Mathematics - Further (9231) | Hooke's law

Further Mechanics (Paper 3) – Chapter 3.4: Hooke's Law

Hello Future Further Mathematician!
Welcome to Hooke's Law! This chapter takes the foundational concepts of force and motion you learned in Mechanics and applies them to objects that can stretch and compress—like springs and elastic strings. This introduces the concept of elastic potential energy, which is essential for solving complex dynamics problems using energy conservation. Don't worry if the formulas look new; they are simple extensions of concepts you already know!

The ability to model systems using Hooke’s Law is crucial for many real-world applications, from designing suspension systems in cars to understanding the physics of molecular bonds.

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1. Understanding Hooke's Law: The Core Relationship

What is Hooke's Law?

Hooke's law is a simple model that describes the behavior of elastic materials (like springs or elastic strings) when they are stretched or compressed.

It states that the force required to stretch an elastic object is directly proportional to the extension (or compression) of that object, provided the elastic limit is not exceeded.

Key Concept: Extension (\(x\))
The extension \(x\) is always the difference between the current length \(L\) and the natural length \(l\).

$$x = |L - l|$$

The Formula for Tension (\(T\))

When we turn the proportionality into an equation, we get the standard form of Hooke's Law for calculating the magnitude of the force (usually tension, \(T\)) in the elastic string or spring:

Hooke's Law Formula:
$$T = \frac{\lambda x}{l}$$

Where:

• \(T\) is the magnitude of the Tension (force) in the string or spring (in Newtons, N).
• \(\lambda\) (lambda) is the Modulus of Elasticity (in Newtons, N). This is a constant determined by the material.
• \(x\) is the extension or compression from the natural length (in meters, m).
• \(l\) is the natural length (or unstretched length) of the string or spring (in meters, m).

The Modulus of Elasticity (\(\lambda\))

The term Modulus of Elasticity (\(\lambda\)) is essential. It is a measure of how "stiff" the material is:

• A large \(\lambda\) means the string or spring is very stiff, requiring a large force \(T\) to achieve a small extension \(x\).
• A small \(\lambda\) means the object is highly elastic and stretches easily.

Did you know? If you set the extension \(x\) equal to the natural length \(l\), the formula simplifies to \(T = \lambda\). This gives a physical meaning to \(\lambda\): it is the force required to double the length of the string!

Important Distinction: Strings vs. Springs

In Further Mechanics, we must be careful about whether the object is a string or a spring:

• Elastic String: Can only exert a force (tension, \(T\)) when it is extended (\(x > 0\)). It goes slack if compressed (\(x < 0\)) or at its natural length (\(x = 0\)), meaning \(T=0\).
• Elastic Spring: Can exert a force (tension/thrust) when extended OR compressed. You must determine the direction of the force based on the context (e.g., compression causes a pushing force).

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2. Elastic Potential Energy (EPE)

When you stretch a spring, you are doing work against the tension. This work is stored as Elastic Potential Energy (\(E\)). This is similar to how work done against gravity is stored as Gravitational Potential Energy.

The Formula for EPE

Since the force \(T\) is not constant (it increases as \(x\) increases), we must integrate the force with respect to the extension to find the work done, \(E = \int_0^x T \, dx\).

Using the syllabus requirement, we focus on recalling and using the final result:

Elastic Potential Energy Formula:
$$E = \frac{\lambda x^2}{2l}$$

Where:

• \(E\) is the Elastic Potential Energy stored (in Joules, J).
• \(\lambda\), \(x\), and \(l\) are the Modulus of Elasticity, extension/compression, and natural length, respectively, as defined above.

Memory Trick: Notice the relationship between the formulas for Force \(T\) and Energy \(E\). \(T\) is proportional to \(x\) (power 1), while \(E\) is proportional to \(x^2\) (power 2). The rest of the terms \(\left(\frac{\lambda}{l}\right)\) are identical coefficients.

Common Mistake to Avoid!

Students often forget the factor of 2 in the denominator of the EPE formula. Remember: EPE is the area under the triangular Force-Extension graph, which is \(\frac{1}{2} \times \text{Base} \times \text{Height}\).

$$E = \frac{1}{2} \times x \times T \quad \implies \quad E = \frac{1}{2} \times x \times \left(\frac{\lambda x}{l}\right) = \frac{\lambda x^2}{2l}$$

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3. Solving Problems Using Work and Energy

The syllabus requires you to solve problems where considerations of work and energy are needed. This means applying the principle of Conservation of Mechanical Energy.

For a particle attached to an elastic string or spring moving under gravity, the total mechanical energy is conserved if there are no external resistive forces (like air resistance).

Total Initial Energy = Total Final Energy

$$\text{KE}_1 + \text{GPE}_1 + \text{EPE}_1 = \text{KE}_2 + \text{GPE}_2 + \text{EPE}_2$$

Where:

• Kinetic Energy (KE): \(\frac{1}{2} m v^2\)
• Gravitational Potential Energy (GPE): \(mgh\) (Remember to define a zero reference level for height \(h\)).
• Elastic Potential Energy (EPE): \(\frac{\lambda x^2}{2l}\) (Remember, \(x\) is zero if the string/spring is at or below its natural length and slack/unstretched).

Step-by-Step Strategy for Energy Problems

Scenario: A particle is released from rest and moves vertically attached to an elastic string.

1. Identify Key Points: Define your initial state (A) and the final state you are interested in (B, usually maximum extension/maximum speed).
2. Establish Reference Levels: Choose a reference height (\(h=0\)) for GPE. A good choice is usually the starting point or the lowest point of the motion.
3. Determine Initial Energy:
   • \(KE_A\): Often 0 if starting from rest.
   • \(GPE_A\): \(mgh_A\).
   • \(EPE_A\): Often 0 if starting from natural length.
4. Determine Final Energy:
   • \(KE_B\): Often 0 if finding the maximum extension (where velocity \(v=0\)).
   • \(GPE_B\): \(mgh_B\).
   • \(EPE_B\): \(\frac{\lambda x^2}{2l}\). (Ensure \(x\) accounts for the total stretch from the natural length).
5. Apply Conservation: Set \(\text{Total Energy}_A = \text{Total Energy}_B\) and solve for the unknown variable (e.g., maximum extension \(x\)).

Application Examples from the Syllabus

Problems involving Hooke's Law often require you to combine GPE, KE, and EPE. You might encounter scenarios like:

1. Vertical Motion:
A mass attached to a spring oscillating up and down.
Forces involved: Gravity (\(mg\)) and Tension (\(T\)).
Energy considerations: Changes in GPE must be balanced by changes in EPE and KE.

2. Inclined Plane:
A particle sliding down a smooth inclined plane attached to a string fixed at the top.
Forces involved: Gravity (\(mg\)), Normal Reaction (\(R\)), and Tension (\(T\)).
Energy considerations: You must use the height component parallel to the plane when calculating GPE changes.

3. Conical Pendulum (Elastic String):
A particle attached to an elastic string rotating in a horizontal circle.
This requires a balance of forces (centripetal force) using Newton's Second Law (\(F=ma\)) AND Hooke's Law.

$$T \cos \theta = mg \quad \text{(Vertical Equilibrium)}$$
$$T \sin \theta = m r \omega^2 \quad \text{or} \quad m \frac{v^2}{r} \quad \text{(Horizontal Centripetal Force)}$$
$$T = \frac{\lambda x}{l}$$
Solving these requires substituting \(T\) from Hooke's law into the force equations. Energy considerations are usually not needed unless the radius of the circle is changing dynamically.