Connect simple harmonic motion, Hooke’s law, and circular motion

Ontario Grade 12 Physics

Energy and Momentum

Ontario Grade 12 Physics — expectation C3.2

A mass attached to a spring can move back and forth in a repeating pattern. This motion is called simple harmonic motion when the restoring force is proportional to displacement and points toward the equilibrium position. Hooke’s law describes that force for an ideal spring. A useful model connects the back-and-forth motion to the shadow, or projection, of an object moving uniformly around a circle. This lesson develops those connections with Grade 12 algebra and vector components.

What you will learn

Describe simple harmonic motion using displacement, restoring force, and acceleration.
Use Hooke’s law to connect a spring’s force to an object’s displacement.
Explain how simple harmonic motion can be modelled as the projection of uniform circular motion.
Use the spring model to relate mass, spring stiffness, and oscillation period.

1. Prerequisite bridge and motion model

In earlier physics, you used displacement, velocity, acceleration, and force. Displacement is a vector: it has both magnitude and direction. Mass, period, and spring stiffness are scalars: they have magnitude but no direction. The SI unit for displacement is the metre, for mass the kilogram, and for time the second.

For a spring-and-mass system, choose the mass as the system. View it from a stationary reference frame, such as the laboratory floor. Let the spring lie along the horizontal

x

-axis. Put the equilibrium position at

x=0

, and choose right as the positive direction. Displacement

x

is measured from equilibrium, not from the spring’s unstretched end.

Equilibrium is the position where the forces on the mass balance. If the mass is pulled to the right, the spring pulls it left; if it is pulled left, the spring pulls it right. The force that points back toward equilibrium is called a restoring force.

Simple harmonic motion (SHM) is repeating motion in which the restoring force, and therefore acceleration, is proportional to displacement and opposite in direction. The mass moves fastest as it passes equilibrium and momentarily stops at each turning point. Amplitude

A

is the greatest distance from equilibrium. Period

T

is the time for one complete cycle, measured in seconds.

F_x\propto -x

Use displacement from equilibrium and state the positive direction.
SHM has a restoring force that grows with displacement and points toward equilibrium.
Amplitude is a distance; displacement and force also require a direction or sign.

2. Hooke’s law and the spring’s restoring force

Hooke’s law models an ideal spring as long as it is not stretched or compressed beyond the range where the relationship applies. The spring force is proportional to the displacement from equilibrium. The spring constant

k

measures stiffness: a larger

k

means more force is needed for the same displacement. Its SI unit is newtons per metre.

With right defined as positive, a positive displacement produces a negative spring force. A negative displacement produces a positive force. The minus sign in Hooke’s law records this direction; it does not mean that the force has a negative size.

If the spring is the only unbalanced force along the motion, Newton’s second law connects net force to acceleration. Comparing that law with Hooke’s law shows that acceleration also points opposite displacement and grows in proportion to it. This is the defining force-and-acceleration pattern of SHM.

A simple labelled motion sketch can help keep the signs clear: the equilibrium point is

x=0

; the right turning point is

x=+A

with spring force left; the left turning point is

x=-A

with spring force right. At equilibrium, the spring force is zero in this ideal model.

F_x=-kx,\qquad F_{\mathrm{net},x}=ma_x

The spring constant has units of $\mathrm{N/m}$ .
The force sign gives direction relative to the chosen axis.
At a turning point, displacement and restoring force have their greatest magnitudes.

3. Circular motion as a model for SHM

Uniform circular motion means motion around a circle at constant speed. Imagine a point moving around a circle of radius

A

at a steady angular rate. Angular position

\theta

describes the point’s location around the circle, in radians. The point’s horizontal coordinate is the projection of its position onto a diameter.

As the point travels around the circle, that horizontal projection moves back and forth between

-A

and

+A

. It is a model of SHM. If the circular motion starts at the rightmost point, the projected displacement is

x=A\cos\theta

. The amplitude of the projection is the circle’s radius.

