AP Physics 1 · Lesson 1

What makes motion simple harmonic

Know Know it

An object oscillates when it moves back and forth about an equilibrium position. Move it away and a restoring force points back toward equilibrium; at equilibrium the restoring force is zero (for a mass on a level spring, the whole net force is zero there).

The motion is simple harmonic motion (SHM) when the restoring force is proportional to the displacement x from equilibrium and opposite to it: Fnet=−kxF_{net} = -kx. Newton’s second law then gives a=−kmxa = -\frac{k}{m}x, so the acceleration is largest at the turning points and zero at equilibrium.

An ideal spring obeys Hooke’s law, Fs=−kΔxF_s = -k\Delta x, so a mass on a spring is the standard example. A pendulum is nearly simple harmonic for small swings: the restoring force along the arc is mgsin⁡θmg\sin\theta, which is close to mgθ=mgLsmg\theta = \frac{mg}{L}s (s is the distance along the arc) when θ, in radians, is small.

Where the oscillator is and what it is doing
PositionSpeedRestoring force and acceleration along the path
Turning point, x = +Azero for an instantlargest, pointing in the −x direction
Equilibrium, x = 0largestzero
Turning point, x = −Azero for an instantlargest, pointing in the +x direction

Worked example Easy

A seat on an amusement ride hangs from a long, stiff spring and bounces straight up and down. At the instant the seat reaches the lowest point of its bounce, which statement is true?

  1. Its velocity is zero and the net force on it is zero.
  2. It is moving at its greatest speed and the net force on it points up.
  3. Its velocity is zero and the net force on it points up.Answer
  4. Its velocity is zero and the net force on it points down.

How to solve it

  1. The lowest point is a turning point, so the velocity is zero for an instant.
  2. The seat is below equilibrium, and the restoring force always points back toward equilibrium, so the net force points up.

Why each choice is right or wrong

  • A. Incorrect. Zero net force happens only at the equilibrium position, in the middle of the bounce, not at the lowest point where the spring is stretched the most.
  • B. Incorrect. The seat moves fastest as it passes equilibrium; at the lowest point it is momentarily at rest before heading back up.
  • C. Correct. The lowest point is a turning point, so the seat stops for an instant, and it is below equilibrium, so the restoring force points up, back toward equilibrium.
  • D. Incorrect. Below equilibrium the spring is stretched more than at rest, so its upward pull is larger than the weight and the net force points up, not down.

Trap Stopped does not mean no force

At a turning point the object is at rest for an instant, so it is tempting to say no force acts. In fact the restoring force, and so the acceleration along the path, is at its largest there. The restoring force is zero only at equilibrium, where the speed is greatest.

Explain Explain it

Task verbs: explain, justify

To decide whether a motion is simple harmonic, look at the force, not just at the repetition. Ask two things: does the net force always point toward one position, and does it grow in proportion to the distance from that position? Doubling the displacement should double the restoring force.

A constant force added to a spring force, such as gravity on a hanging mass, only moves the equilibrium position; measured from the new equilibrium, the net force is still −kx-kx, so the motion is still simple harmonic. If the restoring force does not grow in step with the displacement, the motion can still repeat, but it is not simple harmonic.

Worked example Medium

A 1.2 kg bob hangs on a light string 2.0 m long. It is pulled aside until it is 0.10 m from its lowest point, measured along its arc, and held still. Use g = 9.8 m/s². About how large is the restoring force along the arc?

  1. 0.59 NAnswer
  2. 1.2 N
  3. 12 N
  4. 0.060 N

How to solve it

  1. The angle is small: θ≈s/L=0.10/2.0=0.050\theta \approx s/L = 0.10/2.0 = 0.050 rad.
  2. The restoring force is about mgθ=(1.2)(9.8)(0.050)=0.59mg\theta = (1.2)(9.8)(0.050) = 0.59 N, pointing back toward the lowest point.

Why each choice is right or wrong

  • A. Correct. For a small angle the restoring force is about mgLs=(1.2)(9.8)2.0(0.10)=0.59\frac{mg}{L}s = \frac{(1.2)(9.8)}{2.0}(0.10) = 0.59 N.
  • B. Incorrect. 1.2 N multiplies the weight by the 0.10 m arc distance without dividing by the 2.0 m length, so it is not a force along the arc.
  • C. Incorrect. 12 N is the whole weight of the bob; only the small part of the weight along the arc, about mg times s/L, pulls it back.
  • D. Incorrect. 0.060 N leaves out g: it uses the mass, 1.2 kg, in place of the weight, 11.8 N.

Explain Your turn

A small cart rolls along a level, frictionless track between two rubber bumpers 1.5 m apart. It moves at a steady speed from one bumper to the other, bounces off each without losing speed, and repeats, so the motion has a fixed period. Say what kind of net force simple harmonic motion needs, compare it with the net force on this cart, and explain whether the cart’s motion is simple harmonic.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says simple harmonic motion needs a net force that points toward equilibrium and is proportional to the displacement
  • Says no net force acts on the cart while it rolls between the bumpers
  • Says the cart’s motion is not simple harmonic

Model answer

Simple harmonic motion needs a restoring force that points toward equilibrium and grows in proportion to the displacement, F = −kx. Between the bumpers no net force acts on this cart, so it coasts at a constant speed; it is pushed back only during the moment it touches a bumper. Its net force does not grow with its distance from the middle, so the motion repeats but is not simple harmonic.

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