1. Characteristics of SHM

Learning Outcomes

• I can define simple harmonic motion (SHM).
• I can identify systems that undergo SHM.
• I can explain the role of restoring forces in SHM.
• I can distinguish SHM from other forms of oscillatory motion.
• I can describe the motion of oscillating systems using scientific terminology.

Key Topics:

• Periodic motion vs. SHM.
• Restoring force proportional to displacement: $F=-kx$.

What Is Oscillatory Motion?

Oscillatory motion is repeated movement back and forth around a central position.

Many systems in physics can oscillate.

Examples include:

• A mass attached to a spring
• A swinging pendulum
• A vibrating guitar string
• A diving board moving up and down
• A tuning fork after it is struck

The central position around which the object oscillates is called the equilibrium position.

When an oscillating object reaches one side of its motion, it reverses direction and moves back toward equilibrium.

 

 

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What Is Simple Harmonic Motion?

Simple harmonic motion, or SHM, is a special type of oscillatory motion.

In SHM, the restoring force acting on an object is:

• Directed toward the equilibrium position.
• Directly proportional to the object's displacement from equilibrium.

Mathematically:

F ∝ −x

where:

• F = restoring force
• x = displacement from equilibrium
• The negative sign indicates that the force acts in the opposite direction to the displacement

This relationship is the defining characteristic of simple harmonic motion.

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The Equilibrium Position

The equilibrium position is the position where the system would naturally remain if it were not disturbed.

At equilibrium:

x = 0

For a horizontal mass-spring system, this is the position where the mass rests when no net force is acting on it.

If the mass is pulled away from equilibrium and released, it begins to oscillate.

The object repeatedly moves:

left → centre → right → centre

and the pattern repeats.

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Displacement in SHM

Displacement describes the object's position relative to equilibrium.

We normally represent displacement using:

x

At equilibrium:

x = 0

On one side:

x > 0

On the opposite side:

x < 0

The sign tells us which side of equilibrium the object is on.

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What Is a Restoring Force?

A restoring force is a force that acts to return an object toward its equilibrium position.

Imagine pulling a mass attached to a spring to the right.

The spring pulls the mass back toward the centre.

If the mass moves to the left of equilibrium, the spring pushes or pulls it toward the centre again.

Therefore, the restoring force always acts opposite to the displacement.

Displacement right→Force left​ Displacement left→Force right​

This constant tendency to return toward equilibrium produces the oscillation.

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Restoring Force and Displacement

For SHM, the magnitude of the restoring force increases as the object moves farther from equilibrium.

If the displacement doubles, the restoring force doubles.

If the displacement triples, the restoring force triples.

Therefore:

F ∝ x

in magnitude.

Including the direction gives:

F ∝ −x

This means the restoring force is both:

proportional to displacement and opposite in direction.

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Hooke's Law and SHM

A mass attached to an ideal spring provides an important example of SHM.

For a spring:

F = −kx

where:

• F = restoring force
• k = spring constant
• x = displacement
• The negative sign indicates the restoring direction

This is Hooke's law.

If the spring obeys Hooke's law, the restoring force is proportional to displacement.

Therefore, an ideal mass-spring system can undergo simple harmonic motion.

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Example: A Mass on a Spring

Imagine a mass attached to a horizontal spring.

Position A: Maximum Displacement

The mass is pulled to the right.

Its displacement is maximum.

The restoring force is also maximum and acts toward the left.

The mass is momentarily stationary before reversing direction.

Position B: Equilibrium

The mass passes through the centre.

Its displacement is:

x = 0

The restoring force is zero.

However, the mass is moving at its maximum speed.

Position C: Opposite Maximum Displacement

The mass reaches the other side.

Its displacement is again maximum in magnitude.

Its speed is zero for an instant.

The restoring force is maximum and points back toward equilibrium.

The cycle then repeats.

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Amplitude

The amplitude is the maximum displacement of an oscillating object from its equilibrium position.

It is usually represented by:

A

If an object moves between:

x = −5 cm

and:

x = +5 cm

then its amplitude is:

A = 5 cm​

The total distance between the two extreme positions is 10 cm, but the amplitude is only 5 cm.

Amplitude is always measured from equilibrium to an extreme position.

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One Complete Oscillation

One complete oscillation occurs when an object returns to the same position while moving in the same direction.

For example:

+A → 0 → −A → 0 → +A

is one complete oscillation.

The object has moved from one extreme to the other and back again.

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Period

The period is the time required for one complete oscillation.

It is represented by:

T

and measured in:

seconds (s)

For example, if a pendulum completes one full oscillation in 2.0 seconds:

T = 2.0 s​

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Frequency

The frequency is the number of complete oscillations occurring each second.

It is represented by:

f

and measured in:

hertz (Hz)

One hertz means:

1 oscillation per second

For example:

f = 5 Hz

means that the system completes:

5 oscillations every second

Period and frequency are closely related:

f = \( \frac{1}{T} \)​

and:

T =​ \( \frac{1}{f} \)

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SHM as a Graph

If displacement is plotted against time, ideal SHM produces a repeating wave-shaped pattern.

 

 

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The graph allows us to identify:

• Maximum positive displacement
• Maximum negative displacement
• Equilibrium position
• Amplitude
• Period
• Repeated oscillations

The repeating shape shows that the motion is periodic.

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Speed During SHM

The speed of an object undergoing SHM constantly changes.

At the extreme positions:

x = ±A

the speed is:

v = 0​

At equilibrium:

x = 0

the speed is:

maximum​

This happens because the object accelerates toward equilibrium and slows down as it moves away from equilibrium.

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Acceleration During SHM

Since:

F = ma

and the restoring force in SHM satisfies:

F ∝ −x

the acceleration must also satisfy:

a ∝ −x

In fact:

a = −ω²x

where ω is the angular frequency.

This is another mathematical definition of SHM.

The acceleration always points toward equilibrium.

At equilibrium:

a = 0

At maximum displacement:

∣a∣ = maximum

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Position, Speed, and Acceleration

The behaviour of an object during SHM can be summarized as follows:

Position	Displacement	Speed	Restoring Force	Acceleration
Positive extreme	+A	Zero	Maximum	Maximum
Equilibrium	0	Maximum	Zero	Zero
Negative extreme	−A	Zero	Maximum	Maximum

At both extreme positions, the force and acceleration point toward equilibrium.

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Pendulums and SHM

A swinging pendulum can also approximately undergo SHM.

 

 

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When the pendulum is displaced from its equilibrium position, gravity produces a restoring effect that pulls it back toward the centre.

However, a pendulum behaves like ideal SHM only when the oscillations are relatively small.

For larger angles, the relationship between restoring force and displacement is no longer sufficiently proportional.

Therefore:

A simple pendulum approximately undergoes SHM for small oscillations.​

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Examples of Systems That Can Undergo SHM

Common examples include:

Mass-Spring System

A mass attached to an ideal spring oscillates around equilibrium.

Simple Pendulum

For small angles, a pendulum approximately undergoes SHM.

Vibrating Tuning Fork

The prongs move rapidly back and forth around equilibrium.

Vibrating String

Small sections of a stretched string can oscillate approximately harmonically.

Suspended Mass

A mass hanging vertically from a spring can oscillate vertically around its equilibrium position.

Many more complicated vibrating systems can also be approximated using SHM.

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Is Every Oscillation SHM?

No.

All SHM is oscillatory motion, but not all oscillatory motion is simple harmonic motion.

For motion to be SHM, the restoring force must satisfy:

F ∝ −x

An object may move repeatedly back and forth without satisfying this condition.

For example:

• A pendulum swinging through very large angles
• A bouncing ball
• A strongly driven mechanical system
• Oscillations involving complicated or non-linear forces

These may be oscillatory, but they are not necessarily simple harmonic.

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SHM vs. General Oscillatory Motion

Simple Harmonic Motion	General Oscillatory Motion
Repeats around equilibrium	Repeats around a position
Has a restoring force	May have a restoring force
Restoring force is proportional to displacement	Force may have a more complicated relationship
Force points toward equilibrium	Motion may not satisfy this condition
Produces a sinusoidal displacement-time pattern	Pattern may have a different shape

The defining feature of SHM is therefore not simply that the object moves back and forth.

It is the relationship between restoring force and displacement.

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Describing Oscillations Scientifically

Instead of saying:

The object moves back and forth.

we can describe the motion more precisely:

The object oscillates periodically about its equilibrium position with an amplitude of 4 cm and a period of 2 s.

This statement uses scientific terminology to describe:

• The type of motion
• The equilibrium position
• The amplitude
• The period

Scientific vocabulary allows oscillations to be described clearly and quantitatively.

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A Complete Example

A mass attached to a spring oscillates between:

x = −0.08 m

and:

x = +0.08 m

It completes 10 oscillations in 5.0 seconds.

Amplitude

A = 0.08 m​

Period

T = \( \frac{5.0}{10} \)​

T = 0.50 s​

Frequency

f = \( \frac{1}{T} \)​

 f = \( \frac{1}{0.5} \)​

f = 2.0 Hz​

The system completes two full oscillations every second.

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Did You Know?

Simple harmonic motion appears throughout physics because many systems behave approximately like SHM when they are displaced only a small distance from a stable equilibrium position.

This makes SHM useful for studying:

• Mechanical vibrations
• Sound
• Waves
• Molecular vibrations
• Electrical oscillations
• Seismic motion

Even complicated oscillating systems can often be understood by first studying the much simpler model of simple harmonic motion.

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Key Vocabulary

Oscillation – Repeated motion back and forth around an equilibrium position.

Simple harmonic motion (SHM) – Oscillatory motion in which the restoring force is proportional to displacement and directed toward equilibrium.

Equilibrium position – The central position where the net restoring force is zero.

Displacement – The distance and direction of an object from equilibrium.