The circular-motion model also explains the direction of acceleration. The acceleration of a point in uniform circular motion points toward the centre. Its magnitude is

v^2/A

, where

v

is the circular speed. For a constant angular rate

\omega

, this magnitude is

\omega^2A

. The horizontal component of this inward acceleration is opposite the horizontal displacement, so

a_x=-\omega^2x

. Here

\omega

is angular speed, measured in radians per second.

For a mass on an ideal spring, Newton’s second law and Hooke’s law give

a_x=-(k/m)x

. Comparing this with the projection model gives

\omega^2=k/m

. The period is the time for one full turn in the circular model, so

T=2\pi/\omega

. Combining these Grade 12 relationships gives the period of an ideal mass–spring oscillator. A greater mass gives a longer period; a stiffer spring gives a shorter period.

x=A\cos\theta,\qquad \omega^2=\frac{k}{m},\qquad T=2\pi\sqrt{\frac{m}{k}}

The SHM displacement matches one coordinate of uniform circular motion.
The projection’s acceleration points opposite its displacement.
For an ideal mass–spring system, stiffness and mass determine the period.

4. Reading the relationships and checking a model

These connections are useful because the spring law, Newton’s second law, and circular-motion model describe the same pattern in different ways. Hooke’s law describes the force. Newton’s second law connects force and acceleration. The circular projection describes the repeating displacement pattern.

The model has limits. It assumes an ideal spring and a mass moving along one line, with the spring force providing the net force along that line. The lesson’s equations do not claim that every repeating motion is SHM. A motion matches this model when its restoring force is proportional to displacement and directed toward equilibrium.

When solving a problem, first identify the system, reference frame, positive direction, and known quantities. Decide whether the question asks for force, acceleration, angular speed, or period. Preserve units in substitutions. A final answer should have the expected SI unit and a direction when it describes a vector. Check whether its size agrees with the situation: a larger spring constant should not produce a longer period if mass is unchanged.

a_x=-\frac{k}{m}x=-\omega^2x

State the model assumptions before applying the spring-period relationship.
Keep vector direction separate from scalar magnitude.
Use units and the trend with mass or stiffness as reasonableness checks.

Worked example

Spring force and acceleration at a turning point

A

0.40\,\mathrm{kg}

mass is attached to an ideal horizontal spring with

k=50\,\mathrm{N/m}

. The mass is at

x=+0.060\,\mathrm{m}

from equilibrium. Right is positive. Find the spring force and the acceleration.

Set the system and direction
The system is the mass, viewed from the stationary laboratory frame. The positive direction is right. The known quantities are $m=0.40\,\mathrm{kg}$ , $k=50\,\mathrm{N/m}$ , and $x=+0.060\,\mathrm{m}$ . The unknowns are the spring force and acceleration.
Apply Hooke’s law
Hooke’s law gives the force component along the spring. The negative result means that the force points left, toward equilibrium.
$F_x=-(50\,\mathrm{N/m})(+0.060\,\mathrm{m})=-3.0\,\mathrm{N}$
Find acceleration
The spring force is the net horizontal force in this model. Use Newton’s second law and keep the negative sign to show the direction.
$a_x=\frac{-3.0\,\mathrm{N}}{0.40\,\mathrm{kg}}=-7.5\,\mathrm{m/s^2}$

Answer: The spring force is

3.0\,\mathrm{N}

left, and the acceleration is

7.5\,\mathrm{m/s^2}

left.

Check: The force unit follows from

\mathrm{(N/m)(m)=N}

, and force divided by mass gives

\mathrm{m/s^2}

. Both point opposite the positive displacement, as required for a restoring force.

Worked example

Period of a mass–spring oscillator

An ideal spring has stiffness

k=80\,\mathrm{N/m}

and holds a

0.50\,\mathrm{kg}

mass. Find the oscillation period.

