Springs (VCE SSCE Physics): Revision Notes

Springs

Introduction: Elasticity in nature

Springs are found throughout nature and technology. One remarkable example is in animal tendons, which are connective tissues made mainly of collagen fibres that attach muscles to bones.

When animals move at high speeds, they use elastic energy storage in tendons and ligaments to conserve energy. When a runner's foot strikes the ground, their Achilles tendon stretches and stores elastic potential energy. As the runner pushes off, this stored energy is released, helping propel them forward. This energy recovery system is remarkably efficient, typically returning 5-13% of the mechanical energy used in movement.

What is elasticity?

Elastic materials

When you compress, extend or twist certain materials, they store energy by changing shape. If these materials then return to their original shape when you release them, we call them elastic materials. An elastic material is one that can store elastic potential energy and return to its original shape once the energy is released.

Many solid materials demonstrate elastic behaviour, including rubber, neoprene and steel. These materials are used extensively in engineering and everyday applications because of their ability to store and release energy predictably.

In contrast, plastic materials are those that deform permanently when stressed, even slightly. Mud, clay and plasticine are examples of plastic materials that do not return to their original shape after being deformed.

Elastic limit

Every elastic material has an elastic limit. This is the maximum stress or deformation that the material can withstand without causing permanent deformation. If you stretch or compress an elastic material beyond its elastic limit, it will not return to its original shape and may be permanently damaged.

Hooke's law

Springs have been used in machines and devices for over 300 years. They appear in toys, pens, electronics, mattresses, medical devices, automatic watches, mining equipment and vehicles. We can use springs so effectively because we understand their predictable behaviour.

The relationship between force and extension

In 1676, British physicist Robert Hooke discovered a fundamental property of springs. Hooke's law states that in an ideal spring, the force applied to the spring is directly proportional to its extension or compression.

An ideal spring is a theoretical concept - a spring that is frictionless, has no mass, and perfectly obeys Hooke's law. Whilst no real spring is truly ideal, many springs behave very close to this ideal model within their working range.

To understand Hooke's law, imagine a spring resting horizontally with no force applied. The spring has zero extension. Now apply different forces:

• When you apply 50 N of force, the spring extends by 0.2 m
• When you apply 100 N of force, the spring extends by 0.4 m
• When you apply 150 N of force, the spring extends by 0.6 m

Notice the pattern: doubling the force doubles the extension. This linear relationship is the essence of Hooke's law.

Understanding force-extension graphs

When we plot force against extension for an ideal spring, we get a straight line passing through the origin. This straight line demonstrates that force is directly proportional to extension - the key principle of Hooke's law.

The spring constant

The gradient (slope) of a force-extension graph is called the spring constant, represented by the symbol $k$. The spring constant tells us about the stiffness of a spring.

Since we calculate the gradient by dividing force (in newtons) by extension (in metres), the unit for spring constant is N m$^{-1}$ (newtons per metre).

For the graph shown above, we can calculate the spring constant:

$k = \frac{\text{rise}}{\text{run}} = \frac{200 - 0}{0.8 - 0} = 250 \text{ N m}^{-1}$

What does spring constant tell us?

The spring constant gives us a measure of how stiff a spring is:

• A higher spring constant means the spring is stiffer - it requires more force to produce the same extension
• A lower spring constant means the spring is less stiff - it requires less force to produce the same extension

For example, a spring with $k = 500$ N m$^{-1}$ is twice as stiff as a spring with $k = 250$ N m$^{-1}$. To extend both springs by the same amount, you would need twice as much force for the stiffer spring.

The graph above compares two springs. The steeper blue line represents the stiffer spring ($k = 500$ N m$^{-1}$), whilst the gentler red line represents the less stiff spring ($k = 250$ N m$^{-1}$).

Mathematical formula for Hooke's law

We can express Hooke's law mathematically:

Formula: $F = -kx$

Where:

• $F$ = Force applied to the spring (N)
• $k$ = Spring constant (N m$^{-1}$)
• $x$ = Extension or compression of the spring (m)

Elastic potential energy

What is elastic potential energy?

When you compress, extend or twist an elastic material, you store energy in it. This stored energy is called elastic potential energy (sometimes called spring potential energy). When you release the elastic material, it returns to its original shape and the stored energy is released.

Deriving the elastic potential energy formula

Understanding where the elastic potential energy formula comes from helps you remember and use it correctly. Let's think about a force-compression graph:

The y-axis shows force (measured in newtons), and the x-axis shows distance (compression or extension, measured in metres). This type of graph is actually a special case of a force-distance graph.

For a linear force-extension graph (which we get for an ideal spring obeying Hooke's law), the area under the graph forms a triangle. The area of a triangle is:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times F$

where $x$ is the extension/compression and $F$ is the force.

From Hooke's law, we know that $F = kx$. Substituting this into our area equation:

$E_s = \frac{1}{2} \times x \times (kx)$

$E_s = \frac{1}{2}kx^2$

Formula: $E_s = \frac{1}{2}kx^2$

Where:

• $E_s$ = Elastic potential energy (J)
• $k$ = Spring constant (N m$^{-1}$)
• $x$ = Extension or compression of the spring (m)

Energy required to extend springs

An important concept to understand is that extending a spring requires more energy the more it's already stretched. To extend a spring by 1 cm when it's at rest requires less energy than extending the same spring by 1 cm when it's already stretched.

The graph above shows force versus extension for a spring with $k = 200$ N m$^{-1}$. The shaded regions represent the energy needed to extend the spring by 1 cm at different points. Notice how the shaded area (and therefore the energy required) increases as the spring becomes more extended.

