7.1 Simple harmonic motion

Why does a spring keep bouncing back and forth? Because the force always points home, and it gets stronger the further you go from home.

Little picture stories

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

1. Pull the block, then let go

2. The force always points home

The block is pulled twice as far from home. The spring's pull is:

3. A car over a speed bump

Read the story as text

A car rolls over a speed bump. The body of the car dips down, pops up, dips again, and slowly settles. Under each wheel is a big steel spring. Push the car down and the spring pushes it back up. Let it rise too far and the spring (and gravity) pull it back down. Every time it overshoots, something pulls it back toward the middle.

This back and forth around a middle point is called an oscillation. It shows up in guitar strings, swings, buildings in an earthquake and atoms in a solid.

The question: what kind of force makes an object oscillate in the smooth, regular way we call simple harmonic motion?

Where is a block on a spring moving fastest?

4. Check yourself

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

a) The block is to the left of home. Which way does the spring push it?

Show answer

To the right, toward home. The spring force always points back to equilibrium.

b) Where is the block moving fastest: at an end, or passing home?

Show answer

Passing home. It speeds up all the way in. At home the force is zero, so it shoots past. At the ends it stops for an instant.

c) A spring with k = 50 N/m is stretched 0.20 m. How big is its force? What if it is stretched 0.40 m?

Show answer

F = kx = (50)(0.20) = 10 N, pointing home. Twice the stretch gives twice the force: 20 N.

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Answer these three. All right on the first try? You can skip ahead.

In simple harmonic motion, the restoring force is:

A spring has k = 100 N/m. The block is at x = +0.05 m. What is the spring force?

At each end of its swing, a block on a spring has:

The idea: a force that always points home

In pictures

  • The block's home is where the spring is not stretched or squashed. Physics calls it equilibrium, x = 0.
  • Pull it away and the spring always points it back home: pull right, it pulls left; push left, it pushes right.
  • Further from home means a bigger pull. Twice as far, twice the pull.
  • At home there is no pull, but the block is moving fastest, so it shoots past. At each end it stops for an instant.

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

Picture a block on a smooth table, tied to a spring. With the spring at its natural length, the block can sit still. That spot is the equilibrium position. The net force there is zero. We call it x = 0 and measure every position from it. In this unit, positive is to the right.

Pull the block to the right. The stretched spring pulls it left, back toward equilibrium. Push it to the left. The squashed spring pushes it right, again toward equilibrium. A force that always points back to equilibrium is a restoring force.

For a spring, the size of the force grows in step with the distance (Hooke's law):

Fs = −k x

k is the spring constant in N/m (how stiff the spring is). The minus sign says the force points the opposite way to the displacement x.

Simple harmonic motion (SHM) happens when the restoring force is proportional to the displacement from equilibrium and points the other way: F = −kx. Then, by Newton's second law,

a = F/m = −(k/m) x

The acceleration is always toward equilibrium, and biggest at the ends.

The amplitude A is the largest distance from equilibrium. If you pull the block to 0.10 m and let go, A = 0.10 m and the block swings between x = +0.10 m and x = −0.10 m (no friction).

x = 0 x = −A x = +A F biggest, v = 0 F = 0, speed biggest F biggest, v = 0

With numbers

A 0.50 kg block on a spring with k = 50 N/m is pulled to x = +0.10 m and released. Positive is to the right.

Show all steps as text
  1. Force at release: F = −kx = −(50 N/m)(0.10 m) = −5.0 N (5.0 N to the left). The spring is stretched to the right, so it pulls left, toward equilibrium.
  2. Acceleration at release: a = F/m = (−5.0 N)/(0.50 kg) = −10 m/s². Newton's second law; the spring force is the only horizontal force.
  3. Halfway in, at x = +0.05 m: F = −(50)(0.05) = −2.5 N, a = −5.0 m/s². Half the displacement gives half the force. That is what "proportional" means.
  4. At x = 0: F = 0 and a = 0, but the block is moving fastest. It has been speeding up the whole way in. Now nothing speeds it up or slows it down, so it overshoots.
  5. At x = −0.10 m: F = +5.0 N, a = +10 m/s², and v = 0 for an instant. The amplitude is A = 0.10 m. With no friction, no energy is lost, so it reaches the same distance on the other side.

Play: a mass on a horizontal spring

The block starts at x = +A, at rest. Watch the purple force arrow and the green velocity arrow. Then look at the graphs: when is x biggest? When is v biggest? What does a do?

position xvelocity vacceleration aforce F

Try: double k and watch the force arrow and the graphs. Then double m. Then change only the amplitude. Which one changes how quickly it repeats? (You will meet the formula in 7.2.)

Worked examples

Is it SHM? basic

Which of these is simple harmonic motion? (1) a block on a spring on a smooth table, (2) a ball bouncing up and down on a hard floor, (3) a swing moving through small angles.

