Edexcel A-level Further Maths revision notes

Elastic strings and springs and elastic energy

Section FM1-3

Year 2

2 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section FM1-3

Checked against Edexcel 9FM0 section FM1-3. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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FM1-3.1

Elastic strings and springs. Hooke's law.

Notes

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Explanation

Extension is stretched length minus natural length. A Hookean spring has tension magnitude $T=kx$ , where $k$ is the spring constant in $\text{N m}^{-1}$ .
A taut elastic string of natural length $l$ and modulus $\lambda$ obeys $T=\lambda x/l$ , where $\lambda$ is measured in newtons.
A string cannot push: when its extension is zero or negative it is slack and its tension is zero.
A spring may be extended or compressed, with force opposing the displacement.
Examiners expect natural length, actual length and extension to be distinguished, equilibrium forces to be resolved consistently, and the correct Hooke's-law form chosen for a spring or string.

The linear tension-extension graph for a Hookean spring.

Worked example

An elastic string has natural length $0.9\,\text{m}$ and modulus $54\,\text{N}$ . Find its tension when its length is $1.15\,\text{m}$ .

1.The extension is $x=1.15-0.90=0.25\,\text{m}$ .
2.Use the string form $T=\lambda x/l$ .
3. $T=54(0.25)/0.9=15\,\text{N}$ .

Answer: The tension is $15\,\text{N}$ .

Common mistakes

Don't substitute the stretched length for $x$ instead of calculating extension from the natural length.
Don't use $T=kx$ with the modulus of a string and omit division by natural length.
Don't assign a negative compressive tension to a slack string.

Exam tip

Write the extension explicitly before Hooke's law; this separates the length mark from the force calculation.

Tier 1 · Easy

ORIGINAL

A light spring has spring constant

80\,\text{N m}^{-1}

and is extended by

0.06\,\text{m}

. Find its tension.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

An elastic string has natural length

1.5\,\text{m}

and modulus

120\,\text{N}

. It is stretched to length

1.8\,\text{m}

. Calculate its tension.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.5\,\text{kg}

particle hangs in equilibrium from a vertical elastic string of natural length

0.8\,\text{m}

and modulus

49\,\text{N}

. Taking

g=9.8\,\text{m s}^{-2}

, determine the extension and the equilibrium length of the string.

(4)

(Total for Question 1 is 4 marks)

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FM1-3.2

Energy stored in an elastic string or spring.

Notes

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Your confidence:

Explanation

Elastic energy is the work done in stretching or compressing a Hookean object from zero extension. It is the area under the tension-extension graph, so a spring stores $\tfrac12kx^2$ and a taut elastic string stores $\dfrac{\lambda x^2}{2l}$ .
The factor $\tfrac12$ arises because tension increases linearly from zero to its final value.
In a work-energy problem, calculate extension from natural length at every position and include string energy only while the string is taut.
Gravitational and kinetic energy may appear in the same balance.
Examiners expect a clear initial-equals-final energy equation and reject final tension times extension as the stored energy.

Worked example

A $1.5\,\text{kg}$ particle hangs from a vertical elastic string of natural length $0.8\,\text{m}$ and modulus $48\,\text{N}$ . It is released from rest at natural length. Find its speed after descending $0.4\,\text{m}$ , taking $g=9.8\,\text{m s}^{-2}$ .

1.Loss of gravitational potential energy: $1.5(9.8)(0.4)=5.88\,\text{J}$ .
2.Elastic energy at extension $0.4$ : $\dfrac{48(0.4)^2}{2(0.8)}=4.8\,\text{J}$ .
3. $5.88=\tfrac12(1.5)v^2+4.8$ , so $0.75v^2=1.08$ .

Answer: $v=1.20\,\text{m s}^{-1}$ .

Common mistakes

Don't use $Tx$ for stored energy instead of the triangular area $\tfrac12Tx$ .
Don't measure extension from the release point rather than from natural length.
Don't include elastic-string energy at a position where the string is slack.

Exam tip

In a multi-energy question, label each term as kinetic, gravitational or elastic before simplifying.

Tier 1 · Easy

ORIGINAL

A spring of constant

200\,\text{N m}^{-1}

is compressed by

0.08\,\text{m}

. Calculate the elastic energy stored.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

An elastic string has natural length

1.2\,\text{m}

and modulus

60\,\text{N}

. Find the energy stored when its length is

1.5\,\text{m}

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

2\,\text{kg}

particle is attached to the lower end of a vertical elastic string with natural length

1\,\text{m}

and modulus

49\,\text{N}

. It is released from rest when the string is at natural length. Taking

g=9.8\,\text{m s}^{-2}

, find its speed after descending

0.6\,\text{m}

(5)

(Total for Question 1 is 5 marks)

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