrossing zero, going up0+ (fastest)0

Two rules that answer most questions:

1. Velocity is the slope of the x-t graph: zero at peaks and valleys, biggest where the graph crosses the axis.

2. Acceleration always has the opposite sign to x: biggest at the ends, zero at x = 0. The a-t graph is the x-t graph flipped upside down.

The largest speed and acceleration are vmax = 2πfA and amax = (2πf)²A.

With numbers

A = 0.20 m, f = 0.50 Hz (so T = 2.0 s), starting at +A. Positive is to the right.

Show all steps as text
  1. Equation: x = 0.20 cos(2π · 0.50 · t) = 0.20 cos(πt), in metres. Put A and f into x = A cos(2πft).
  2. At t = 0.25 s: x = 0.20 cos(π · 0.25) = 0.20 cos(0.785 rad) = 0.20 × 0.707 = 0.14 m. Radians! In degrees mode you would get a wrong answer.
  3. At t = 0.50 s (T/4): x = 0.20 cos(π/2) = 0. It is passing through equilibrium, moving left at top speed.A quarter period after the start.
  4. At t = 1.0 s (T/2): x = 0.20 cos(π) = −0.20 m, at rest for an instant. Half a period gets it to the other side.
  5. Top speed: vmax = 2πfA = 2π(0.50)(0.20) = 0.63 m/s. Top acceleration: amax = (2πf)²A = (π)²(0.20) = 1.97 m/s². vmax happens at x = 0, amax at x = ±A.

Play: change the starting phase

The sim uses x = A cos(2πft + φ). φ = 0° starts at +A; 90° starts at 0 moving left; 180° starts at −A; 270° starts at 0 moving right. Predict the velocity graph first, then step frame by frame and check the rules: v = 0 at the peaks of x, a opposite to x.

position xvelocity vacceleration a

Worked examples

Write the equation from a graph basic

An x-t graph starts at its highest point, x = 0.30 m, and the next highest point is at t = 0.80 s. Write x(t).

Show solution
Show all steps as text
  1. A = 0.30 m (height of the peak above the middle line).Amplitude is measured from equilibrium, not peak to valley.
  2. T = 0.80 s (peak to next peak), so f = 1/0.80 = 1.25 Hz.One full cycle between matching points.
  3. It starts at the top, so use cosine: x = 0.30 cos(2π · 1.25 · t) = 0.30 cos(2.5πt).Cosine starts at its maximum.

Signs at a point basic

Using the graph above, give the signs of x, v and a at point M.

Show solution
Show all steps as text
  1. x: the point is above the axis, so x is positive.Read the height.
  2. v: the graph is sloping down, so v is negative.v is the slope of x-t.
  3. a: opposite to x, so a is negative. The object is speeding up (v and a same sign) as it heads to equilibrium.a = −(k/m)x.

Using the equation medium

x = 0.15 cos(4πt), with x in metres and t in seconds. Find A, f, T and x at t = 0.10 s.

Show solution
Show all steps as text
  1. A = 0.15 m. 2πf = 4π, so f = 2.0 Hz and T = 0.50 s.Match to x = A cos(2πft).
  2. x(0.10) = 0.15 cos(0.4π) = 0.15 cos(1.257 rad) = 0.15 × 0.309 = 0.046 m.0.10 s is a fifth of a period: past the top, not yet at the middle.

Top speed and top acceleration medium

For the same motion, find the largest speed and acceleration, and where they happen.

Show solution
Show all steps as text
  1. vmax = 2πfA = 2π(2.0)(0.15) = 1.9 m/s, at x = 0.Fastest at equilibrium.
  2. amax = (2πf)²A = (4π)²(0.15) = 158 × 0.15 = 24 m/s², at x = ±0.15 m.Biggest force at the ends.

Two blocks a quarter period apart AP

Blocks 1 and 2 are on identical springs. Block 1 is released from +A at t = 0. Block 2 passes x = 0 moving in the +x direction at t = 0. Describe x, v and a for each block at t = T/4.

Show solution
Show all steps as text
  1. Block 1: x = A cos(2πft). At T/4 the angle is π/2: x = 0, moving in the −x direction at top speed, a = 0.A quarter cycle takes it from the end to the middle.
  2. Block 2: x = A sin(2πft). At T/4: x = +A, v = 0, a = −amax (pointing back toward x = 0).A quarter cycle takes it from the middle to the end.
  3. Block 2's graph is block 1's graph shifted right by T/4: they have the same A and T but different phase.Same springs and masses → same period; only the start differs.

Practice

  1. On the position-time graph of an oscillating block, where is the block's speed greatest?

    Show answer

    (C). Speed is the size of the slope. The graph is steepest where it crosses x = 0 and flat at peaks and valleys.

  2. x = 0.10 cos(6πt), in metres and seconds. What is the period?

    Show answer

    (C). 2πf = 6π, so f = 3.0 Hz, and T = 1/f = 0.33 s. (B) is the frequency in the wrong unit.