Restoring force – A force that acts toward the equilibrium position.

Amplitude – The maximum displacement from equilibrium.

Period – The time required for one complete oscillation.

Frequency – The number of complete oscillations per second.

Hertz (Hz) – The unit of frequency, equal to one oscillation per second.

Periodic motion – Motion that repeats after equal intervals of time.

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Key Takeaways

• Oscillatory motion is repeated movement around an equilibrium position.
• Simple harmonic motion is a special type of oscillation.
• In SHM, the restoring force is proportional to displacement and acts in the opposite direction: F ∝ −x
• A mass attached to an ideal spring is a standard example of SHM.
• A simple pendulum approximately undergoes SHM when its oscillations are small.
• Amplitude is the maximum displacement from equilibrium.
• Period is the time for one complete oscillation.
• Frequency is the number of oscillations completed each second.
• At maximum displacement, speed is zero while restoring force and acceleration are greatest.
• At equilibrium, speed is greatest while restoring force and acceleration are zero.
• Not every oscillation is SHM.
• The defining feature of SHM is the relationship between restoring force and displacement.

2. Springs and Hooke’s Law

Springs and Hooke’s Law

A spring is an elastic object that can be stretched or compressed. When a spring is displaced from its equilibrium position, it produces a restoring force that acts back toward equilibrium.

This behaviour is described by Hooke’s Law and provides one of the simplest and most important examples of simple harmonic motion (SHM).

Springs are used in:

• Vehicle suspension systems
• Spring balances
• Shock absorbers
• Mechanical clocks
• Mattresses
• Vibration-control systems
• Scientific instruments

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Learning Targets

• I can state and apply Hooke’s Law.
• I can calculate restoring forces using spring constants.
• I can interpret force-extension graphs.
• I can explain the relationship between force and displacement in SHM.
• I can analyze oscillations of spring-mass systems.

Elastic Behaviour of Springs

When a force is applied to a spring, the spring changes length.

A pulling force causes the spring to:

extend

A pushing force causes the spring to:

compress

When the applied force is removed, an elastic spring returns to its original length, provided it has not been stretched beyond its elastic limit.

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The change in length from the equilibrium position is called the displacement or extension.

Hooke’s Law

For an ideal spring, the restoring force is proportional to its displacement from equilibrium.

F = -kx

where:

F = restoring force in newtons (N)

k = spring constant in newtons per metre (N/m)

x = displacement from equilibrium in metres (m)

The negative sign tells us that the restoring force acts in the opposite direction to the displacement.

Understanding the Negative Sign

Suppose a spring-mass system is at equilibrium.

If the mass is pulled to the right:

Displacement → right

but:

Restoring force → left

If the mass is pushed to the left:

Displacement → left

but:

Restoring force → right

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The spring always attempts to return the mass toward its equilibrium position.

This opposition between force and displacement is fundamental to simple harmonic motion.

The Spring Constant

The spring constant, k, measures the stiffness of a spring.

Its unit is:

N/m

A large spring constant means the spring is relatively:

stiff

A small spring constant means the spring is relatively:

flexible

For example:

Spring A: k = 50 N/m

Spring B: k = 500 N/m

Spring B requires much more force to produce the same extension.

Worked Example — Calculating Restoring Force

A spring has:

k = 200 N/m

It is stretched:

x = 0.05 m

The magnitude of the restoring force is:

F = kx

F = (200)(0.05)

F = 10 N

Therefore, the spring produces a restoring force of:

10 N

directed opposite to the displacement.

Worked Example — Compression

A spring has:

k = 80 N/m

It is compressed by:

0.15 m

The magnitude of the restoring force is:

F = kx

F = (80)(0.15)

F = 12 N

Therefore:

Restoring force = 12 N

The force acts toward the equilibrium position.

Rearranging Hooke’s Law

We can use the magnitude form:

F = kx

to calculate other quantities.

To calculate spring constant:

k = F/x

To calculate displacement:

x = F/k

Always check that displacement is expressed in metres when k is measured in N/m.

Worked Example — Finding the Spring Constant

A force of:

15 N

extends a spring by:

0.060 m

Calculate the spring constant.

k = F/x

k = 15 / 0.060

k = 250 N/m

Therefore:

Spring constant = 250 N/m

Worked Example — Finding Extension

A spring has:

k = 400 N/m

A force of:

20 N

is applied.

Calculate the extension.

x = F/k

x = 20/400

x = 0.050 m

Therefore:

Extension = 0.050 m

or:

5.0 cm

Force-Extension Graphs

Hooke’s Law can be represented using a force-extension graph.

For an ideal spring, force and extension are directly proportional.

Therefore, the graph is a:

straight line through the origin

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Doubling the extension requires twice the force.

Tripling the extension requires three times the force.

This proportional relationship applies while the spring obeys Hooke’s Law.

Finding the Spring Constant from a Graph

If the graph has:

Force on the vertical axis

and:

Extension on the horizontal axis

then:

Gradient = change in force / change in extension

Therefore:

Gradient = k

A steeper line represents a spring with a larger spring constant.

Worked Example — Reading a Graph

Suppose a force-extension graph contains the point:

Extension = 0.20 m

Force = 30 N

Calculate the spring constant.

k = F/x

k = 30/0.20

k = 150 N/m

Therefore:

Spring constant = 150 N/m

Comparing Springs Using Graphs

Suppose two springs have straight-line force-extension graphs.

Spring A has a steeper graph than Spring B.

This means:

kA > kB

Therefore:

Spring A is stiffer than Spring B

because a greater force is needed to produce the same extension.

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Limit of Proportionality

Real springs do not obey Hooke’s Law indefinitely.

At first:

Force ∝ Extension

The graph is straight.

Eventually, the relationship may stop being proportional.

The point where this begins is called the:

limit of proportionality

Beyond this point, doubling the force no longer necessarily doubles the extension.

Elastic Limit

The elastic limit is the point beyond which an object may not return completely to its original shape when the force is removed.

If a spring is stretched too far, it may become permanently deformed.

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For many introductory problems, we assume the spring remains within the region where Hooke’s Law applies.

Springs and Simple Harmonic Motion

Attach a mass to a spring on a low-friction horizontal surface.

At equilibrium, the spring is neither stretched nor compressed.

Pull the mass away from equilibrium and release it.

The mass begins to oscillate back and forth.

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This motion can approximate simple harmonic motion.

Force and Displacement in SHM

For a spring:

Restoring force ∝ −displacement

This is the defining condition for simple harmonic motion.

The further the mass moves from equilibrium, the larger the restoring force becomes.

At equilibrium:

x = 0

therefore:

Restoring force = 0

At maximum displacement:

|x| = maximum

therefore:

Restoring force magnitude = maximum

Three Important Positions

Consider a mass oscillating horizontally on a spring.

Maximum Extension

The spring is stretched the most.

Displacement = maximum

Restoring force = maximum toward equilibrium

Speed = 0

Equilibrium Position

The spring passes through its equilibrium position.

Displacement = 0

Spring restoring force = 0

Speed = maximum

Maximum Compression

The spring is compressed the most.

Displacement = maximum in the opposite direction

Restoring force = maximum toward equilibrium

Speed = 0

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Acceleration in a Spring-Mass System

Newton’s Second Law tells us:

F = ma

For a spring:

F = −kx

Therefore:

ma = −kx

So:

a = −(k/m)x

This shows that acceleration is proportional to displacement and acts in the opposite direction.

Therefore:

a ∝ −x

This is another way to recognize simple harmonic motion.

Acceleration at Different Positions

At equilibrium:

x = 0

so:

a = 0

At maximum displacement:

|x| = maximum

so:

|a| = maximum

This may initially seem surprising.

The mass moves fastest at equilibrium, even though its acceleration is zero at that instant.

At the ends of the oscillation, its speed is zero but its acceleration is greatest.

Worked Example — Acceleration

A:

0.50 kg

mass is attached to a spring with:

k = 100 N/m

The mass is displaced:

0.10 m

Calculate the magnitude of its acceleration.

First calculate the restoring force:

F = kx

F = (100)(0.10)

F = 10 N

Now:

a = F/m

a = 10/0.50

a = 20 m/s²

Therefore:

Acceleration magnitude = 20 m/s²

directed toward equilibrium.

Oscillation Cycle

Suppose a mass begins at maximum extension.

It then moves:

Maximum extension → equilibrium → maximum compression → equilibrium → maximum extension

This is:

one complete oscillation

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The motion then repeats if energy losses are negligible.

Period of a Spring-Mass System

The period, T, is the time required for one complete oscillation.

For an ideal mass-spring system:

T = 2π√(m/k)

where:

T = period in seconds

m = mass in kilograms

k = spring constant in N/m

This equation tells us how the mass and spring stiffness affect the oscillation.

Effect of Mass

From:

T = 2π√(m/k)

increasing the mass increases the period.

Therefore:

Larger mass → Longer period → Slower oscillation

A heavier mass takes longer to complete each oscillation.

Effect of Spring Constant

Increasing k decreases the period.

Therefore:

Stiffer spring → Shorter period → Faster oscillation

A stiff spring produces a stronger restoring force for the same displacement, causing the system to oscillate more rapidly.

Worked Example — Period

A:

0.50 kg

mass is attached to a spring with:

k = 200 N/m

Calculate the period.

T = 2π√(m/k)

Substitute:

T = 2π√(0.50/200)

T = 2π√0.0025

T = 2π(0.05)

T ≈ 0.314 s

Therefore:

Period ≈ 0.31 s

Frequency

The frequency, f, tells us how many complete oscillations occur each second.

Frequency and period are related by:

f = 1/T

For the previous example:

T ≈ 0.314 s

Therefore:

f = 1/0.314

f ≈ 3.18 Hz

The mass completes approximately:

3.18 oscillations per second

Energy in a Spring

A stretched or compressed spring stores elastic potential energy.

