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What condition on the restoring force makes a system oscillate in simple harmonic motion?

Topic 7.1 Defining Simple Harmonic Motion: identify simple harmonic motion by the linear restoring force F = -kx and describe the resulting oscillation.

A focused answer to AP Physics 1 Topic 7.1, covering simple harmonic motion as oscillation driven by a restoring force proportional to displacement, the condition F = -kx, the role of the equilibrium position, and how the mass-spring and pendulum systems meet this condition, with full worked examples.

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  1. What this topic is asking
  2. The defining condition
  3. Why the motion oscillates
  4. What systems show SHM
  5. Try this

What this topic is asking

The College Board (Topic 7.1) wants you to identify simple harmonic motion (SHM) by its defining condition: a restoring force proportional to the displacement from equilibrium and directed back toward it, F=βˆ’kxF = -kx. This single condition, built on the spring force of Topic 2.5 and Newton's second law, is what produces the smooth back-and-forth oscillation studied in the rest of the unit.

The defining condition

This is the whole topic in one line. The minus sign in F=βˆ’kxF = -kx is essential: it makes the force a restoring force, always pointing back toward the equilibrium position. A spring stretched to the right pulls left; compressed to the left, it pushes right. This linear relationship between force and displacement is what distinguishes SHM from other oscillations and is exactly the Hooke's-law spring force you met in Topic 2.5.

Why the motion oscillates

Tracing one cycle makes the mechanism clear: released from a turning point, the object accelerates toward equilibrium, gaining speed; it overshoots through equilibrium at top speed (no force there to stop it); the restoring force then decelerates it on the far side until it stops at the opposite turning point, and the cycle repeats. This is why the system oscillates rather than settling at equilibrium.

What systems show SHM

The two model systems of the unit both satisfy the F=βˆ’kxF = -kx condition. A mass on a spring does so exactly, with kk the spring constant. A simple pendulum does so approximately, for small angles, where the restoring component of gravity is very nearly proportional to the displacement; for large swings the relationship is no longer linear and the motion departs from true SHM. Recognizing the defining condition is the strategic key to the whole unit: once you know a system has a linear restoring force, you immediately know it oscillates in SHM, that its acceleration tracks its displacement, and that the energy, period and graphical relations of the following topics apply. The deeper idea is that SHM is the universal small-oscillation behavior of any stable equilibrium, which is why the same mathematics describes springs, pendulums, vibrating molecules and many other systems.

Try this

Q1. A spring of constant 3030 N/m is stretched 0.200.20 m. Calculate the restoring force. [2 points]

  • Cue. F=βˆ’kx=βˆ’(30)(0.20)=βˆ’6.0F = -kx = -(30)(0.20) = -6.0 N; magnitude 6.06.0 N toward equilibrium.

Q2. State where in the motion of an SHM oscillator the acceleration is greatest. [1 point]

  • Cue. At the extremes (turning points), where the displacement is largest.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)5 marksSection II (short FRQ). A block on a frictionless surface is attached to a spring of constant k=40k = 40 N/m and displaced 0.100.10 m from equilibrium. (a) State the condition a force must satisfy for the resulting motion to be simple harmonic. (b) Calculate the magnitude and direction of the restoring force at the moment of release. (c) Explain why the acceleration is largest at the extremes of the motion and zero at equilibrium.
Show worked answer β†’

A 5-point FRQ on the defining condition of simple harmonic motion.

(a) Condition (1 point): the net force must be a restoring force proportional to the displacement and directed back toward equilibrium, F=βˆ’kxF = -kx.
(b) Restoring force (2 points): F=βˆ’kx=βˆ’(40)(0.10)=βˆ’4.0F = -kx = -(40)(0.10) = -4.0 N. The magnitude is 4.04.0 N, directed back toward equilibrium (opposite the displacement).
(c) Explain (2 points): since F=βˆ’kxF = -kx and a=F/ma = F/m, the acceleration is proportional to the displacement. At the extremes xx is largest, so the restoring force and acceleration are largest; at equilibrium x=0x = 0, so the force and acceleration are zero (though the speed is greatest there).

Markers reward stating the F=βˆ’kxF = -kx condition, computing the restoring force with its direction, and linking the acceleration to displacement.

AP 2023 (style)1 marksSection I (multiple choice). Which condition must hold for an object's motion to be simple harmonic? (A) constant force toward equilibrium (B) force proportional to displacement and directed toward equilibrium (C) force proportional to velocity (D) zero net force. Justify your reasoning.
Show worked answer β†’

A 1-point MCQ on the defining condition. The answer is (B).

Simple harmonic motion requires a restoring force proportional to displacement and always directed back toward equilibrium, F=βˆ’kxF = -kx. A constant force gives constant acceleration, not oscillation; a force proportional to velocity is damping. The trap is (A): the restoring force is not constant, it grows with displacement.

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