  3. On the graph above, what are the signs of (x, v, a) at point Q?

    Show answer

    (B). At Q the graph crosses zero (x = 0, so a = 0) while sloping down (v negative, at its largest size). (D) describes point M.

  4. At t = 0 a block is at equilibrium moving in the +x direction. Which equation describes it?

    Show answer

    (C). sin(0) = 0, so it starts at x = 0, and sin increases at first, so it moves in +x. (D) starts at 0 but moves in −x.

  5. The velocity of an oscillating block is at its most positive value when the x-t graph is

    Show answer

    (D). The slope is steepest and positive there (point S on the graph).

  6. A block has x = 0.050 cos(2π · 4.0 · t). Where is it at t = 1/16 s?

    Show answer

    (A). T = 1/4.0 = 0.25 s, and 1/16 s = 0.0625 s = T/4. The angle is 2π(4.0)(0.0625) = π/2, and cos(π/2) = 0. (D) is what you get with cos(45°), from using the wrong angle.

  7. Short answer. A student says: "The acceleration is greatest where the speed is greatest." Use the x-t graph to explain why this is wrong.

    Show answer

    Speed is greatest where the x-t graph is steepest, which is where it crosses x = 0. There the spring is at equilibrium length, so the force and acceleration are zero (a = −(k/m)x = 0). The acceleration is greatest at the peaks and valleys, where x = ±A, and there the slope (velocity) is zero. So the two are greatest at different places, a quarter period apart.

    Scoring idea: 1 point: speed max at x = 0 (steepest slope); 1 point: a ∝ −x so a = 0 there and max at ±A.

  8. Short answer. x = 0.080 cos(πt) (metres, seconds). Find (a) x at t = 0.50 s and 1.0 s, (b) the first time the speed is greatest and that speed.

    Show answer
    1. f = 0.50 Hz, T = 2.0 s. x(0.50) = 0.080 cos(π/2) = 0; x(1.0) = 0.080 cos(π) = −0.080 m.Quarter and half period.
    2. Speed is greatest at x = 0, first at t = 0.50 s, moving in the −x direction.It starts at +A and heads toward the middle.
    3. vmax = 2πfA = 2π(0.50)(0.080) = 0.25 m/s.Top speed formula.

AP question types for this topic

Here are four short free response questions, one of each AP Physics 1 type, all about this topic. Try each one on paper first, then open the worked answer.

1. Mathematical Routines

Owen pulls a glider on an air track (attached to a spring) to x = +A and releases it at t = 0. The glider oscillates with period T and no friction.

  1. Write x(t). Then derive an expression, in terms of T, for the first time the glider reaches x = +A/2.
  2. For A = 0.080 m and T = 1.6 s, calculate that time and the glider's largest speed.
  3. Where is the glider at t = 1.0 s, and which way is it moving?
Show worked answer and scoring

(a)

  1. It starts at +A, so x(t) = A cos(2πt/T).
  2. Set x = A/2: cos(2πt/T) = ½. The first angle with cosine ½ is π/3, so 2πt/T = π/3 and t = T/6.

(b)

  1. t = T/6 = 1.6/6 = 0.27 s.The glider is slow near the end, so covering the first half of the distance takes more than half of the quarter period: T/6 is more than T/8.
  2. vmax = 2πfA = 2πA/T = 2π(0.080)/1.6 = 0.314 m/s.

(c)

  1. Angle = 2π(1.0)/1.6 = 3.927 rad (calculator in radians). x = 0.080 cos(3.927) = -0.057 m.
  2. t = 1.0 s is between T/2 = 0.8 s (at −A) and 3T/4 = 1.2 s (at 0), so the glider is left of equilibrium and moving right, back toward x = 0.

AP-style scoring

  • 1 point for x = A cos(2πt/T) (cosine because it starts at +A), and 1 point for t = T/6.
  • 1 point for t ≈ 0.27 s and vmax ≈ 0.31 m/s.
  • 1 point for x ≈ −0.057 m with the direction (moving right) justified by where t falls in the cycle.

2. Translation Between Representations

Sana records a block on a spring with a motion sensor. The position-time graph for one period, starting from release at +A, is shown first. Positive is to the right.

The x-t graph (one period), then four possible velocity-time graphs for the same time:

tx
(x) position x against t over one period
tv
(A) starts at 0, dips down first
tv
(B) starts at 0, rises first
tv
(C) starts at the top
tv
(D) starts at the bottom
  1. Which graph, A to D, is the velocity-time graph?
  2. Justify your choice using slopes of the x-t graph at t = 0 and at t = T/4.
  3. Describe the shape of the a-t graph in words. If A = 0.12 m and f = 0.50 Hz, give the largest speed and the largest acceleration.
Show worked answer and scoring

(a)

  1. Graph A: starts at zero and goes negative first.

(b)

  1. Velocity is the slope of x-t. At t = 0 the x-t graph is at its peak, where the slope is zero, so v starts at 0. That rules out C and D.
  2. Just after t = 0 the x-t graph slopes down, and at T/4 it crosses zero with its steepest negative slope, so v is most negative at T/4. That is graph A, not B.