For an ideal spring:

Elastic potential energy = ½kx²

At maximum displacement:

• Elastic potential energy is maximum.
• Speed is zero.
• Kinetic energy is minimum.

At equilibrium:

• Elastic potential energy associated with displacement from equilibrium is minimum.
• Speed is maximum.
• Kinetic energy is maximum.

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Energy continually transfers between elastic potential energy and kinetic energy.

Energy During One Oscillation

At maximum extension:

Elastic PE = maximum

KE = 0

As the mass approaches equilibrium:

Elastic PE decreases

KE increases

At equilibrium:

KE = maximum

The mass then compresses the spring.

As compression increases:

KE decreases

Elastic PE increases

At maximum compression:

KE = 0

and elastic potential energy is again maximum.

Force-Displacement Graph

For an ideal spring, restoring force and displacement have a linear relationship.

Because the force is opposite to displacement, a graph of restoring force against displacement has a negative gradient.

The gradient is:

−k

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This graph directly illustrates the SHM condition:

Restoring force is proportional and opposite to displacement.

Vertical Spring-Mass Systems

A mass can also oscillate vertically on a spring.

When a mass hangs from a vertical spring, gravity stretches the spring until equilibrium is reached.

At equilibrium:

Spring force upward = Weight downward

Therefore:

kx = mg

where x is the equilibrium extension caused by the hanging mass.

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If the mass is then displaced from this equilibrium position and released, it can oscillate vertically.

Worked Example — Hanging Mass

A:

2.0 kg

mass hangs from a spring with:

k = 400 N/m

Using:

g = 9.8 m/s²

the weight is:

mg = (2.0)(9.8)

mg = 19.6 N

At equilibrium:

kx = mg

Therefore:

x = mg/k

x = 19.6/400

x = 0.049 m

So the spring extends approximately:

4.9 cm

Real Spring-Mass Systems

Real oscillating systems do not continue forever.

Energy is gradually transferred to the surroundings through:

• Air resistance
• Friction
• Internal heating of the spring
• Sound

As energy is dissipated, the amplitude gradually decreases.

This is called damping.

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An ideal SHM model often assumes these energy losses are negligible.

Springs in Vehicle Suspension

Vehicle suspension systems use springs to allow wheels and the vehicle body to respond to bumps in the road.

However, springs alone would cause the vehicle to continue oscillating.

Vehicles therefore also use dampers or shock absorbers.

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The spring provides a restoring force.

The damper removes mechanical energy from the oscillation.

Together, they provide a smoother and more controlled ride.

Investigating Hooke’s Law Experimentally

A simple Hooke’s Law investigation can use:

• Clamp stand
• Spring
• Mass hanger
• Slotted masses
• Metre ruler

First measure the spring's original length.

Add masses gradually and record the new length.

Calculate:

Extension = New length − Original length

Calculate the applied force using:

F = mg

Then plot:

Force against extension

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A straight-line region through the origin provides evidence that:

F ∝ x

Improving the Experiment

For more reliable results:

• Measure from the same reference point each time.
• Read the ruler at eye level to reduce parallax error.
• Allow the spring to stop moving before measuring.
• Add masses in regular intervals.
• Repeat measurements.
• Avoid exceeding the elastic limit.
• Use a wide range of safe forces.
• Plot a best-fit line rather than joining individual points.

These procedures help improve the quality of the force-extension data.

Common Mistakes

Mistake: Forgetting to Convert Centimetres to Metres

If:

x = 5 cm

then:

x = 0.05 m

when using k in N/m.

Mistake: Ignoring the Direction of Restoring Force

The spring force acts opposite to the displacement.

This is why the full form of Hooke’s Law contains a negative sign.

Mistake: Assuming Hooke’s Law Always Applies

Hooke’s Law applies only while force and extension remain proportional.

Real springs eventually stop behaving linearly.

Mistake: Confusing Stiffness

A larger k means:

stiffer spring

not a more flexible spring.

Mistake: Thinking Speed Is Greatest at Maximum Displacement

At maximum displacement:

speed = 0

The speed is greatest at:

equilibrium

Mistake: Thinking Acceleration Is Greatest at Equilibrium

At equilibrium:

restoring force = 0

therefore:

acceleration = 0

Acceleration magnitude is greatest at maximum displacement.

Key Terms

Spring – elastic object that can be stretched or compressed.

Equilibrium position – position where the net restoring force associated with the oscillation is zero.

Displacement – distance and direction from equilibrium.

Extension – increase in length from the spring's original or reference length.

Restoring force – force directed toward equilibrium.

Spring constant – measure of spring stiffness, measured in N/m.

Hooke’s Law – relationship between spring restoring force and displacement.

Limit of proportionality – point beyond which force and extension are no longer directly proportional.

Elastic limit – point beyond which permanent deformation may occur.

Simple harmonic motion – oscillation in which restoring force or acceleration is proportional and opposite to displacement.

Period – time for one complete oscillation.

Frequency – number of complete oscillations per second.

Amplitude – maximum displacement from equilibrium.

Damping – reduction in oscillation amplitude due to energy dissipation.

Check Your Understanding

1. What does the spring constant tell us about a spring?

2. A spring has k = 150 N/m and is stretched 0.08 m. Calculate the magnitude of the restoring force.

3. A 12 N force extends a spring by 0.040 m. Calculate its spring constant.

4. A spring has k = 500 N/m. Calculate the extension caused by a 25 N force.

5. Explain the meaning of the negative sign in Hooke’s Law.

6. What does the gradient of a force-extension graph represent when force is on the vertical axis?

7. Two springs have force-extension graphs. Spring A has the steeper line. Which spring is stiffer? Explain.

8. What happens when a spring is stretched beyond its elastic limit?

9. Where is the speed of a spring-mass oscillator greatest?

10. Where is its acceleration magnitude greatest?

11. Explain why a spring-mass system can undergo simple harmonic motion.

12. A 0.40 kg mass is attached to a spring with k = 100 N/m. Calculate its period.

13. What happens to the period if the mass is increased?

14. What happens to the period if a stiffer spring is used?

15. Describe the energy transfers that occur as a mass moves from maximum extension through equilibrium.

Key Takeaways

• Springs produce a restoring force when stretched or compressed.
• Hooke’s Law describes the relationship between restoring force and displacement.
• The restoring force acts in the direction opposite to displacement.
• The spring constant k measures the stiffness of a spring.
• A larger spring constant means a stiffer spring.
• Within the proportional region, force and extension are directly proportional.
• A force-extension graph is therefore a straight line through the origin.
• If force is plotted against extension, the gradient gives the spring constant.
• Hooke’s Law does not apply indefinitely to real springs.
• A spring-mass system can approximate simple harmonic motion.
• In SHM, restoring force and acceleration are proportional and opposite to displacement.
• Speed is greatest at equilibrium.
• Restoring force and acceleration magnitude are greatest at maximum displacement.
• Increasing mass increases the period of a spring-mass oscillator.
• Increasing spring stiffness decreases the period.
• Energy transfers between elastic potential and kinetic energy during oscillation.
• Real oscillations gradually lose energy through damping.

Did You Know?

A spring-mass oscillator is one of the most useful models in physics because the same mathematics can describe many systems that do not look like springs at all. Small vibrations of molecules, parts of buildings, vehicle suspension systems, electrical circuits, and even some models of atomic behaviour can display patterns closely related to simple harmonic motion.

3. Pendulums

Pendulums

A simple pendulum consists of a small mass called a bob suspended from a fixed point by a light string or rod.

When the bob is displaced from its equilibrium position and released, gravity causes it to swing back and forth.

For relatively small angles, this repeated motion closely approximates simple harmonic motion (SHM).

Pendulums provide an important connection between:

• Forces
• Oscillations
• Energy
• Period and frequency
• Simple harmonic motion

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Learning Targets

• I can describe the motion of a simple pendulum.
• I can explain why small-angle pendulums approximate SHM.
• I can identify factors affecting pendulum period.
• I can compare pendulum motion with spring oscillations.
• I can solve problems involving pendulum motion.

The Simple Pendulum

An ideal simple pendulum consists of:

• A point-like mass called the bob
• A light, inextensible string
• A fixed pivot
• Negligible air resistance and friction

The distance from the pivot to the centre of the bob is the pendulum length, L.

The lowest position is the equilibrium position.

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Real pendulums do not perfectly satisfy these assumptions, but the simple-pendulum model provides a very useful approximation.

Motion of a Pendulum

Imagine pulling a pendulum bob to one side and releasing it.

The bob moves:

Maximum displacement → Equilibrium → Maximum displacement on opposite side → Equilibrium → Starting position

This represents:

one complete oscillation

The motion then repeats.

The maximum displacement from equilibrium is called the:

amplitude

Important Positions During the Swing

Consider three important positions:

A — Maximum displacement to the left

B — Equilibrium position

C — Maximum displacement to the right

At A:

Speed = 0

At B:

Speed = maximum

At C:

Speed = 0

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The bob continually speeds up and slows down as it oscillates.

Forces Acting on the Pendulum

Two main forces act on the bob:

Weight (mg) acting vertically downward.

Tension (T) acting along the string toward the pivot.

The weight can be resolved into components.

One component acts along the string.

Another component acts tangentially to the bob's path.

The tangential component of gravity acts as the restoring force.

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The Restoring Force

When the bob is displaced from equilibrium, gravity produces a force component that pulls it back toward equilibrium.

The tangential restoring force is:

F = −mg sin θ

where:

m = mass of the bob

g = gravitational field strength

θ = angular displacement

The negative sign indicates that the force acts opposite to the displacement.

This is similar to the restoring force produced by a spring.

