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Hooke's Law Simplified Revision Notes

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15.1.1 Hooke's Law

Introduction

Hooke's law describes the relationship between the tension in a stretched elastic string or spring and its extension. It is a fundamental concept in mechanics and elasticity, used to model the behaviour of springs and elastic materials.

Hooke's law states:

T=λxlT = \frac{\lambda x}{l}

where:

  • TT is the tension in the string or spring (N\text{N})
  • λ\lambda is the modulus of elasticity (N\text{N})
  • xx is the extension of the string or spring (m\text{m})
  • ll is the natural length of the string or spring (m\text{m})

Natural Length (ll)

  • The natural length is the length of the string or spring when it is unstretched and not under tension.
  • Its units are meters (m\text{m})

Modulus of Elasticity (λ\lambda)

  • The modulus of elasticity is a measure of how "stretchy" the string or spring is.
  • A larger λ\lambda means the string is less stretchy, requiring greater force to achieve the same extension.
  • Its units are newtons (N\text{N})

Extension (xx)

  • The extension is the additional length the string or spring gains when stretched.
  • It is calculated as:
x=stretched lengthnatural length.x = \text{stretched length} - \text{natural length}.
  • Its units are meters (m\text{m})

Assumptions of Hooke's Law

  • Hooke's law only applies within the elastic limit, where the string or spring returns to its original length when the tension is removed.
  • For problems in this topic, assume Hooke's law is valid unless stated otherwise.

Worked Example

infoNote

Example: Tension in a Spring


Problem

A spring with a natural length of 0.5m0.5 \, \text{m} and a modulus of elasticity λ=100N\lambda = 100 \, \text{N} is stretched to a length of 0.7m0.7 \, \text{m}. Find:

  1. The extension of the spring.
  2. The tension in the spring.

Part 1: Calculate the Extension


Step 1: Use the formula for extension:

x=stretched lengthnatural lengthx = \text{stretched length} - \text{natural length}

Step 2: Substitute the values:

x=0.70.5=0.2mx = 0.7 - 0.5 = 0.2 \, \text{m}

Part 2: Calculate the Tension


Step 1: Use Hooke's law:

T=λxlT = \frac{\lambda x}{l}

Step 2: Substitute the values (λ=100N,x=0.2m,l=0.5m\lambda = 100 \, \text{N}, x = 0.2 \, \text{m}, l = 0.5 \, \text{m}):

T=100×0.20.5T = \frac{100 \times 0.2}{0.5}

Step 3: Simplify:

T=200.5=40NT = \frac{20}{0.5} = 40 \, \text{N}

Final Answer:

Extension: 0.2m0.2 \, \text{m}

Tension: 40N40 \, \text{N}

Note Summary

infoNote

Common Mistakes

  1. Forgetting to calculate the extension: Always find x=stretched lengthnatural lengthx = \text{stretched length} - \text{natural length}
  2. Misinterpreting λ\lambda: Remember that λ\lambda is a constant specific to the material.
  3. Exceeding the elastic limit: Hooke's law only applies when the string or spring is not permanently deformed.
infoNote

Key Formula

  1. Hooke's Law:
T=λxlT = \frac{\lambda x}{l}
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