(c)

  1. The a-t graph is the x-t graph flipped upside down (a = −(k/m)x): it starts at its most negative value, crosses zero at T/4, and is most positive at T/2.
  2. vmax = 2πfA = 2π(0.50)(0.12) = 0.38 m/s; amax = (2πf)²A = (3.142)²(0.12) = 1.18 m/s².

AP-style scoring

  • 1 point for graph A.
  • 1 point for v = 0 at t = 0 (zero slope at the peak), and 1 point for negative v just after release (downward slope).
  • 1 point for the inverted-cosine a-t description, with vmax ≈ 0.38 m/s and amax ≈ 1.2 m/s².

3. Experimental Design and Analysis

Hana and Leo have a cart on a spring on a level track, a motion sensor connected to a computer that draws x-t and v-t graphs, and a metre stick. They want to test the claim that the largest speed is proportional to the amplitude, and use it to find the period without a stopwatch.

Their data:

Amplitude A (m)Largest speed vmax (m/s)
0.0200.106
0.0400.208
0.0600.316
0.0800.417
0.1000.525
  1. Describe how they get A and vmax for each trial from the sensor's graphs.
  2. What should they graph so it is linear, and what does the slope mean?
  3. Use the data to find the slope and the period T.
Show worked answer and scoring

(a)

  1. Release the cart from several different distances. On the x-t graph, A is half the distance from a peak to the next valley (or peak minus the middle line).
  2. On the v-t graph, vmax is the height of a peak (it happens where the x-t graph crosses the middle). Average several peaks to reduce error.

(b)

  1. vmax = 2πfA = (2π/T)·A. Graph vmax against A: a straight line through the origin if the claim is right.
  2. Slope = 2π/T = 2πf, so T = 2π/slope.

(c)

  1. Best-fit line: slope ≈ 5.24 s⁻¹ (intercept ≈ +0.000 m/s, essentially zero), so the points do lie on a line through the origin.
  2. T = 2π/slope = 6.283/5.24 ≈ 1.20 s.

AP-style scoring

  • 1 point for reading A from the x-t graph and vmax from the v-t graph peaks, and 1 point for repeats or averaging.
  • 1 point for vmax against A with slope 2π/T.
  • 1 point for slope ≈ 5.3 s⁻¹ and T ≈ 1.2 s from the best-fit line.

4. Qualitative/Quantitative Translation

Two identical blocks on identical springs are released at the same instant: Ines pulls hers out to A, Felix pulls his out to 2A. Felix claims: "Both blocks pass through equilibrium at the same moment, but mine is going twice as fast when it gets there."

  1. Explain in words, using the shapes of the two x-t graphs, why the claim is correct.
  2. Use x = A cos(2πft) to show the times at which each block is at x = 0, and the speed there.
  3. Check with T = 0.80 s, A = 0.050 m. Which part of your derivation matches each part of the claim?
Show worked answer and scoring

(a)

  1. The period does not depend on amplitude, so both x-t graphs are cosines with the same period: Felix's is the same wave stretched to twice the height. They cross zero at the same times.
  2. Stretching a graph vertically by 2 doubles every slope, and the slope at the crossing is the speed there, so Felix's block is twice as fast.

(b)

  1. x = 0 when cos(2πft) = 0, i.e. 2πft = π/2, so t = 1/(4f) = T/4 for both (A does not appear).
  2. Speed at x = 0 is vmax = 2πfA: proportional to A, so 2A gives twice the speed.

(c)

  1. Both at x = 0 at t = T/4 = 0.20 s. Ines: v = 2π(0.050)/0.80 = 0.39 m/s; Felix: 2π(0.100)/0.80 = 0.79 m/s, twice as fast. ✓
  2. “Same moment” is t = T/4 with no A in it; “twice as fast” is the factor A in vmax = 2πfA.

AP-style scoring

  • 1 point for “same period, so same zero crossings” and 1 point for “twice the height, twice the slope”.
  • 1 point for t = T/4 independent of A, and vmax = 2πfA.
  • 1 point for the numbers (0.20 s, 0.39 m/s, 0.79 m/s) linked back to the claim.

Common mistakes

The mistake: calculator in degrees.

Why it is wrong: 2πft is an angle in radians. cos(π/4) in degree mode gives cos(0.785°) ≈ 1.

How to spot it: check that x at t = T/4 comes out as 0.

The mistake: thinking the wave is the path of the object.

Why it is wrong: a spring-block only moves back and forth on a line. The wave is position plotted against time.

How to spot it: read the axis labels: x up the side, t along the bottom.

The mistake: "biggest x means biggest v".

Why it is wrong: at the peaks the slope is flat, so v = 0. v is biggest where x = 0.

How to spot it: velocity is the slope, not the height.

The mistake: reading the period as peak to valley, or the amplitude as peak to valley.

Why it is wrong: peak to valley is half a period in time and 2A in height.

How to spot it: period = peak to next peak; amplitude = middle line to peak.