Unit 7.1 | Defining Simple Harmonic Motion

Problem Type 1: Is it simple harmonic motion?

How to spot it: The question gives a force law, a graph of force against position, or several motions and asks which one is simple harmonic.

Setup: Check two things. Does the force point back toward equilibrium (restoring)? Is it proportional to the displacement? On a force vs. position graph, SHM is a straight line through the origin with a negative slope, and the slope is [katex] -k [/katex].

An object moves under one of three forces: (a) [katex] F = -20x [/katex], (b) [katex] F = -5x^3 [/katex], (c) [katex] F = +8x [/katex], with [katex] F [/katex] in newtons and [katex] x [/katex] in meters. Which give simple harmonic motion?

Only (a). It points back toward equilibrium and is proportional to [katex] x [/katex], like a spring with [katex] k = 20 \text{ N/m} [/katex]. Force (b) is restoring, so the object still oscillates, but doubling [katex] x [/katex] makes the force 8 times bigger, not 2 times, so it is not SHM. Force (c) points away from equilibrium: the object never comes back.

Plenty of motions repeat without being SHM. A ball bouncing on the floor repeats, but while it is in the air the only force is its constant weight, which does not depend on how high it is.

AP trap: Calling every repeating motion simple harmonic. Periodic is not enough: the restoring force has to be proportional to the displacement.

Problem Type 2: Force and acceleration at a point in the cycle

How to spot it: The question names a point (“at the turning point”, “at x = [katex] 0.05\,\text{m} [/katex]”, “halfway to equilibrium”) and asks for the force, acceleration or speed there, or asks where one of them is largest or zero.

Setup: Use [katex] F = -kx [/katex] and [katex] a = -\frac{k}{m}x [/katex] with the displacement from equilibrium. Both are largest at [katex] x = \pm A [/katex] and zero at [katex] x = 0 [/katex]. Speed does the opposite.

A [katex] 0.50\,\text{kg} [/katex] block on a frictionless table is attached to a spring with [katex] k = 200 \text{ N/m} [/katex]. It is pulled [katex] 0.10\,\text{m} [/katex] to the right and released. Find the largest force and acceleration on the block, and the force and acceleration when it is [katex] 0.05\,\text{m} [/katex] to the left of equilibrium.

The largest force is at the turning points: [katex] F_{max} = kA = (200)(0.10) = 20 \text{ N} [/katex], so [katex] a_{max} = \frac{20}{0.50} = 40 \text{ m/s}^2 [/katex]. At [katex] x = -0.05 \text{ m} [/katex]: [katex] F = -(200)(-0.05) = +10 \text{ N} [/katex] and [katex] a = \frac{10}{0.50} = +20 \text{ m/s}^2 [/katex], pointing right, back toward equilibrium. Half the displacement, half the force.

AP trap: Saying the acceleration is zero at the turning points because the block stops there. The velocity is zero for an instant, but the spring is stretched the most, so the force and acceleration are at their largest. Zero velocity does not mean zero acceleration.

Problem Type 3: Vertical springs

How to spot it: A mass hangs from a spring, or sits on top of one, and bobs up and down.

Setup: Gravity only moves the equilibrium. Find the new equilibrium from [katex] kd = mg [/katex], then measure [katex] x [/katex] from there: the net force is still [katex] -kx [/katex], with the same [katex] k [/katex].

A [katex] 0.40\,\text{kg} [/katex] mass hangs from a spring with [katex] k = 98 \text{ N/m} [/katex]. How far does it stretch the spring at equilibrium? The mass is then pulled down another [katex] 3.0\,\text{cm} [/katex] and released. Find the amplitude, and the net force on the mass at its lowest point.

Equilibrium: [katex] d = \frac{mg}{k} = \frac{(0.40)(9.8)}{98} = 0.040 \text{ m} [/katex], or [katex] 4.0\,\text{cm} [/katex]. It is released from rest [katex] 3.0\,\text{cm} [/katex] below that point, so the amplitude is [katex] 3.0\,\text{cm} [/katex]: it moves between [katex] 1.0\,\text{cm} [/katex] and [katex] 7.0\,\text{cm} [/katex] below the natural length.

At the lowest point the spring is stretched [katex] 7.0\,\text{cm} [/katex], so it pulls up with [katex] (98)(0.070) = 6.86 \text{ N} [/katex], while gravity pulls down with [katex] (0.40)(9.8) = 3.92 \text{ N} [/katex]. Net force: [katex] 2.94 \text{ N} [/katex] upward. The shortcut gives the same answer: [katex] kA = (98)(0.030) = 2.94 \text{ N} [/katex].

▶ Watch: a [katex] 1.0\,\text{kg} [/katex] block on a [katex] 50\,\text{N/m} [/katex] spring that settles [katex] 0.20\,\text{m} [/katex] lower, right at its equilibrium, and the energy change of the block-spring-Earth system (starts at 14:48).

AP trap: Measuring the amplitude from the spring’s natural length. A hanging mass oscillates around its new equilibrium, so the amplitude is how far it was pulled from there.

Problem Type 4: Pendulum forces

How to spot it: A pendulum bob is pulled to an angle. The question asks for the restoring force, the force at the bottom, or when a pendulum counts as SHM.

Setup: The restoring force is the part of the weight along the arc, [katex] mg\sin\theta [/katex]. For small angles [katex] \sin\theta \approx \theta [/katex] (in radians), so the restoring force is proportional to the displacement and the motion is close to SHM.

The tension and the part of the weight along the string, [katex] mg\cos\theta [/katex], point along the string, so they cannot push the bob along its arc. Only [katex] mg\sin\theta [/katex] does. With [katex] \theta = \frac{s}{L} [/katex] for an arc length [katex] s [/katex]:

[katex] F \approx -mg\,\theta = -\frac{mg}{L}\,s [/katex]

That is Hooke’s law with [katex] \frac{mg}{L} [/katex] in place of [katex] k [/katex]. A pendulum at small angles is a simple harmonic oscillator.

A [katex] 2.0\,\text{kg} [/katex] bob hangs on a [katex] 1.0\,\text{m} [/katex] string. Find the restoring force when it is pulled to [katex] 10^\circ [/katex], and compare it with the small-angle estimate. Then do the same at [katex] 60^\circ [/katex].

At [katex] 10^\circ [/katex]: [katex] mg\sin\theta = (2.0)(9.8)\sin 10^\circ = 3.40 \text{ N} [/katex]. The estimate uses the angle in radians, [katex] 10^\circ = 0.1745 \text{ rad} [/katex]: [katex] (2.0)(9.8)(0.1745) = 3.42 \text{ N} [/katex], within 1%. At [katex] 60^\circ [/katex]: [katex] (19.6)\sin 60^\circ = 17.0 \text{ N} [/katex], but [katex] (19.6)(1.047) = 20.5 \text{ N} [/katex], about 21% too big. At [katex] 60^\circ [/katex] the pendulum still swings, but its motion is not simple harmonic.

AP trap: Saying the net force on the bob is zero at the bottom of the swing. The restoring force is zero there, but the bob is moving in a circle, so the tension is bigger than [katex] mg [/katex] and the net force points up, toward the pivot. That is the centripetal force.

Lesson 7.2 Preview

Now you know what SHM is. In the next lesson we time it: the period of a mass on a spring and of a pendulum, and the surprising list of things that do not change it. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-7-1-defining-simple-harmonic-motion/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-7-1-defining-simple-harmonic-motion/