Springs (VCE SSCE Physics): Revision Notes

Springs

Introduction

When objects are bent, stretched, twisted, or compressed, work must be done on them. If the object returns to its original shape once the external force is removed, it stores energy during deformation. This stored energy is called elastic potential energy. Springs are excellent examples of objects that store elastic potential energy when compressed or stretched.

Hooke's law

Understanding spring behaviour

When you apply force to stretch or compress a spring, the amount of force needed depends on the spring's stiffness. Different springs require different forces to achieve the same extension. When we plot the applied force against the extension for various springs, we get straight-line graphs passing through the origin.

The graph shows three different springs (A, B, and C) and their force-extension relationships. Notice that:

• All lines pass through the origin (zero force means zero extension)
• The lines have different gradients (steepness)
• Steeper gradients indicate stiffer springs

The spring constant

The gradient of a force-extension graph represents the spring's stiffness, called the spring constant ($k$). This value tells us how much force is needed to stretch or compress the spring by one metre.

To calculate the gradient:

$\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{\Delta F}{\Delta x}$

The unit is newtons per metre ($\text{N} \cdot \text{m}^{-1}$).

A stiffer spring has a larger spring constant and a steeper gradient. For example, if spring A in the graph requires 8 N to extend by 0.3 m:

$k = \frac{8 - 0}{0.3 - 0} = 26.7 \text{ N} \cdot \text{m}^{-1}$

Hooke's law formula

Robert Hooke discovered the linear relationship between force and extension for ideal springs. Hooke's law states:

$F = -kx$

Where:

• $F$ = force applied to the spring (N)
• $k$ = spring constant ($\text{N} \cdot \text{m}^{-1}$)
• $x$ = compression or extension of the spring (m)

Elastic potential energy

Area under the force-extension graph

The work done on a spring equals the area under the force-extension graph. Since force varies linearly with extension for an ideal spring, this area forms a triangle.

Deriving the elastic potential energy formula

The area under the force-extension graph equals half the base multiplied by the height:

$\text{area} = \frac{1}{2}Fx$

Since $F = kx$ (from Hooke's law), we can substitute:

$\text{elastic potential energy} = \frac{1}{2}(kx)x = \frac{1}{2}kx^2$

This gives us the formula for elastic potential energy:

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

Where:

• $E_s$ = elastic potential energy (J)
• $k$ = spring constant ($\text{N} \cdot \text{m}^{-1}$)
• $x$ = compression or extension of the spring (m)

Why more energy is needed at greater extensions

According to Hooke's law, the more a spring is extended or compressed, the greater the restoring force becomes. This means progressively more energy is required to extend a spring by the same amount as the extension increases.

The graph illustrates that extending a spring from 1 m to 2 m requires much less energy than extending it from 4 m to 5 m. The energy required is shown by the shaded areas under the graph.

Change in elastic potential energy

When calculating the change in elastic potential energy, we can either find the area under the relevant section of the graph or use the formula:

$\Delta E_s = \frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2$

Where:

• $x_f$ = final extension (m)
• $x_i$ = initial extension (m)

Horizontal spring systems

Energy transformations in horizontal springs

Consider a spring attached to a wall at one end with a mass attached to the other end, resting on a frictionless surface. When the spring oscillates, elastic potential energy and kinetic energy constantly transform into each other.

The spring oscillates between three key positions:

• Point A: Maximum compression
• Point B: Equilibrium position (natural length)
• Point C: Maximum extension

The equilibrium position

The equilibrium position is the point where the net force on the oscillating mass is zero. For a horizontal spring, this occurs when the spring is at its natural length (neither compressed nor extended).

