Welcome to Hooke's Law!
Ever bounced on a trampoline, used a hair tie, or watched a bungee jumper leap off a platform? All of these rely on materials that stretch and snap back into shape. In this chapter of AS 2 Mechanics 1, we will explore Hooke's Law, which describes the forces and energy stored in elastic strings and springs.
Don't worry if mechanics sometimes feels daunting! We will break everything down step-by-step using clear definitions, diagrams in words, and structured methods to solve any exam problem.
Did you know? Robert Hooke first published his law in 1676 as a Latin anagram: "ceiiinosssttuv". Two years later, he revealed the solution: "Ut tensio, sic vis", which translates to "As the extension, so the force"!
1. Fundamental Definitions
Before jumping into the mathematics, let's make sure we understand the key terminology used in CCEA Mechanics:
• Natural Length (\(l\)): The original, unstretched (and uncompressed) length of the string or spring when no external forces act upon it, measured in metres (\(\text{m}\)).
• Extended Length (\(L\)): The new length of the string or spring when it has been stretched, measured in metres (\(\text{m}\)).
• Extension (\(x\)): The extra distance the string or spring has stretched beyond its natural length. It is calculated as:
\(x = L - l\)
• Compression (\(x\)): For a spring only, when pushed to a length shorter than its natural length, the compression is \(x = l - L\).
• Modulus of Elasticity (\(\lambda\)): A measure of how stiff or stretchy a material is, measured in Newtons (\(\text{N}\)). A higher \(\lambda\) means a stiffer string that is harder to stretch.
• Stiffness / Spring Constant (\(k\)): An alternative way to express stiffness, where \(k = \frac{\lambda}{l}\), measured in \(\text{N m}^{-1}\).
Key Takeaway: Extension \(x\) is never the total length! It is always the difference between the stretched length and the natural length.
2. Hooke's Law Formula
Hooke's Law states that the tension \(T\) in an elastic string or spring is directly proportional to its extension \(x\) and inversely proportional to its natural length \(l\).
The Standard CCEA Formula:
\(T = \frac{\lambda x}{l}\)
Where:
• \(T\) is the tension in Newtons (\(\text{N}\))
• \(\lambda\) is the modulus of elasticity in Newtons (\(\text{N}\))
• \(x\) is the extension in metres (\(\text{m}\))
• \(l\) is the natural length in metres (\(\text{m}\))
If the question uses the spring constant \(k\), the formula can simply be written as:
\(T = kx\)
Strings vs. Springs: A Vital Distinction
• Light Elastic String: A string can only be stretched. When stretched, it experiences a pulling force called tension (\(T\)). When compressed (\(x < 0\)), a string becomes slack, meaning \(T = 0\). It cannot push!
• Light Elastic Spring: A spring can be stretched or compressed. When stretched, it pulls inward with tension. When compressed, it pushes outward with a compressive force called thrust (\(C\) or \(T\)). The formula for thrust is still \(T = \frac{\lambda x}{l}\), where \(x\) is the compression.
Memory Trick: Think of a piece of cooked spaghetti vs. a metal coil. You can pull spaghetti and it resists (tension), but if you push the ends together, it goes floppy (slack). The metal coil pushes back (thrust)!
Quick Example: Finding Tension
An elastic string has a natural length of \(0.8\text{ m}\) and a modulus of elasticity of \(40\text{ N}\). It is stretched to a length of \(1.1\text{ m}\). Find the tension in the string.
Step 1: Calculate the extension \(x\):
\(x = 1.1 - 0.8 = 0.3\text{ m}\)
Step 2: Apply Hooke's Law:
\(T = \frac{\lambda x}{l} = \frac{40 \times 0.3}{0.8} = \frac{12}{0.8} = 15\text{ N}\)
Key Takeaway: Check whether the question describes a string or a spring. If a string goes slack, tension is immediately zero!
3. Elastic Potential Energy (EPE)
When you stretch a rubber band, you do work on it. That work is stored inside the band as Elastic Potential Energy (EPE). When released, this stored energy converts into kinetic energy or gravitational potential energy.