Why Pendulums Can Approximate SHM

For an object to undergo simple harmonic motion, the restoring force must be:

proportional to displacement

and:

opposite to displacement

For a pendulum:

F = −mg sin θ

Normally:

sin θ ≠ θ

so the relationship is not perfectly proportional.

However, for small angles measured in radians:

sin θ ≈ θ

Therefore:

F ≈ −mgθ

The restoring force becomes approximately proportional to angular displacement.

This is why a pendulum at small angles approximates SHM.

The Small-Angle Approximation

The approximation:

sin θ ≈ θ

works best when θ is small and measured in radians.

For example:

5° ≈ 0.087 rad

and:

sin 5° ≈ 0.087

These values are extremely close.

At larger angles, the difference becomes more significant.

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For introductory pendulum calculations, angles below approximately 10–15° are commonly treated as small enough for the SHM approximation.

Period of a Pendulum

The period, T, is the time required for one complete oscillation.

For a simple pendulum undergoing small oscillations, explore how length and release angle affect its motion here:

where:

T = period in seconds (s)

L = pendulum length in metres (m)

g = gravitational field strength in m/s²

Notice something interesting:

Mass does not appear in the equation.

Pendulum Length

The period depends on the pendulum's length.

A longer pendulum has a:

longer period

A shorter pendulum has a:

shorter period

Therefore:

Long pendulum → Slower oscillation

Short pendulum → Faster oscillation

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The relationship is not directly proportional.

Instead:

T ∝ √L

What Happens if Length Is Doubled?

Suppose:

L → 2L

Since:

T ∝ √L

the new period becomes:

Tnew = √2 × Told

Since:

√2 ≈ 1.41

doubling the length increases the period by a factor of approximately:

1.41

It does not double the period.

What Happens if Length Is Quadrupled?

Suppose the length becomes:

4L

Then:

√4 = 2

Therefore:

Tnew = 2Told

So:

Quadrupling the length doubles the period.

This square-root relationship is important when analyzing pendulum systems.

Worked Example — Calculating Period

A pendulum has length:

L = 1.0 m

Use:

g = 9.8 m/s²

Calculate the period.

T = 2π√(L/g)

T = 2π√(1.0/9.8)

T = 2π√0.102

T ≈ 2π(0.319)

T ≈ 2.01 s

Therefore:

Period ≈ 2.0 s

The pendulum takes approximately two seconds to complete one full oscillation.

Worked Example — A Shorter Pendulum

Calculate the period of a:

0.25 m

pendulum.

Use:

g = 9.8 m/s²

T = 2π√(0.25/9.8)

T ≈ 2π(0.160)

T ≈ 1.00 s

Therefore:

Period ≈ 1.0 s

Notice that reducing the length from 1.0 m to 0.25 m reduces the period from approximately 2.0 s to 1.0 s.

The length was divided by 4, so the period was divided by 2.

Gravitational Field Strength

The period also depends on gravitational field strength.

Since:

T ∝ 1/√g

stronger gravity produces a:

shorter period

Weaker gravity produces a:

longer period

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A pendulum would therefore swing more slowly on the Moon than on Earth because the Moon has a weaker gravitational field.

Worked Example — Pendulum on the Moon

Suppose:

L = 1.0 m

On the Moon:

g ≈ 1.62 m/s²

Calculate the period.

T = 2π√(1.0/1.62)

T ≈ 2π(0.786)

T ≈ 4.94 s

So a 1.0 m pendulum would have a period of approximately:

4.9 s

on the Moon.

On Earth, the same pendulum has a period of approximately:

2.0 s

Does Mass Affect the Period?

Suppose one pendulum has a:

50 g bob

and another identical pendulum has a:

500 g bob

If they have the same length and experience the same gravitational field, the ideal small-angle model predicts:

the same period

Mass does not appear in the period equation.

This is an important and sometimes surprising result.

Does Amplitude Affect the Period?

For small angles, the period is approximately independent of amplitude.

A pendulum released from:

5°

and one released from:

10°

will have very similar periods if their lengths are equal.

However, at larger amplitudes, the simple equation becomes less accurate.

The period becomes slightly longer as the starting angle becomes large.

Therefore:

Small amplitude → SHM approximation works well

Large amplitude → Motion deviates from ideal SHM

Frequency of a Pendulum

The frequency, f, is the number of complete oscillations per second.

Frequency and period are related by:

f = 1/T

If:

T = 2.0 s

then:

f = 1/2.0

f = 0.50 Hz

The pendulum completes:

0.50 oscillations each second

or one oscillation every two seconds.

Worked Example — Frequency

A pendulum completes one oscillation every:

1.25 s

Therefore:

T = 1.25 s

Calculate frequency.

f = 1/T

f = 1/1.25

f = 0.80 Hz

Therefore:

Frequency = 0.80 Hz

Energy in Pendulum Motion

A pendulum continually transfers energy between:

gravitational potential energy

and:

kinetic energy

At maximum displacement:

GPE = maximum

KE = minimum

Speed = 0

As the bob falls toward equilibrium:

GPE decreases

KE increases

At equilibrium:

GPE = minimum

KE = maximum

Speed = maximum

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As the bob rises on the opposite side, kinetic energy is transferred back into gravitational potential energy.

Energy During One Oscillation

Starting at maximum displacement:

Maximum GPE, zero KE

↓

Moving toward equilibrium:

GPE → KE

↓

At equilibrium:

Maximum KE, minimum GPE

↓

Moving upward:

KE → GPE

↓

At opposite maximum displacement:

Maximum GPE, zero KE

In an ideal system:

Total mechanical energy remains constant.

Speed and Acceleration

At maximum displacement:

Speed = 0

but:

Restoring acceleration magnitude = maximum

At equilibrium:

Speed = maximum

but:

Tangential restoring acceleration = 0

This is the same basic pattern observed in a spring-mass oscillator.

Pendulum Displacement-Time Graph

For small oscillations, pendulum displacement varies approximately sinusoidally with time.

The displacement-time graph therefore resembles a sine or cosine wave.

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Important features include:

Amplitude → maximum displacement

Period → time for one complete cycle

Equilibrium → displacement = 0

Comparing Pendulums and Springs

Pendulums and spring-mass systems can both approximate simple harmonic motion.

For a spring:

Restoring force ∝ −displacement

For a small-angle pendulum:

Restoring force ≈ proportional to −displacement

Both systems:

• Oscillate around an equilibrium position.
• Have maximum speed at equilibrium.
• Have zero speed at maximum displacement.
• Have maximum restoring acceleration at maximum displacement.
• Transfer energy between potential and kinetic energy.
• Can produce approximately sinusoidal displacement-time graphs.

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Important Difference Between Springs and Pendulums

The period of a spring-mass system is:

T = 2π√(m/k)

Therefore, it depends on:

• Mass
• Spring constant

The period of a simple pendulum is:

T = 2π√(L/g)

Therefore, it depends on:

• Pendulum length
• Gravitational field strength

So the factors controlling their periods are different.

Worked Comparison

Consider two systems.

Spring

m = 1.0 kg

k = 100 N/m

The period depends on m and k.

Pendulum

L = 1.0 m

g = 9.8 m/s²

The period depends on L and g.

Changing the mass affects the spring oscillator's period.

Changing the pendulum bob's mass does not affect the ideal pendulum period.

Finding Pendulum Length from Period

The period equation can be rearranged.

Start with:

T = 2π√(L/g)

Divide by 2π:

T/(2π) = √(L/g)

Square both sides:

T²/(4π²) = L/g

Therefore:

L = gT²/(4π²)

This allows us to calculate the length required for a particular period.

Worked Example — Finding Length

A pendulum has:

T = 1.5 s

Use:

g = 9.8 m/s²

Calculate its length.

L = gT²/(4π²)

L = (9.8)(1.5²)/(4π²)

L = (9.8)(2.25)/(39.48)

L ≈ 0.56 m

Therefore:

Pendulum length ≈ 0.56 m

Measuring g with a Pendulum

The pendulum equation can also be rearranged to estimate gravitational field strength:

g = 4π²L/T²

Therefore, if we accurately measure:

• Pendulum length
• Pendulum period

we can experimentally estimate g.

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This is a classic physics experiment.

Measuring the Period Experimentally

Timing only one oscillation can produce a large percentage uncertainty because human reaction time is significant.

A better method is to time several oscillations.

For example, measure the time for:

10 oscillations

Suppose:

10 oscillations = 18.4 s

Then:

T = 18.4/10

T = 1.84 s

This reduces the effect of reaction-time uncertainty.

Investigating Length and Period

A useful experiment is to investigate:

How does pendulum length affect period?

Change the pendulum length while keeping other variables controlled.

For each length:

• Use a small release angle.
• Release without pushing.
• Time 10 or 20 oscillations.
• Calculate the period.
• Repeat measurements.
• Calculate an average.

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Variables in a Pendulum Experiment

If investigating the effect of length:

Independent variable → pendulum length

Dependent variable → period

Important control variables include:

• Same bob
• Same release angle
• Same gravitational field
• Same release technique

The angle should remain relatively small so that the SHM approximation remains valid.

Plotting Pendulum Data

Because:

T = 2π√(L/g)

T is not directly proportional to L.

A graph of:

T against L

will therefore be curved.

However, squaring the equation gives:

T² = (4π²/g)L

Therefore:

T² ∝ L

A graph of:

T² against L

should be a straight line through approximately the origin.

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This provides a useful experimental test of the pendulum model.

Real Pendulums and Damping

A real pendulum eventually stops swinging.

This happens because mechanical energy is transferred to the surroundings through:

• Air resistance
• Friction at the pivot
• Sound
• Heating

The amplitude gradually decreases.

This is called:

damping

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An ideal pendulum model assumes these energy losses are negligible.

Pendulum Clocks

Pendulums became extremely important in accurate mechanical clocks because their periods can be highly regular.