Change in elastic potential energy

When calculating the change in elastic potential energy as a spring moves from one extension to another, you can either:

1. Find the area under the graph between the two positions, or
2. Calculate the difference: $\Delta E_s = \frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2$

where $x_f$ is the final extension and $x_i$ is the initial extension.

Energy transformations in horizontal spring systems

The equilibrium position

Imagine an ideal spring lying horizontally on a frictionless surface. One end is fixed to a wall, and the other end is attached to an object that can move freely. If you pull the object back and release it, the system will oscillate.

The equilibrium position is where the object has no net force acting on it. For a horizontal spring, this occurs when the spring is at its natural length (neither compressed nor extended). At this position, there is no spring force, and any other forces (like friction, if present) are balanced.

Energy during oscillation

As the spring oscillates, energy continuously transforms between kinetic energy and elastic potential energy. The total energy remains constant (assuming no friction or air resistance).

At any point during oscillation:

$E_T = E_k + E_s$

or in full:

$E_T = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

Where:

• $E_T$ = Total energy (J)
• $m$ = Mass of the object (kg)
• $v$ = Velocity of the object (m $\cdot$ s$^{-1}$)
• $k$ = Spring constant (N m$^{-1}$)
• $x$ = Extension or compression from the natural length (m)

The oscillation cycle

Let's follow one complete oscillation, starting when the spring is fully extended:

Position 1 (Maximum extension):

• Spring is fully extended
• Elastic potential energy is maximum
• Kinetic energy is zero (object momentarily at rest)
• Spring force pulls object back towards equilibrium

Moving from Position 1 to Position 2:

• Spring begins to contract
• Elastic potential energy decreases
• Kinetic energy increases
• Object accelerates towards equilibrium

Position 2 (Equilibrium):

• Spring at natural length
• Elastic potential energy is zero
• Kinetic energy is maximum (fastest speed)
• Object continues moving due to momentum

Moving from Position 2 to Position 3:

• Spring begins to compress
• Elastic potential energy increases
• Kinetic energy decreases
• Object slows down

Position 3 (Maximum compression):

• Spring is fully compressed
• Elastic potential energy is maximum
• Kinetic energy is zero (object momentarily at rest)
• Spring force pushes object back towards equilibrium

Returning to Position 1: The cycle then repeats in reverse, with the object accelerating back through the equilibrium position and returning to maximum extension.

Energy transformations in vertical spring systems

Equilibrium in a vertical spring

For a vertical spring with one end fixed to a ceiling and a mass hanging from the other end, the equilibrium position is different from a horizontal spring. At equilibrium, the spring must be extended enough so that the upward spring force exactly balances the downward weight force.

At equilibrium, the forces balance:

$F_{\text{spring}} = F_{\text{weight}}$

Using Hooke's law ($F = kx$) and the weight formula ($F = mg$):

$kx = mg$

Where:

• $k$ = Spring constant (N m$^{-1}$)
• $x$ = Extension of spring at equilibrium (m)
• $m$ = Mass (kg)
• $g$ = Gravitational field strength (9.8 N kg$^{-1}$ on Earth)

Energy during vertical oscillation

In a vertical spring system, three forms of energy are important:

1. Gravitational potential energy
2. Elastic potential energy
3. Kinetic energy

As the spring oscillates, energy transforms between these three forms, but the total energy remains constant.

The oscillation cycle in a vertical spring

Position 1 (Starting point, natural length): When the mass is initially dropped from the spring's natural length:

• Spring potential energy = zero (spring at natural length)
• Kinetic energy = zero (object starts from rest)
• Gravitational potential energy = maximum
• Weight force pulls object downward

Positions 1 to 2 (Moving towards equilibrium):

• Object accelerates downward due to gravity
• Spring potential energy increases (spring extends)
• Kinetic energy increases (object speeds up)
• Gravitational potential energy decreases (object moves down)

Position 2 (Equilibrium):

• Kinetic energy is maximum (fastest speed)
• Spring force balances weight force
• Object continues moving downward due to momentum

Positions 2 to 3 (Beyond equilibrium):

• Object continues moving down but slowing
• Spring potential energy continues to increase
• Kinetic energy decreases
• Gravitational potential energy continues to decrease
• Net force is now upward (spring force > weight)

Position 3 (Lowest point):

• Kinetic energy = zero (object momentarily at rest)
• Gravitational potential energy = zero (by our chosen reference)
• Spring potential energy = maximum
• Spring force pulls object upward

Positions 3 to 4 (Moving upward):

• Spring contracts
• Spring potential energy decreases
• Kinetic energy increases
• Gravitational potential energy increases

Position 4 (Equilibrium again):

• Kinetic energy is maximum
• Object passes through equilibrium moving upward

Returning to Position 1: The object continues upward, slowing down, until it returns to the starting position, and the cycle repeats.

Mathematical relationship for vertical springs

The gravitational potential energy lost by the object transforms into elastic potential energy and kinetic energy:

$mgx = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where:

• $m$ = Mass (kg)
• $g$ = Gravitational field strength (9.8 N kg$^{-1}$)
• $x$ = Change in height = change in spring length (m)
• $k$ = Spring constant (N m$^{-1}$)
• $v$ = Velocity (m $\cdot$ s$^{-1}$)

This formula allows you to calculate velocities, positions, or spring constants in vertical spring systems.

Exam skills: How to 'explain' your answer

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