The EPE Formula:
\(\text{EPE} = \frac{\lambda x^2}{2l}\) or \(\text{EPE} = \frac{1}{2}kx^2\)
Where:
• \(\text{EPE}\) is measured in Joules (\(\text{J}\))
• \(\lambda\) is the modulus of elasticity in \(\text{N}\)
• \(x\) is the extension (or compression) in \(\text{m}\)
• \(l\) is the natural length in \(\text{m}\)
Where does this come from? (Work Done by a Variable Force)
Because the tension increases as the string stretches, the force is not constant. The work done in stretching an elastic string from extension \(0\) to extension \(x\) is the integral of tension with respect to extension:
\(\text{Work Done} = \int_{0}^{x} T \, \text{d}u = \int_{0}^{x} \frac{\lambda u}{l} \, \text{d}u = \left[ \frac{\lambda u^2}{2l} \right]_0^x = \frac{\lambda x^2}{2l}\)
Common Mistake Alert: When finding the work done to stretch a string from extension \(x_1\) to extension \(x_2\), do NOT calculate \(\frac{\lambda (x_2 - x_1)^2}{2l}\)!
Instead, calculate the difference between the two energies:
\(\text{Work Done} = \frac{\lambda x_2^2}{2l} - \frac{\lambda x_1^2}{2l}\)
Key Takeaway: \(\text{EPE}\) depends on the square of the extension (\(x^2\)). If you double the extension, the stored energy increases by a factor of 4!
4. Equilibrium Problems
In equilibrium, all forces acting on a particle are balanced, so the resultant force is zero (\(\sum F = 0\)).
Horizontal Equilibrium
A particle of mass \(m\) rests on a smooth horizontal table, connected to fixed points by one or more elastic strings. To solve these problems:
1. Identify the natural lengths and current lengths to find each extension: \(x = L - l\).
2. Write expressions for the tension in each string using \(T = \frac{\lambda x}{l}\).
3. Resolve forces horizontally: Forces to the left = Forces to the right.
Vertical Equilibrium
A particle of mass \(m\) is suspended vertically from a fixed support by an elastic string or spring.
When hanging in equilibrium at rest:
\(\text{Upward Tension} = \text{Downward Weight}\)
\(T = mg\)
\(\frac{\lambda x}{l} = mg\)
Worked Example: Vertical Hanging Particle
A particle of mass \(3\text{ kg}\) is suspended from a ceiling by a light elastic string of natural length \(0.5\text{ m}\) and modulus of elasticity \(60\text{ N}\). Find the extension of the string and the total length at equilibrium. (Take \(g = 9.8\text{ m s}^{-2}\)).
Step 1: Set up the balance equation:
\(T = mg\)
Step 2: Substitute Hooke's Law into the equation:
\(\frac{\lambda x}{l} = mg\)
\(\frac{60 x}{0.5} = 3 \times 9.8\)
\(120x = 29.4\)
\(x = \frac{29.4}{120} = 0.245\text{ m}\)
Step 3: Find total length \(L\):
\(L = l + x = 0.5 + 0.245 = 0.745\text{ m}\)
Key Takeaway: For vertical hanging strings, start directly with \(T = mg\) and substitute \(T = \frac{\lambda x}{l}\).
5. Conservation of Energy Problems
When no external resistance forces (like friction or air resistance) do work, the Total Mechanical Energy of a system is conserved:
\(\text{Initial Total Energy} = \text{Final Total Energy}\)
The Three Forms of Mechanical Energy:
• Kinetic Energy (KE): \(\text{KE} = \frac{1}{2}mv^2\)
• Gravitational Potential Energy (GPE): \(\text{GPE} = mgh\) (where \(h\) is the vertical height above a chosen reference level)
• Elastic Potential Energy (EPE): \(\text{EPE} = \frac{\lambda x^2}{2l}\) (when \(x > 0\))
The General Energy Equation:
\(\text{KE}_1 + \text{GPE}_1 + \text{EPE}_1 = \text{KE}_2 + \text{GPE}_2 + \text{EPE}_2\)
If friction or another resistive force is present, we use the Work-Energy Principle:
\(\text{Initial Energy} - \text{Work Done against Resistance} = \text{Final Energy}\)
where \(\text{Work Done against Friction} = F_r \times d = \mu R \times d\).
