Hooke’s law, elastic strings and springs
Relate tension to extension using modulus of elasticity and track elastic energy and work in particle motion.
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Use Hooke’s law and modulus of elasticity for strings or springs
- Calculate elastic potential energy
- Solve force and energy problems with elastic elements
Modulus relates tension to extension
For a light elastic string with natural length l and modulus λ, T=λx/l while it is stretched by x. A spring may be modelled similarly when the syllabus defines its force-extension relation.
The string is slack when its length is below natural length and cannot exert a compressive force. Keep extension separate from total length: total length is l+x.
Elastic energy is work stored during extension
The elastic potential energy is ½λx²/l. It equals the area under the linear tension-extension graph from zero extension to x.
Use energy conservation only when the other forces do no net work or their work is included. On an incline, include the change in gravitational potential energy and any applied work as well.
Pause and explain
Which habit best checks your work on hooke’s law, elastic strings and springs?
Worked example
A string has natural length 2 m and modulus 120 N. Find its tension and elastic energy when its extension is 0.1 m.
Show the worked solution
- The ratio extension/natural length is 0.1/2=0.05.
- T=120(0.05)=6 N.
- Elastic energy is ½(120)(0.1²)/2=0.3 J.
Answer T=6 N and elastic energy=0.3 J.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
An elastic string of natural length 1.5 m and modulus 90 N is stretched to 1.8 m. Find its tension.
Give me a hint
Find extension first, then use T=λx/l.
Compare my reasoning
- Extension x=1.8−1.5=0.3 m.
- T=90(0.3)/1.5.
- Evaluate the tension.
18 N.
Look for these in your work
- Extension correct
- Natural length used in denominator
- Tension calculated
A spring has force-extension relation T=40x N. Find the energy stored when its extension is 0.25 m.
Give me a hint
The force-extension graph is a straight line from zero to T=40(0.25).
Compare my reasoning
- The final tension is 10 N.
- The stored energy is the triangular area ½×extension×final tension.
- Calculate ½(0.25)(10).
1.25 J.
Look for these in your work
- Final force correct
- Area under graph used
- Energy in joules stated
A particle is released from rest while attached to a stretched elastic string on a vertical line. List the energy changes you must include when finding its speed after moving downward.
Give me a hint
Account for elastic energy, gravitational potential energy and kinetic energy between the two positions.
Compare my reasoning
- Calculate the initial and final elastic energies from their extensions.
- Include the decrease in gravitational potential energy as the particle descends.
- Set the total energy change equal to the gain in kinetic energy, including any other work if present.
Use change in elastic energy plus change in gravitational potential energy to determine the kinetic-energy gain, with signs and any external work included.
Look for these in your work
- Both stored-energy terms named
- Change rather than absolute levels used
- Work by other forces considered
Common mistakes
- Using total length instead of extension
- Allowing an elastic string to push when slack
- Forgetting the factor ½ in elastic energy
What can you explain now?
What is the extension of a string with natural length l and current length L?
Compare with the explanation
L−l
Hooke’s law uses extension from natural length, not the whole length.
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Make it stick
How do natural length, extension, tension and stored elastic energy fit together?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.