5

The period can be controlled mainly by adjusting the pendulum length.

A longer pendulum swings more slowly.

A shorter pendulum swings more quickly.

This allowed pendulums to provide a regular timing mechanism.

Common Mistakes

Mistake: Measuring Pendulum Length Incorrectly

Pendulum length should normally be measured from:

the pivot to the centre of the bob

not merely the length of the visible string.

Mistake: Thinking a Heavier Bob Has a Longer Period

For an ideal simple pendulum:

Period does not depend on bob mass.

Mistake: Assuming Period Is Directly Proportional to Length

The relationship is:

T ∝ √L

not:

T ∝ L

Mistake: Using the Small-Angle Equation at Very Large Angles

The standard period equation is an approximation for relatively small oscillations.

Mistake: Timing Only One Oscillation

Reaction-time uncertainty becomes significant.

Time several oscillations and divide by the number of oscillations.

Mistake: Confusing Half an Oscillation with a Full Oscillation

Moving from one extreme to the opposite extreme is:

half an oscillation

Returning to the original extreme completes:

one full oscillation

Mistake: Thinking Speed Is Greatest at Maximum Displacement

At maximum displacement:

speed = 0

Speed is greatest at:

equilibrium

Key Terms

Pendulum – mass suspended from a pivot that can swing back and forth.

Bob – mass at the end of a pendulum.

Pivot – fixed point from which the pendulum is suspended.

Equilibrium position – lowest central position of the pendulum.

Amplitude – maximum displacement from equilibrium.

Oscillation – one complete repeated cycle of motion.

Period – time required for one complete oscillation.

Frequency – number of complete oscillations per second.

Restoring force – force acting toward equilibrium.

Small-angle approximation – approximation that sin θ ≈ θ when θ is small and measured in radians.

Simple harmonic motion – oscillatory motion in which restoring force or acceleration is proportional and opposite to displacement.

Damping – gradual decrease in amplitude due to energy transfer to the surroundings.

Check Your Understanding

1. Describe one complete oscillation of a pendulum.

2. Where is the speed of a pendulum bob greatest?

3. Where is its speed zero?

4. What provides the restoring force for a pendulum?

5. Explain why a small-angle pendulum approximates simple harmonic motion.

6. A pendulum has a length of 0.80 m. Calculate its period using g = 9.8 m/s².

7. What happens to the period if the pendulum length increases?

8. If the pendulum length is quadrupled, what happens to its period?

9. Does increasing the mass of the bob change the ideal pendulum period?

10. Explain why a pendulum swings more slowly on the Moon than on Earth.

11. A pendulum has a period of 2.5 s. Calculate its frequency.

12. A pendulum has a period of 2.0 s. Approximately how long will 20 complete oscillations take?

13. Describe the energy changes as the bob moves from maximum displacement toward equilibrium.

14. Give two similarities between a pendulum and a spring-mass oscillator.

15. Give one important difference between the factors controlling the periods of pendulums and spring-mass oscillators.

Key Takeaways

• A simple pendulum consists of a bob suspended from a fixed pivot.
• The bob oscillates around an equilibrium position.
• Gravity provides the restoring effect.
• For small angles, the restoring force is approximately proportional and opposite to displacement.
• Therefore, a small-angle pendulum approximates simple harmonic motion.
• The period depends mainly on pendulum length and gravitational field strength.
• Increasing pendulum length increases the period.
• Increasing gravitational field strength decreases the period.
• Bob mass does not affect the period of an ideal simple pendulum.
• For small oscillations, amplitude has very little effect on period.
• Speed is greatest at equilibrium.
• Speed is zero at maximum displacement.
• Gravitational potential energy and kinetic energy continually transfer between each other.
• Pendulums and spring-mass systems can both approximate SHM.
• Pendulum period depends on L and g, while spring period depends on m and k.
• Timing several oscillations improves experimental measurements.
• A graph of T² against L should be approximately linear.
• Real pendulums experience damping because energy is transferred to the surroundings.

Did You Know?

Pendulums can actually be used to measure gravity. Because their period depends on gravitational field strength, scientists can measure the length and period of a carefully controlled pendulum and calculate the local value of g. Small variations in gravitational field strength from one location to another can even provide information about differences in the density of rocks beneath Earth's surface.

4. SHM Graphs and Equations

SHM Graphs and Equations

Simple harmonic motion (SHM) can be described using graphs and mathematical equations. These representations allow us to determine where an oscillating object is, how fast it is moving, and how it is accelerating at any point during its motion.

Three graphs are especially important:

• Displacement–time graphs
• Velocity–time graphs
• Acceleration–time graphs

All three have repeating, sinusoidal shapes, but they do not reach their maximum and minimum values at the same time.

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Learning Targets

• I can interpret displacement-time graphs for SHM.
• I can interpret velocity-time and acceleration-time graphs for SHM.
• I can determine amplitude, period, and frequency from graphs.
• I can relate graphical representations to physical motion.
• I can apply mathematical models of SHM to solve problems.

Displacement in SHM

Displacement, x, tells us the position of an oscillating object relative to its equilibrium position.

We usually define:

x = 0

as the equilibrium position.

One side of equilibrium is assigned positive displacement and the other negative displacement.

At the two extremes:

x = +A

and:

x = −A

where A is the amplitude.

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The amplitude is therefore the maximum displacement from equilibrium.

The Displacement-Time Graph

A displacement-time graph for ideal SHM is sinusoidal.

It may look like a sine or cosine curve depending on where the object begins its motion.

5

The vertical axis represents:

Displacement, x

The horizontal axis represents:

Time, t

The graph tells us where the object is relative to equilibrium at every instant.

Reading Physical Motion from a Displacement-Time Graph

Suppose an oscillator begins at:

x = +A

It then moves toward equilibrium.

The sequence is:

+A → 0 → −A → 0 → +A

This represents one complete oscillation.

At:

x = +A

the object is at one extreme.

At:

x = 0

the object is passing through equilibrium.

At:

x = −A

the object is at the opposite extreme.

Finding Amplitude from a Graph

Amplitude is the maximum displacement from equilibrium.

Suppose a graph reaches:

+0.12 m

and:

−0.12 m

Then:

Amplitude = 0.12 m

Do not measure from the minimum to the maximum.

The distance from −A to +A is:

2A

So in this example:

Peak-to-peak displacement = 0.24 m

but:

Amplitude = 0.12 m

Finding the Period

The period, T, is the time required for one complete oscillation.

On a displacement-time graph, measure the time between two equivalent points in consecutive cycles.

For example:

maximum → next maximum

or:

minimum → next minimum

or:

same equilibrium crossing in the same direction → next identical crossing

5

Worked Example — Finding Period

Suppose a displacement-time graph has maxima at:

t = 1.0 s

and:

t = 5.0 s

Therefore:

T = 5.0 − 1.0

T = 4.0 s

The period is:

4.0 s

Finding Frequency

Frequency is the number of complete oscillations per second.

Frequency and period are related by:

f = 1/T

where:

f = frequency in hertz (Hz)

T = period in seconds (s)

For the previous example:

T = 4.0 s

Therefore:

f = 1/4.0

f = 0.25 Hz

Angular Frequency

SHM equations commonly use angular frequency, represented by:

ω

Angular frequency is related to ordinary frequency by:

ω = 2πf

Since:

f = 1/T

we can also write:

ω = 2π/T

The unit of angular frequency is:

radians per second (rad/s)

Worked Example — Angular Frequency

An oscillator has:

T = 2.0 s

First calculate frequency:

f = 1/T

f = 1/2.0

f = 0.50 Hz

Now calculate angular frequency:

ω = 2πf

ω = 2π(0.50)

ω = π rad/s

or approximately:

ω = 3.14 rad/s

Mathematical Model for Displacement

One common equation for SHM is:

x = A cos(ωt)

where:

x = displacement

A = amplitude

ω = angular frequency

t = time

This equation describes an oscillator that begins at:

x = +A

when:

t = 0

because:

cos 0 = 1

Therefore:

x = A

Sine and Cosine Models

SHM can also be written using a sine function:

x = A sin(ωt)

The difference depends on the object's starting position.

For:

x = A cos(ωt)

the object begins at maximum positive displacement.

For:

x = A sin(ωt)

the object begins at equilibrium moving in the positive direction.

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Both equations describe simple harmonic motion.

Phase

The position of an oscillator within its cycle is called its phase.

Two oscillators can have:

• The same amplitude
• The same frequency
• The same period

but still be at different points in their cycles.

They are then said to have a:

phase difference

Phase can be measured in:

• Degrees
• Radians

One complete cycle corresponds to:

360°

or:

2π radians

Velocity in SHM

Velocity tells us how quickly displacement is changing.

Therefore:

Velocity = gradient of the displacement-time graph

When the displacement-time graph is steepest, the magnitude of velocity is greatest.

When the displacement-time graph has a horizontal tangent, velocity is zero.

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Velocity at Maximum Displacement

At:

x = +A

or:

x = −A

the object momentarily stops before reversing direction.

Therefore:

v = 0

So:

Maximum displacement → Zero velocity

Velocity at Equilibrium

At:

x = 0

the oscillator is passing through equilibrium.

Its speed is greatest here.

Therefore:

Equilibrium → Maximum speed

The velocity may be positive or negative depending on which direction the object is moving.

Maximum Velocity

For SHM:

Maximum speed = Aω

Therefore:

vmax = Aω

This tells us that maximum speed increases if:

• Amplitude increases.
• Angular frequency increases.

Worked Example — Maximum Speed

An oscillator has:

A = 0.20 m

and:

f = 2.0 Hz

First calculate angular frequency:

ω = 2πf

ω = 2π(2.0)

ω = 4π rad/s

Now:

vmax = Aω

vmax = (0.20)(4π)

vmax ≈ 2.51 m/s

Therefore:

Maximum speed ≈ 2.5 m/s

Velocity-Time Graph

Velocity also varies sinusoidally with time.