Step-by-Step Guide for Energy Problems:
1. Choose a Zero GPE Datum Level: Pick the lowest point mentioned in the problem as \(h = 0\). This ensures all heights \(h\) are positive!
2. Write Down Initial State (Position 1): Find \(\text{KE}_1\), \(\text{GPE}_1\), and \(\text{EPE}_1\).
3. Write Down Final State (Position 2): Find \(\text{KE}_2\), \(\text{GPE}_2\), and \(\text{EPE}_2\).
4. Form and Solve the Equation: Equate initial and final total energies.
Worked Example: Dropping a Mass on an Elastic String
A particle of mass \(2\text{ kg}\) is attached to one end of a light elastic string of natural length \(1.2\text{ m}\) and modulus of elasticity \(48\text{ N}\). The other end is attached to a fixed ceiling point \(O\). The particle is held at \(O\) and released from rest. Find the maximum distance the particle falls below \(O\). (Take \(g = 9.8\text{ m s}^{-2}\)).
Step 1: Understand the motion
The particle falls under gravity for \(1.2\text{ m}\) with the string slack (\(\text{EPE} = 0\)). Once it falls past \(1.2\text{ m}\), the string stretches by extension \(x\), creating tension that slows the particle down until it momentarily comes to rest at the lowest point.
Step 2: Define positions and datum
• Let the lowest point be our datum line (\(h = 0\)).
• At the release point \(O\), height \(h = 1.2 + x\).
• At the lowest point, velocity \(v = 0\) (momentarily at rest), height \(h = 0\), and extension is \(x\).
Step 3: Energy at Release (Position 1)
• \(\text{KE}_1 = 0\) (released from rest)
• \(\text{GPE}_1 = mgh = 2 \times 9.8 \times (1.2 + x) = 19.6(1.2 + x) = 23.52 + 19.6x\)
• \(\text{EPE}_1 = 0\) (string is not stretched)
Step 4: Energy at Lowest Point (Position 2)
• \(\text{KE}_2 = 0\) (at maximum distance, particle is momentarily at rest)
• \(\text{GPE}_2 = 0\) (at datum line)
• \(\text{EPE}_2 = \frac{\lambda x^2}{2l} = \frac{48 x^2}{2(1.2)} = \frac{48 x^2}{2.4} = 20x^2\)
Step 5: Apply Conservation of Energy
\(\text{Initial Energy} = \text{Final Energy}\)
\(23.52 + 19.6x = 20x^2\)
\(20x^2 - 19.6x - 23.52 = 0\)
Step 6: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{19.6 \pm \sqrt{(-19.6)^2 - 4(20)(-23.52)}}{2(20)}\)
\(x = \frac{19.6 \pm \sqrt{384.16 + 1881.6}}{40}\)
\(x = \frac{19.6 \pm \sqrt{2265.76}}{40} = \frac{19.6 \pm 47.6}{40}\)
Since extension must be positive: \(x = \frac{19.6 + 47.6}{40} = \frac{67.2}{40} = 1.68\text{ m}\).
Step 7: Answer the specific question
The question asks for the maximum distance below \(O\):
\(\text{Total Distance} = l + x = 1.2 + 1.68 = 2.88\text{ m}\)
Key Takeaway: At maximum extension, velocity is zero! Use this fact to eliminate the \(\text{KE}\) term at the lowest point.
6. Summary & Top Exam Tips
• Check units: Natural length \(l\) and extension \(x\) must be in metres (\(\text{m}\)). If given in \(\text{cm}\), divide by 100 first!
• Tension vs Extension: Hooke's Law is \(T = \frac{\lambda x}{l}\). Remember that \(x\) is extension, not the total length.
• Squared terms in EPE: \(\text{EPE} = \frac{\lambda x^2}{2l}\). Always square the extension before multiplying.
• Work Done to stretch from \(x_1\) to \(x_2\): Always use \(\frac{\lambda x_2^2}{2l} - \frac{\lambda x_1^2}{2l}\).
• Maximum speed occurs at equilibrium: When a falling particle attached to an elastic string reaches its maximum speed, the acceleration is zero (\(a = 0\)), which means \(T = mg\). You can find the extension at this point and use conservation of energy to find the maximum speed!