Define the system and identify the quantities
The system is the mass and spring in the stationary laboratory frame. Choose the spring’s line as the motion axis; the period does not depend on which direction is called positive. The known quantities are $m=0.50\,\mathrm{kg}$ and $k=80\,\mathrm{N/m}$ . The unknown is $T$ .
Choose the spring-period model
For an ideal mass–spring oscillator, the period depends on mass and spring stiffness. Substitute values in SI units.
$T=2\pi\sqrt{\frac{0.50\,\mathrm{kg}}{80\,\mathrm{N/m}}}$
Evaluate and round
Since $\mathrm{N/m}$ is equivalent to $\mathrm{kg/s^2}$ , the quantity under the square root has units of seconds squared. Round to two significant figures, matching the given values.
$T=0.50\,\mathrm{s}$

Answer: The period is

0.50\,\mathrm{s}

.

Check: The units reduce to seconds. The period is reasonable for the given combination: greater stiffness or smaller mass would make it shorter.

Worked example

Position from the circular-motion projection

A point moves uniformly around a circle of radius

0.12\,\mathrm{m}

. Its angular position is

60^\circ

from the positive horizontal axis. Find the horizontal projection’s displacement from the centre.

Set the geometry and direction
Use the circle’s centre as the origin and the horizontal axis as the displacement axis. Right is positive. The projection is the SHM model’s displacement; the known radius is $A=0.12\,\mathrm{m}$ and the angle is $60^\circ$ .
Use the horizontal component
The horizontal component of the radius is the radius multiplied by the cosine of its angle from the positive horizontal axis.
$x=A\cos\theta=(0.12\,\mathrm{m})\cos 60^\circ$
Evaluate and state direction
The cosine is positive at this angle, so the projection lies to the right of the centre.
$x=+0.060\,\mathrm{m}$

Answer: The projected displacement is

0.060\,\mathrm{m}

to the right.

Check: The result is a length, and it is smaller than the circle’s radius. A positive component agrees with an angle measured from the positive horizontal axis.

Common mistakes and how to avoid them

Treating the minus sign in Hooke’s law as a negative force magnitude.

Correction: The sign gives the force’s direction on the chosen axis. The force magnitude is non-negative.

Measuring spring displacement from the unstretched length instead of equilibrium.

Correction: For the SHM model,

x

is the displacement from the equilibrium position.

Confusing the circle’s radius with the SHM displacement at every instant.

Correction: The radius is the amplitude. The instantaneous displacement is the radius’s component along the chosen axis.

Assuming a larger spring constant makes the period longer.

Correction: The period is proportional to the square root of mass and inversely proportional to the square root of stiffness.

Lesson summary

Hooke’s law gives a spring force proportional to displacement and opposite in direction.
When the spring force is the net force, the mass has the restoring acceleration pattern of SHM.
SHM can be represented by projecting uniform circular motion onto a diameter.
For an ideal mass–spring oscillator, $T=2\pi\sqrt{m/k}$ .

Check your understanding

Question 1

A mass is displaced to the left of equilibrium. Which way does the ideal spring’s restoring force point?

Left, away from equilibrium
Right, toward equilibrium
There is no force until the mass moves
Downward, regardless of the spring’s orientation

Show answer and explanation

Right, toward equilibrium

The restoring force points opposite the displacement. A leftward displacement therefore gives a rightward force.

Question 2

If the mass is unchanged and the spring constant increases, what happens to the ideal oscillator’s period?

It increases
It decreases
It stays the same
It becomes zero

Show answer and explanation

It decreases

The period varies inversely with the square root of spring stiffness, so increasing stiffness makes the period shorter.

Question 3

In the circular-projection model, what does the circle’s radius represent?

The period
The spring constant
The amplitude of the projected motion
The mass

Show answer and explanation

The amplitude of the projected motion

The projection reaches the circle’s leftmost and rightmost points, so its greatest displacement from the centre equals the radius.

Key terms

Equilibrium position: The position where the forces on the object balance.
Restoring force: A force that points toward an equilibrium position.
Simple harmonic motion: Repeating motion with a restoring force proportional to displacement and opposite in direction.
Amplitude: The greatest displacement from equilibrium.
Period: The time required for one complete cycle.
Spring constant: A measure of spring stiffness, expressed in newtons per metre.
Projection: The component of a position along a chosen axis; in this model it gives the SHM displacement.
Angular speed: The rate at which angular position changes, measured in radians per second.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Physics (SPH4U), expectation C3.2. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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