However, the velocity graph is shifted relative to the displacement graph.

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When displacement is maximum:

Velocity = 0

When displacement is zero:

|Velocity| = maximum

Therefore, displacement and velocity are:

one-quarter of a cycle out of phase

This corresponds to:

90°

or:

π/2 radians

Direction from the Velocity Graph

The sign of velocity tells us the direction of motion.

v > 0

means motion in the positive direction.

v < 0

means motion in the negative direction.

Suppose:

x > 0

and:

v < 0

The object is on the positive side of equilibrium but moving back toward equilibrium.

Graphs allow us to determine both position and direction of motion.

Acceleration in SHM

Acceleration is the rate of change of velocity.

Therefore:

Acceleration = gradient of the velocity-time graph

For SHM, acceleration has a particularly important relationship with displacement:

a = −ω²x

This equation is one of the defining mathematical relationships of simple harmonic motion.

Understanding a = −ω²x

The equation tells us two important things.

First:

|a| ∝ |x|

The further the object is from equilibrium, the greater its acceleration magnitude.

Second:

The negative sign means acceleration acts in the direction opposite to displacement.

Therefore, acceleration always points:

toward equilibrium

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Acceleration at Equilibrium

At equilibrium:

x = 0

Therefore:

a = −ω²(0)

so:

a = 0

At this instant:

• Displacement is zero.
• Acceleration is zero.
• Speed is maximum.

Acceleration at Maximum Displacement

At:

x = +A

acceleration is:

a = −ω²A

At:

x = −A

acceleration is:

a = +ω²A

Therefore:

Maximum displacement → Maximum acceleration magnitude

The acceleration always points toward equilibrium.

Maximum Acceleration

The maximum acceleration magnitude is:

amax = Aω²

This occurs at maximum displacement.

Worked Example — Maximum Acceleration

An oscillator has:

A = 0.050 m

and:

ω = 8.0 rad/s

Calculate its maximum acceleration.

amax = Aω²

amax = (0.050)(8.0²)

amax = (0.050)(64)

amax = 3.2 m/s²

Therefore:

Maximum acceleration = 3.2 m/s²

Acceleration-Time Graph

Acceleration varies sinusoidally.

It has the same period as displacement and velocity.

However, acceleration and displacement always have opposite signs.

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When:

x = maximum positive

then:

a = maximum negative

When:

x = maximum negative

then:

a = maximum positive

Therefore, acceleration and displacement are:

180° out of phase

or:

π radians out of phase

Comparing the Three Graphs

For a complete understanding of SHM, displacement, velocity, and acceleration should be considered together.

At maximum positive displacement:

x = +A

v = 0

a = maximum negative

At equilibrium while moving in the negative direction:

x = 0

v = maximum negative

a = 0

At maximum negative displacement:

x = −A

v = 0

a = maximum positive

At equilibrium while moving in the positive direction:

x = 0

v = maximum positive

a = 0

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A Quarter-Cycle Pattern

A useful way to understand SHM is to divide one oscillation into four equal parts.

Starting at maximum positive displacement:

0T

x = +A
v = 0
a = maximum negative

¼T

x = 0
v = maximum negative
a = 0

½T

x = −A
v = 0
a = maximum positive

¾T

x = 0
v = maximum positive
a = 0

1T

x = +A
v = 0
a = maximum negative

Then the cycle repeats.

Relating Graphs to a Spring

Imagine a mass attached to a horizontal spring.

At maximum extension:

• Displacement is maximum.
• Velocity is zero.
• Restoring force is maximum toward equilibrium.
• Acceleration magnitude is maximum.

As the mass approaches equilibrium:

• Displacement decreases.
• Speed increases.
• Acceleration magnitude decreases.

At equilibrium:

• Displacement is zero.
• Speed is maximum.
• Acceleration is zero.

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The process then reverses as the mass moves toward maximum compression.

Relating Graphs to a Pendulum

A small-angle pendulum follows the same general pattern.

At maximum displacement:

• Speed = 0
• Restoring acceleration magnitude = maximum

At equilibrium:

• Speed = maximum
• Tangential restoring acceleration = 0

5

The same mathematical model can therefore describe both systems approximately.

Velocity as a Function of Displacement

Velocity can also be related directly to displacement:

v² = ω²(A² − x²)

or:

v = ±ω√(A² − x²)

This equation allows us to calculate the velocity at a particular displacement without knowing the time.

The positive or negative sign depends on the direction of motion.

Worked Example — Velocity at a Position

An oscillator has:

A = 0.10 m

ω = 5.0 rad/s

At one instant:

x = 0.060 m

Calculate the speed.

v = ω√(A² − x²)

v = 5.0√(0.10² − 0.060²)

v = 5.0√(0.0100 − 0.0036)

v = 5.0√0.0064

v = 5.0(0.080)

v = 0.40 m/s

Therefore:

Speed = 0.40 m/s

The velocity could be +0.40 m/s or −0.40 m/s depending on the direction of motion.

Worked Example — Using the SHM Equation

An oscillator follows:

x = 0.15 cos(4t)

where x is measured in metres and t in seconds.

Compare this with:

x = A cos(ωt)

Therefore:

A = 0.15 m

and:

ω = 4 rad/s

The period is:

T = 2π/ω

T = 2π/4

T ≈ 1.57 s

The frequency is:

f = 1/T

f ≈ 0.637 Hz

So:

Amplitude = 0.15 m

Period ≈ 1.57 s

Frequency ≈ 0.64 Hz

Worked Example — Finding Acceleration

For the same oscillator:

x = 0.15 cos(4t)

suppose the displacement at a particular instant is:

x = 0.10 m

Use:

a = −ω²x

a = −(4²)(0.10)

a = −1.6 m/s²

The negative sign means the acceleration points in the negative direction.

Because the object currently has positive displacement, this acceleration points:

toward equilibrium

Graph of Acceleration Against Displacement

Instead of plotting acceleration against time, we can plot:

acceleration against displacement

Since:

a = −ω²x

the graph is a straight line through the origin with a negative gradient.

The gradient is:

−ω²

This provides another way to identify SHM experimentally.

Using the Gradient to Find Angular Frequency

Suppose an acceleration-displacement graph has gradient:

−25 s⁻²

Since:

Gradient = −ω²

then:

ω² = 25

Therefore:

ω = 5 rad/s

The period is:

T = 2π/ω

T = 2π/5

T ≈ 1.26 s

Energy and the SHM Graphs

The graphical relationships also help us understand energy.

At maximum displacement:

v = 0

Therefore:

Kinetic energy = minimum

and potential energy is maximum.

At equilibrium:

|v| = maximum

Therefore:

Kinetic energy = maximum

and potential energy is minimum.

In an ideal oscillator:

Total mechanical energy remains constant.

Recognizing SHM from Graphs

A sinusoidal displacement-time graph is strong evidence of oscillatory motion.

However, the defining condition for SHM is:

a ∝ −x

Therefore, an acceleration-displacement graph that forms a straight line through the origin with negative gradient is particularly useful evidence of SHM.

The negative gradient demonstrates that:

• Acceleration is proportional to displacement.
• Acceleration acts opposite to displacement.
• The acceleration therefore acts toward equilibrium.

Common Mistakes

Mistake: Measuring Peak-to-Peak Distance as Amplitude

Amplitude is measured:

equilibrium → maximum displacement

not:

minimum → maximum

Peak-to-peak displacement is 2A.

Mistake: Measuring Half a Cycle as the Period

The period is the time required to return to the same position moving in the same direction.

Maximum positive displacement to maximum negative displacement is only:

T/2

Mistake: Thinking Velocity Is Greatest at Maximum Displacement

At maximum displacement:

v = 0

Velocity magnitude is greatest at equilibrium.

Mistake: Thinking Acceleration Is Greatest at Equilibrium

At equilibrium:

a = 0

Acceleration magnitude is greatest at maximum displacement.

Mistake: Thinking Velocity and Acceleration Always Point Together

They do not.

An oscillator can be moving toward equilibrium while accelerating in the same direction, or moving away from equilibrium while acceleration opposes its motion.

Mistake: Forgetting the Negative Sign

In:

a = −ω²x

the negative sign is essential.

It shows that acceleration acts opposite to displacement.

Mistake: Confusing Frequency and Angular Frequency

Frequency:

f, measured in Hz.

Angular frequency:

ω, measured in rad/s.

They are related by:

ω = 2πf

Key Terms

Simple harmonic motion (SHM) – oscillation in which acceleration is proportional and opposite to displacement.

Displacement – position relative to equilibrium.

Amplitude – maximum displacement from equilibrium.

Period – time for one complete oscillation.

Frequency – number of complete oscillations per second.

Angular frequency – rate of change of phase, measured in rad/s.

Phase – position within an oscillation cycle.

Phase difference – difference between the positions of two oscillations within their cycles.

Equilibrium position – central position where displacement and restoring acceleration are zero.

Sinusoidal – having the mathematical form of a sine or cosine curve.