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Show all steps as text
  1. Block on a spring: F = −kx. Yes, SHM.The force is proportional to displacement and points to equilibrium.
  2. Bouncing ball: no. In the air the force is just gravity, mg, the same size everywhere.The force does not grow with distance from a middle point, so it repeats but it is not SHM.
  3. Swing at small angles: very nearly SHM.The restoring force is mg sin θ ≈ mg θ, which is proportional to the displacement when θ is small.

Force from a squashed spring basic

A spring with k = 200 N/m is pushed so the block sits at x = −0.030 m. What force does the spring exert?

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  1. F = −kx = −(200 N/m)(−0.030 m) = +6.0 N.Two negatives make a positive: the block is left of equilibrium, so the force points right.

Where is the acceleration 3 m/s²? medium

A 2.0 kg block oscillates on a spring with k = 80 N/m and amplitude 0.20 m. At what positions is the size of its acceleration 3.0 m/s²?

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Show all steps as text
  1. Size of acceleration: |a| = (k/m)|x|.From a = −(k/m)x.
  2. |x| = m|a|/k = (2.0 kg)(3.0 m/s²)/(80 N/m) = 0.075 m.Rearrange and put in the numbers.
  3. So x = +0.075 m or −0.075 m. Both are inside the amplitude (0.20 m), so the block does pass these points.At +0.075 m the acceleration is −3.0 m/s²; at −0.075 m it is +3.0 m/s².

A hanging spring medium

A 0.40 kg mass hangs from a vertical spring and stretches it 0.098 m before coming to rest. It is then pulled down another 0.050 m and released. Take up as positive. Find k and the acceleration just after release.

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  1. At rest, spring force balances gravity: k(0.098 m) = (0.40 kg)(9.8 m/s²) = 3.92 N, so k = 3.92/0.098 = 40 N/m.Net force is zero at equilibrium.
  2. Measure x from the new equilibrium, not from the unstretched spring. Now x = −0.050 m.Gravity is already balanced at equilibrium, so the extra spring force is the restoring force.
  3. Net force: F = −kx = −(40)(−0.050) = +2.0 N (up).Same F = −kx as the horizontal spring.
  4. a = F/m = 2.0/0.40 = +5.0 m/s² (up).A vertical spring does SHM around its hanging equilibrium point.

From data to a spring constant AP

A student hangs different loads and measures how far a spring stretches past equilibrium:

Stretch x (m)0.0200.0400.0600.080
Spring force F (N)0.621.181.832.41

(a) What graph should be plotted to find k? (b) Find k. (c) Will a block on this spring do SHM? Justify.

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  1. (a) Plot F on the vertical axis against x on the horizontal axis.F = kx is a straight line through the origin, so the slope is k.
  2. (b) Slope of the best fit line through the origin: ≈ (2.41 − 0.62)/(0.080 − 0.020) = 1.79/0.060 ≈ 30 N/m. (A least squares fit gives 30.2 N/m.)Use points far apart on the best fit line, not just one data pair.
  3. (c) Yes. The data lie close to a straight line through the origin, so F is proportional to x, which is the condition for SHM.AP wants the link: "force proportional to displacement and directed toward equilibrium".

Practice

  1. A block oscillates on a horizontal spring with no friction. Where is its speed greatest?

    Show answer

    (B). The block speeds up all the way to equilibrium because the force points toward x = 0. After it passes x = 0 the force points backward and slows it down. So the top speed is at x = 0. At ±A it is momentarily at rest.

  2. The block is at x = +A/2 and moving to the left. Which way does its acceleration point?

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    (A). a = −(k/m)x. At x = +A/2, x is positive so a is negative (left). The direction of motion does not matter: acceleration depends only on where the block is.

  3. A 0.30 kg block is on a spring with k = 120 N/m. At one moment it is at x = −0.050 m. What is its acceleration?

    Show answer

    (B). F = −kx = −(120)(−0.050) = +6.0 N. a = F/m = 6.0/0.30 = +20 m/s². (D) is the force, with the wrong sign, not the acceleration.

  4. Which force law, acting on an object around x = 0, produces simple harmonic motion?

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    (C). SHM needs a restoring force proportional to displacement. (A) is the same size everywhere, (B) is not proportional to x and points the same way on both sides, and (D) pushes the object away from x = 0, so it would not come back.

  5. The amplitude of a block on a spring is doubled. What happens to the largest force the spring exerts on the block?

    Show answer

    (B). The largest force is at x = ±A: Fmax = kA. Double A, double Fmax. (Four times is what happens to the energy, ½kA², in 7.4.)