Energy changes during oscillation

The total energy remains constant throughout the oscillation (assuming no friction):

$\frac{1}{2}kx_{\text{before}}^2 + \frac{1}{2}mv_{\text{before}}^2 = \frac{1}{2}kx_{\text{after}}^2 + \frac{1}{2}mv_{\text{after}}^2$

At point A (maximum compression):

• Elastic potential energy: 100% of total energy
• Kinetic energy: 0% (momentarily stationary)
• Net force and acceleration: Maximum (directed towards B)

At point B (equilibrium position):

• Elastic potential energy: 0% (no compression or extension)
• Kinetic energy: 100% of total energy (maximum velocity)
• Net force and acceleration: Zero

At point C (maximum extension):

• Elastic potential energy: 100% of total energy
• Kinetic energy: 0% (momentarily stationary)
• Net force and acceleration: Maximum (directed towards B, opposite to point A)

Vertical spring systems

Energy transformations in vertical springs

In a vertical spring system, three forms of energy transform into each other: gravitational potential energy, elastic potential energy, and kinetic energy.

The energy conservation equation for vertical springs is:

$mg\Delta h_1 + \frac{1}{2}kx_1^2 + \frac{1}{2}mv_1^2 = mg\Delta h_2 + \frac{1}{2}kx_2^2 + \frac{1}{2}mv_2^2$

Typical vertical spring setup

A common example involves a mass hanging from a spring attached to a fixed support. When released from rest, the system oscillates between a top position (A), middle position (B), and bottom position (C).

At point A (top position):

• Spring: Unstretched (zero elastic potential energy)
• Gravitational potential energy: Maximum (if we take point C as zero)
• Kinetic energy: Zero (momentarily stationary)
• Acceleration: $9.8$ $\text{m} \cdot \text{s}^{-2}$ downwards (due to gravity only)

At point B (equilibrium position):

• Spring force balances gravitational force ($mg = kx$)
• Kinetic energy: Maximum (approximately 25% of total energy)
• Gravitational potential energy: Half of maximum
• Elastic potential energy: Increasing from zero

At point C (bottom position):

• Elastic potential energy: Maximum (100% of total energy)
• Gravitational potential energy: Zero (by definition)
• Kinetic energy: Zero (momentarily stationary)
• Acceleration: $9.8$ $\text{m} \cdot \text{s}^{-2}$ upwards (net force from spring exceeds weight)

Key insight about total energy

Worked example: Vertical spring system

Understanding force-extension graphs

Reading gradients

The gradient (steepness) of a force-extension graph represents the spring constant. To calculate it:

$\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{\Delta F}{\Delta x}$

The unit is always the y-axis unit divided by the x-axis unit. For force-extension graphs:

$\text{unit} = \frac{\text{force}}{\text{extension}} = \frac{\text{N}}{\text{m}} = \text{N} \cdot \text{m}^{-1}$

This confirms the gradient equals the spring constant: $k = \frac{F}{x}$

Calculating area under graphs

The area under a force-extension graph represents the work done on the spring, which equals the stored elastic potential energy.

For a triangular area:

$\text{area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}Fx$

The unit is found by multiplying the axis units:

$\text{unit} = F \times x = \text{N} \times \text{m} = \text{J}$

Experimental graphs

When plotting elastic potential energy against extension squared ($x^2$), we get a straight line through the origin.

The gradient of this graph is:

$\text{gradient} = \frac{E_s}{x^2} = \frac{\text{J}}{\text{m}^2}$

Since $E_s = \frac{1}{2}kx^2$, we can rearrange to find:

$k = \frac{2E_s}{x^2} = 2 \times \text{gradient}$

Real-world applications

Springs are fundamental components in many everyday applications that demonstrate the principles of elastic potential energy:

• Vehicle suspension systems: Springs absorb energy from bumps, providing smooth rides and protecting passengers
• Bungee jumping: Elastic cords store gravitational potential energy safely, converting it gradually to prevent injury
• Athletic shoes: Elastic materials in soles return energy to improve performance, as seen in modern running shoes
• Trampolines: Springs convert gravitational potential energy to kinetic energy, allowing repeated bouncing
• Mechanical watches: Springs store energy to power mechanisms over extended periods

Remember!