Check Your Understanding

1. What does the amplitude of a displacement-time graph represent?

2. A graph varies between +8 cm and −8 cm. What is its amplitude?

3. Successive maxima occur at 2 s and 7 s. Calculate the period.

4. Calculate the frequency for the graph in Question 3.

5. Where in an SHM cycle is speed greatest?

6. Where is acceleration magnitude greatest?

7. What is the velocity when x = +A?

8. What is the acceleration when x = 0?

9. Explain why displacement and acceleration are 180° out of phase.

10. An oscillator has a period of 0.50 s. Calculate its frequency.

11. Calculate its angular frequency.

12. An oscillator has A = 0.080 m and ω = 6.0 rad/s. Calculate its maximum speed.

13. Calculate its maximum acceleration.

14. An oscillator follows x = 0.20 cos(5t). Identify its amplitude and angular frequency.

15. Explain how an acceleration-displacement graph can demonstrate that an object is undergoing SHM.

Key Takeaways

• SHM can be represented using displacement, velocity, and acceleration graphs.
• Displacement varies sinusoidally with time.
• Amplitude is the maximum displacement from equilibrium.
• Period is the time for one complete oscillation.
• Frequency is the reciprocal of period.
• Angular frequency is related to frequency by ω = 2πf.
• Velocity is the gradient of the displacement-time graph.
• Acceleration is the gradient of the velocity-time graph.
• Speed is maximum at equilibrium.
• Velocity is zero at maximum displacement.
• Acceleration is zero at equilibrium.
• Acceleration magnitude is maximum at maximum displacement.
• Displacement and velocity differ in phase by one-quarter of a cycle.
• Displacement and acceleration are 180° out of phase.
• SHM satisfies the relationship a = −ω²x.
• Maximum speed is Aω.
• Maximum acceleration is Aω².
• An acceleration-displacement graph for SHM is a straight line with negative gradient.
• Mathematical models allow the position, velocity, acceleration, period, and frequency of an oscillator to be calculated.

Did You Know?

SHM graphs are closely related to circular motion. Imagine watching a point moving at constant speed around a circle, but looking only at its horizontal position. The point's horizontal displacement moves back and forth in exactly the same sinusoidal pattern as ideal simple harmonic motion. This connection explains why sine, cosine, angular frequency, and phase appear so naturally in the mathematics of oscillations.

5. Energy in Oscillatory Systems

Energy in Oscillatory Systems

An oscillating system continually transfers energy between different energy stores. In an ideal system, the total mechanical energy remains constant even though the amount stored as kinetic or potential energy changes throughout each oscillation.

Two important examples are:

• A mass attached to a spring
• A simple pendulum

Understanding these energy transfers helps explain why oscillating objects speed up, slow down, stop momentarily, and reverse direction.

6

Mechanical Energy in Oscillations

The mechanical energy of an oscillating system is the combination of its kinetic and potential energy.

During ideal oscillation, energy continuously transfers between these stores.

The important idea is:

Potential energy decreases → Kinetic energy increases

and later:

Kinetic energy decreases → Potential energy increases

In an ideal system, no mechanical energy is lost.

Energy and Position

Consider a mass oscillating horizontally on a spring.

There are three particularly important positions:

• Maximum extension
• Equilibrium
• Maximum compression

5

The energy distribution is different at each position.

Maximum Extension

At maximum extension:

x = +A

where A is the amplitude.

The mass momentarily stops before changing direction.

Therefore:

v = 0

and:

Kinetic energy = 0

The spring is stretched by its maximum amount.

Therefore:

Elastic potential energy = maximum

Moving Toward Equilibrium

As the spring pulls the mass toward equilibrium:

• Displacement decreases.
• Elastic potential energy decreases.
• Speed increases.
• Kinetic energy increases.

Energy is transferred from the spring's elastic potential energy store to the mass's kinetic energy store.

At Equilibrium

At equilibrium:

x = 0

The spring is at its equilibrium position.

The mass has its greatest speed.

Therefore:

Kinetic energy = maximum

and, for a horizontal spring measured relative to equilibrium:

Elastic potential energy = minimum

The mass does not stop at equilibrium because it has kinetic energy.

It continues past equilibrium and begins compressing the spring.

Maximum Compression

As the spring compresses:

• Speed decreases.
• Kinetic energy decreases.
• Elastic potential energy increases.

At maximum compression:

x = −A

and:

v = 0

Therefore:

Kinetic energy = 0

and:

Elastic potential energy = maximum

The mass then reverses direction and the process repeats.

Energy During One Complete Oscillation

Starting at maximum extension:

Maximum extension

Elastic PE = maximum
KE = 0

↓

Moving toward equilibrium

Elastic PE decreases
KE increases

↓

Equilibrium

Elastic PE = minimum
KE = maximum

↓

Moving toward maximum compression

Elastic PE increases
KE decreases

↓

Maximum compression

Elastic PE = maximum
KE = 0

The process then occurs in reverse.

4

Elastic Potential Energy

The energy stored in a stretched or compressed spring is:

Elastic PE = ½kx²

where:

k = spring constant in N/m

x = displacement from equilibrium in m

Elastic potential energy depends on the square of displacement.

This means that both:

x = +A

and:

x = −A

produce the same maximum elastic potential energy.

Worked Example — Energy Stored in a Spring

A spring has:

k = 200 N/m

and is stretched:

x = 0.10 m

Calculate its elastic potential energy.

Elastic PE = ½kx²

Elastic PE = ½(200)(0.10²)

Elastic PE = 100(0.010)

Elastic PE = 1.0 J

Therefore:

Elastic potential energy = 1.0 J

Maximum Energy in a Spring Oscillator

At maximum displacement:

x = A

Therefore, the total mechanical energy of an ideal horizontal spring-mass oscillator is:

Total energy = ½kA²

This energy remains constant throughout the oscillation.

At maximum displacement, it is all elastic potential energy.

At equilibrium, it is all kinetic energy.

Worked Example — Total Mechanical Energy

A spring has:

k = 80 N/m

and the oscillation amplitude is:

A = 0.25 m

Calculate the total mechanical energy.

E = ½kA²

E = ½(80)(0.25²)

E = 40(0.0625)

E = 2.5 J

Therefore:

Total mechanical energy = 2.5 J

At every point in the ideal oscillation, the total mechanical energy remains:

2.5 J

Kinetic Energy

The kinetic energy of the moving mass is:

KE = ½mv²

where:

m = mass in kg

v = velocity in m/s

At equilibrium, speed is maximum.

Therefore:

KE is maximum at equilibrium.

At maximum displacement:

v = 0

so:

KE = 0

Worked Example — Maximum Speed from Energy

A:

0.50 kg

mass oscillates on a spring.

The total mechanical energy is:

4.0 J

At equilibrium, all of this energy is kinetic.

Therefore:

KE = ½mv²

4.0 = ½(0.50)v²

4.0 = 0.25v²

v² = 16

v = 4.0 m/s

Therefore:

Maximum speed = 4.0 m/s

Energy at an Intermediate Position

Between maximum displacement and equilibrium, the system has both:

Elastic potential energy

and:

Kinetic energy

For an ideal horizontal spring:

Total energy = Elastic PE + KE

Therefore:

½kA² = ½kx² + ½mv²

This equation can be used to calculate the speed at any position.

5

Worked Example — Energy at a Particular Position

A spring has:

k = 100 N/m

and amplitude:

A = 0.20 m

Calculate the total mechanical energy.

E = ½kA²

E = ½(100)(0.20²)

E = 2.0 J

Now suppose the mass is at:

x = 0.10 m

Elastic potential energy is:

Elastic PE = ½kx²

Elastic PE = ½(100)(0.10²)

Elastic PE = 0.50 J

Therefore:

KE = Total energy − Elastic PE

KE = 2.0 − 0.50

KE = 1.50 J

At this position:

Elastic PE = 0.50 J

KE = 1.50 J

Total = 2.0 J

Energy Graphs for SHM

Energy can also be plotted against displacement.

Elastic potential energy is:

PE = ½kx²

so its graph is a curved parabola.

Kinetic energy is:

KE = Etotal − ½kx²

so it is greatest at:

x = 0

and falls to zero at:

x = ±A

5

The total mechanical energy appears as a horizontal line because it remains constant in an ideal system.

An Important Energy Pattern

At:

x = −A

PE = maximum
KE = 0

At:

x = 0

PE = minimum
KE = maximum

At:

x = +A

PE = maximum
KE = 0

Notice that energy depends on x², so the energy distribution is the same at +x and −x.

Where Are KE and PE Equal?

For a spring oscillator:

Total energy = ½kA²

Potential energy at displacement x is:

PE = ½kx²

KE and PE are equal when each contains half the total energy.

Therefore:

½kx² = ¼kA²

This gives:

x² = A²/2

and therefore:

|x| = A/√2

So kinetic and potential energy are equal when the oscillator is approximately:

0.707A

from equilibrium.

Worked Example — Equal Energy Position

Suppose:

A = 0.20 m

Then:

|x| = A/√2

|x| = 0.20/1.414

|x| ≈ 0.141 m

Therefore, kinetic and elastic potential energy are equal at approximately:

x = +0.141 m

and:

x = −0.141 m

Energy in a Pendulum

Energy transformations also occur in a simple pendulum.

The important energy stores are:

• Gravitational potential energy
• Kinetic energy

4

At the highest points:

GPE = maximum

KE = minimum

At the lowest point:

GPE = minimum

KE = maximum

Pendulum at Maximum Displacement

At either extreme of the swing:

v = 0

Therefore:

KE = 0

The bob is at its greatest height relative to the lowest point.

Therefore:

GPE = maximum

Gravity then accelerates the bob toward equilibrium.

Pendulum Moving Downward

As the bob moves toward its lowest position:

Height decreases

so:

GPE decreases

At the same time:

Speed increases

so:

KE increases

Therefore:

Gravitational PE → Kinetic energy

Pendulum at Equilibrium

At the lowest point:

Speed = maximum

Therefore:

KE = maximum

If the lowest position is chosen as the zero-height reference:

GPE = minimum

The bob continues through equilibrium because of its kinetic energy.

5

Pendulum Moving Upward

As the bob climbs the opposite side:

Speed decreases

so:

KE decreases

At the same time:

Height increases

so:

GPE increases

Therefore:

Kinetic energy → Gravitational PE

At maximum height, the bob momentarily stops and reverses direction.