  6. A block hangs at rest from a vertical spring (spring constant k). At this equilibrium position the spring is

    Show answer

    (B). At equilibrium the net force is zero, so the spring pulls up with kd = mg, giving d = mg/k. (2mg/k is the lowest point if you let the block drop from the natural length, not the equilibrium.)

  7. Short answer. A student says: "At the turning points the block stops for an instant, so the net force on it must be zero there." Explain what is wrong.

    Show answer

    Zero velocity does not mean zero force. At a turning point the block is at x = ±A, as far as possible from equilibrium, so the spring force kA is the largest it ever gets. That big force is exactly what turns the block around. Net force is zero at x = 0, where the block is moving fastest.

    Scoring idea: 1 point for "force is maximum at the turning points (F = kA)", 1 point for linking force to acceleration (changing velocity), not to velocity itself.

  8. Short answer. A 0.25 kg block on a spring with k = 25 N/m oscillates with amplitude 0.12 m. Find (a) the largest force on the block, (b) the largest acceleration, (c) the acceleration at x = −0.040 m.

    Show answer
    1. Fmax = kA = (25)(0.12) = 3.0 N.Largest displacement gives the largest force.
    2. amax = Fmax/m = 3.0/0.25 = 12 m/s².Newton's second law.
    3. a = −(k/m)x = −(25/0.25)(−0.040) = −(100)(−0.040) = +4.0 m/s².Left of equilibrium, so the acceleration points right.

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

Nadia hangs a block of mass m from a vertical spring with spring constant k. The block settles at rest with the spring stretched by a distance d. She then pulls the block down an extra distance A and lets go, and it oscillates up and down.

  1. Derive an expression for d in terms of m, k and physical constants.
  2. Derive an expression for the magnitude of the block's acceleration just after it is released, in terms of k, A and m.
  3. Nadia uses m = 0.40 kg, k = 80 N/m and A = 0.050 m. Calculate d, the acceleration just after release, and the force the spring exerts on the block at the lowest point.
Show worked answer and scoring

(a)

  1. At rest the net force is zero, so the spring pulls up as hard as gravity pulls down: kd = mg, so d = mg/k.Equilibrium means the forces balance.

(b)

  1. At the lowest point the spring is stretched by d + A, so it pulls up with k(d + A). Gravity pulls down with mg. Taking up as positive, Fnet = k(d + A) − mg = kd + kA − mg = kA, because kd = mg.
  2. a = Fnet/m, so a = kA/m, pointing up.Measured from the new equilibrium, gravity drops out and the spring acts like F = −kx.

(c)

  1. d = mg/k = (0.40)(9.8)/80 = 0.049 m.
  2. a = kA/m = (80)(0.050)/0.40 = 10 m/s², upward.
  3. At the lowest point the stretch is d + A = 0.049 + 0.050 = 0.099 m, so the spring force is k(d + A) = (80)(0.099) = 7.9 N, upward. Check: 7.92 − mg = 7.92 − 3.92 = 4.00 N = m·a. ✓

AP-style scoring

  • 1 point for setting the spring force equal to the weight at equilibrium (kd = mg).
  • 1 point for a net force of k(d + A) − mg (or for measuring from equilibrium so Fnet = kA), and 1 point for a = kA/m.
  • 1 point for numbers consistent with (a) and (b), with units: d ≈ 0.049 m, a = 10 m/s², spring force ≈ 7.9 N.

2. Translation Between Representations

Tomás pulls a cart on a smooth horizontal track to x = +A, where it is attached to a spring, and releases it. Positive x is to the right of equilibrium. He wants a graph of the cart's acceleration a against its position x for every point between −A and +A.

Four possible graphs of acceleration a (up) against position x (across):

xa
(A) straight line through the origin, sloping down
xa
(B) straight line through the origin, sloping up
xa
(C) flat line below the axis
xa
(D) upside-down U shape
  1. Which graph, A, B, C or D, shows a against x?
  2. Justify your choice using Newton's second law and the spring force. Do not just restate the shape.
  3. The cart has mass 0.50 kg and k = 20 N/m. What is the slope of the correct graph, with units, and what is a when x = +0.10 m?
Show worked answer and scoring

(a)

  1. Graph A: a straight line through the origin with a negative slope.

(b)

  1. The only horizontal force is the spring: F = −kx. Newton's second law: a = F/m = −(k/m)x.
  2. k/m is a constant, so a is proportional to x: a straight line through the origin. The minus sign makes a negative when x is positive, so the line slopes down.B has the wrong sign (it would push the cart away), C ignores that the force grows with x, D is not proportional to x.

(c)

  1. Slope = −k/m = −20/0.50 = -40 s⁻² (that is (m/s²) per m).
  2. At x = +0.10 m: a = (-40)(0.10) = -4.0 m/s², i.e. 4.0 m/s² to the left.

AP-style scoring

  • 1 point for choosing graph A.
  • 1 point for using F = −kx with Newton's second law to show a ∝ x (straight line through the origin), and 1 point for explaining the negative slope by the restoring direction.
  • 1 point for slope −40 s⁻² and a = −4.0 m/s².

3. Experimental Design and Analysis

Keiko has a spring, a hanger, a set of slotted masses, a metre stick and a clamp stand. She wants to find the spring constant k and check that the spring obeys F = −kx, the force law behind simple harmonic motion.

Her data (stretch measured from the spring's natural length):

Hanging mass m (kg)Stretch x (m)
0.0500.021
0.1000.039
0.1500.061
0.2000.079
0.2500.101
  1. Describe a procedure she could use, naming what she measures and with what, and one way to reduce error.
  2. What quantities should she graph so that the graph is a straight line? What does the slope mean?
  3. Use the data in the table to find k.
Show worked answer and scoring

(a)

  1. Clamp the spring vertically next to the metre stick. Read the position of the bottom of the spring with no mass (or just the hanger) as the zero.
  2. Hang a known mass, wait for it to be at rest, and read the new position. The stretch x is the difference. Repeat for five or more masses.
  3. To reduce error: read the scale at eye level, and repeat each reading while unloading as well as loading, then average.Eye level avoids parallax; repeats average out random error.

(b)

  1. At rest kx = mg, so x = (g/k)·m. Graph stretch x (vertical) against mass m (horizontal): a straight line through the origin.
  2. The slope is g/k, so k = g/slope. A straight line also confirms the force is proportional to the stretch.

(c)

  1. Best-fit line through the five points: slope ≈ 0.400 m/kg (intercept ≈ +0.2 mm, close to zero as expected).
  2. k = g/slope = 9.8/0.400 ≈ 24.5 N/m.

AP-style scoring

  • 1 point for a procedure that measures stretch for several known masses with the named equipment, and 1 point for a valid way to reduce error.
  • 1 point for graphing x against m (or mg against x) and stating what the slope equals (g/k, or k).
  • 1 point for using the slope of a best-fit line (not one data pair) to get k ≈ 24 to 25 N/m.

4. Qualitative/Quantitative Translation

Ravi and Elena each pull a block on an identical horizontal spring back by the same distance A and release it. Elena's block has twice the mass of Ravi's. Elena claims: "Right after release my block's acceleration is half of Ravi's, because the spring pulls both blocks equally hard but mine is harder to speed up."

  1. Without equations, explain whether Elena's reasoning is correct.
  2. Derive an expression for the acceleration right after release in terms of k, A and m, and use it to find the ratio of Elena's acceleration to Ravi's.
  3. Point to the part of your derivation that matches each idea in your answer to (a). Then check with k = 50 N/m, A = 0.040 m, m = 0.20 kg for Ravi.
Show worked answer and scoring

(a)

  1. Yes. The spring force depends only on how far the spring is stretched, and both are stretched by A, so the forces are equal. The same force on twice the mass gives half the acceleration.

(b)

  1. Spring force at release: |F| = kA. Newton's second law: a = kA/m.
  2. Ratio: aElena/aRavi = (kA/2m)/(kA/m) = ½.

(c)

  1. “Same force” is the numerator kA (same k, same A). “Harder to speed up” is m in the denominator: doubling it halves a.
  2. Ravi: a = (50)(0.040)/0.20 = 10 m/s². Elena: (50)(0.040)/0.40 = 5 m/s², half as much. ✓

AP-style scoring

  • 1 point for stating the claim is correct with both reasons: equal spring force (same stretch) and larger mass.
  • 1 point for a = kA/m and 1 point for the ratio ½.
  • 1 point for linking kA to “same force” and m to “harder to speed up”.

Common mistakes

The mistake: "v = 0 at the ends, so a = 0 there too."

Why it is wrong: acceleration is how fast velocity is changing, not the velocity. At the ends the velocity is flipping direction, and the force kA is biggest.

How to spot it: ask "where is the spring stretched most?" That is where F and a are biggest.

The mistake: "The force is biggest at the middle, because that is where it moves fastest."

Why it is wrong: at x = 0 the spring is at its natural (or equilibrium) length, so F = 0. The block is fast there because it has been pushed all the way in.

How to spot it: fast does not mean a big force. Look at x, not v, to find F.

The mistake: measuring x from the wall, or from the unstretched length of a hanging spring.

Why it is wrong: in F = −kx and a = −(k/m)x, x is the displacement from equilibrium. For a hanging spring, equilibrium is already stretched by mg/k.

How to spot it: at your x = 0, is the net force zero? If not, you picked the wrong zero.

The mistake: dropping the minus sign and giving the force the same direction as x.

Why it is wrong: a restoring force always points back toward equilibrium, opposite to the displacement.

How to spot it: draw a quick arrow: block right of centre → force arrow points left.