Gravitational Potential Energy

Near Earth's surface:

GPE = mgh

where:

m = mass in kg

g = gravitational field strength in m/s²

h = height above the chosen reference level in m

For pendulum problems, the lowest point is often chosen as:

h = 0

---

Worked Example — Pendulum Energy

A:

0.50 kg

pendulum bob rises:

0.20 m

above its lowest position.

Use:

g = 9.8 m/s²

Calculate its gravitational potential energy.

GPE = mgh

GPE = (0.50)(9.8)(0.20)

GPE = 0.98 J

If the pendulum is released from rest and energy losses are negligible, this becomes its maximum kinetic energy at the lowest point.

Therefore:

Maximum KE = 0.98 J

---

Finding Pendulum Speed from Energy

At the lowest point:

KE = initial GPE

Therefore:

½mv² = mgh

The mass appears on both sides and cancels.

This gives:

v = √(2gh)

Notice that the speed does not depend on the mass of the bob.

---

Worked Example — Pendulum Speed

A pendulum bob is released from a position:

0.45 m

above its lowest point.

Calculate its maximum speed.

Use:

g = 9.8 m/s²

v = √(2gh)

v = √(2 × 9.8 × 0.45)

v = √8.82

v ≈ 2.97 m/s

Therefore:

Maximum speed ≈ 3.0 m/s

---

Spring and Pendulum Energy Compared

Both systems show the same general pattern.

Spring

At extremes:

Elastic PE = maximum

KE = zero

At equilibrium:

Elastic PE = minimum

KE = maximum

Pendulum

At extremes:

Gravitational PE = maximum

KE = zero

At equilibrium:

Gravitational PE = minimum

KE = maximum

5

The type of potential energy differs, but the overall pattern is very similar.

---

Conservation of Mechanical Energy

In an ideal oscillating system:

Total mechanical energy remains constant.

Energy does not disappear when kinetic energy decreases.

Instead, it is transferred into potential energy.

Similarly, when potential energy decreases, energy is transferred into kinetic energy.

This is an application of the law of conservation of energy.

---

Ideal vs Real Oscillations

Real oscillators eventually lose amplitude.

A real pendulum gradually stops swinging.

A real spring-mass system gradually stops oscillating.

Does this mean energy has been destroyed?

No.

5

Mechanical energy is transferred to other energy stores.

---

Energy Dissipation

Energy may be transferred to the surroundings through:

• Air resistance
• Friction
• Heating
• Sound
• Internal deformation

As mechanical energy decreases:

Amplitude decreases

This is called:

damping

The total energy of the complete system is still conserved, but the mechanical energy available for oscillation decreases.

---

Energy and Amplitude

For a spring oscillator:

E = ½kA²

Therefore:

E ∝ A²

This has an important consequence.

If amplitude doubles:

A → 2A

then:

E → 4E

So doubling the amplitude requires four times as much energy.

---

Worked Example — Changing Amplitude

A spring oscillator has total energy:

3.0 J

at amplitude:

A

If the amplitude is doubled:

New energy = 4 × 3.0

New energy = 12 J

Therefore:

Doubling amplitude quadruples the mechanical energy.

---

Energy and Time

Kinetic and potential energy also vary periodically with time.

However, there is an important difference between the displacement and energy graphs.

If:

x = A cos(ωt)

then potential energy depends on:

x²

Therefore, the energy repeats twice during each complete displacement cycle.

5

This means the energy graphs have a period of:

T/2

even though the oscillator itself has period T.

---

Analyzing an Oscillator from Energy

Suppose an oscillator is moving away from equilibrium.

As it moves outward:

• Displacement magnitude increases.
• Speed decreases.
• Kinetic energy decreases.
• Potential energy increases.

Suppose instead it is moving toward equilibrium.

Then:

• Displacement magnitude decreases.
• Speed increases.
• Kinetic energy increases.
• Potential energy decreases.

This pattern allows us to infer motion from energy changes.

---

Worked Example — Analyzing Motion

A spring-mass oscillator has:

Total energy = 10 J

At one instant:

Elastic PE = 7 J

Therefore:

KE = 10 − 7

KE = 3 J

A moment later:

Elastic PE = 5 J

The potential energy has decreased.

Therefore, the mass must be moving:

toward equilibrium

Its kinetic energy has increased to:

5 J

and its speed has increased.

---

Energy Bar Diagrams

Energy transformations can also be represented using energy bar diagrams.

At maximum displacement:

PE ██████████

KE

At an intermediate position:

PE ██████

KE ████

At equilibrium:

PE

KE ██████████

The total height represented by PE + KE remains constant in an ideal system.

These diagrams are useful because they emphasize energy conservation rather than just motion.

---

Real-World Oscillatory Systems

Oscillatory energy transformations occur in many systems.

Examples include:

• Playground swings
• Vehicle suspension
• Diving boards
• Trampolines
• Vibrating guitar strings
• Pendulum clocks
• Buildings during earthquakes
• Mechanical vibration systems
• Atomic and molecular vibrations

5

The exact energy stores may differ, but the principle of energy conservation remains fundamental.

---

Shock Absorbers and Energy

Vehicle suspension provides an important example where we want oscillatory mechanical energy to decrease.

Springs allow the suspension to oscillate after the vehicle encounters a bump.

Without damping, the vehicle could continue bouncing.

Shock absorbers transfer mechanical energy mainly into thermal energy.

6

This reduces the amplitude of the oscillations and helps keep the vehicle stable.

---

Common Mistakes

Mistake: Thinking Energy Disappears at the Turning Points

At maximum displacement:

KE = 0

but potential energy is maximum.

The energy has not disappeared.

Mistake: Thinking Maximum Speed Occurs at Maximum Displacement

At maximum displacement:

v = 0

Maximum speed occurs at equilibrium.

Mistake: Thinking Potential Energy Is Always Zero at Equilibrium

This depends on the chosen reference.

For many SHM calculations, potential energy is defined to be minimum or zero at equilibrium for convenience.

Mistake: Forgetting to Square the Displacement

Spring potential energy is:

½kx²

not:

½kx

Mistake: Using the Full Peak-to-Peak Distance as the Amplitude

Amplitude is the distance:

equilibrium → extreme

not:

extreme → opposite extreme

Mistake: Saying Mechanical Energy Is Conserved in Every Real Oscillator

Mechanical energy is conserved only in an ideal system with negligible dissipative effects.

In real systems, some mechanical energy is transferred into thermal energy, sound, and other stores.

Mistake: Thinking Energy Has the Same Period as Displacement

Because energy depends on displacement squared, the PE and KE patterns repeat every:

T/2

---

Key Terms

Oscillatory system – system that repeatedly moves around an equilibrium position.

Mechanical energy – energy associated with motion and position.

Kinetic energy – energy associated with motion.

Elastic potential energy – energy stored when an elastic object is stretched or compressed.

Gravitational potential energy – energy associated with position in a gravitational field.

Equilibrium – central position around which oscillation occurs.

Amplitude – maximum displacement from equilibrium.

Turning point – position where the oscillator momentarily stops and reverses direction.

Damping – reduction in oscillation amplitude due to energy transfer to the surroundings.

Conservation of energy – principle that energy cannot be created or destroyed, only transferred or transformed.

---

Check Your Understanding

1. Where is the kinetic energy of a spring oscillator greatest?

2. Where is its elastic potential energy greatest?

3. What is the kinetic energy at maximum displacement?

4. Explain why a mass does not stop when it reaches equilibrium.

5. A spring has k = 120 N/m and is stretched 0.15 m. Calculate its elastic potential energy.

6. A spring oscillator has k = 200 N/m and amplitude 0.10 m. Calculate its total mechanical energy.

7. The system in Question 6 has 0.40 J of elastic potential energy at one instant. Calculate its kinetic energy.

8. A 0.25 kg mass has 2.0 J of kinetic energy. Calculate its speed.

9. Describe the energy transformations as a spring oscillator moves from maximum extension to equilibrium.

10. Describe the energy transformations as a pendulum moves from its lowest point toward maximum height.

11. A 0.60 kg pendulum bob is raised 0.30 m above its lowest point. Calculate its increase in gravitational potential energy using g = 9.8 m/s².

12. Explain why an ideal pendulum reaches approximately the same height on the opposite side.

13. If the amplitude of a spring oscillator doubles, by what factor does its total energy change?

14. Why does the amplitude of a real oscillator gradually decrease?

15. Explain how conservation of energy applies to a damped oscillator.

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Key Takeaways

• Energy continually changes form during oscillatory motion.
• In an ideal oscillator, total mechanical energy remains constant.
• At maximum displacement, potential energy is maximum and kinetic energy is zero.
• At equilibrium, kinetic energy is maximum and potential energy is minimum.
• A spring oscillator transfers energy between elastic potential energy and kinetic energy.
• A pendulum transfers energy between gravitational potential energy and kinetic energy.
• Spring potential energy is proportional to the square of displacement.
• Total energy of an ideal spring oscillator depends on the square of its amplitude.
• Doubling amplitude produces four times the mechanical energy.
• Energy can be used to calculate an oscillator's speed at different positions.
• At equal positive and negative displacements, the energy distribution is the same.
• Kinetic and potential energy are equal at approximately 0.707A from equilibrium in an ideal spring oscillator.
• Real oscillators experience damping.
• During damping, mechanical energy is transferred to thermal energy, sound, and other stores.
• Energy is still conserved when the entire system and surroundings are considered.
• Energy graphs repeat twice during each complete SHM displacement cycle.

Did You Know?

The energy of an ideal oscillator has an interesting rhythm. During one complete oscillation, the object reaches maximum kinetic energy twice and maximum potential energy twice. This means the kinetic and potential energy patterns repeat at twice the frequency of the displacement. The object might complete 2 oscillations each second, for example, while its energy distribution cycles 4 times each second.