MathsFurther Mechanics 1 › Hooke's law and elastic strings

Hooke's law and elastic strings

Stretch a string and it pulls back in proportion to how far it has been stretched. The constant of proportionality depends on both the material and the natural length, which is what the modulus of elasticity keeps track of.

Builds on Forces and Newton's laws and Statics of a particle.

IN THIS TOPIC

  • Find a tension, an extension or a modulus given the other two.
  • Solve equilibrium problems involving one or more elastic strings.

COMMON MISCONCEPTION

Two strings of the same material with the same modulus give the same tension at the same extension.

Tension proportional to extension

Hooke's law relates the tension T in a stretched elastic string or spring to its extension x.

T=λxlT = \frac{λ x}{l}NOT IN THE BOOKLET — LEARN IT

Hooke's law is not in the booklet, so learn it. Here l is the natural length, the length when unstretched, and λ is the modulus of elasticity, measured in newtons. The gradient of the graph of T against x is λ/l, so a long string of the same material is slacker than a short one. The natural length is half the formula, not a detail.

A string can only pull, so its tension is zero whenever the distance between its ends is less than the natural length. A spring can also push, giving a thrust with the same formula and x taken as the compression. That difference decides whether a case needs checking.

Hooke's law: tension against extension is a straight line through the origin with gradient λ/l0.5 m20 NT = λx / lextensiongradient λ/l = 40 N/m
FIG. 1Tension against extension: a straight line through the origin whose gradient is the modulus divided by the natural length.

WORKED EXAMPLE

Reading the formula three ways

An elastic string of natural length 1.5 m and modulus 60 N is stretched to a length of 2 m. Find the tension.

The extension is 2 − 1.5 = 0.5 m.

T = 60 × 0.5/1.5 = 20 N.

Doubling the natural length while keeping the modulus would halve this tension, since the same stretch is then a smaller fraction of the string.

Equilibrium with elastic strings

Once the tension is written in terms of the extension, an elastic problem becomes an ordinary statics problem. Resolve, write the equilibrium equations, and substitute λx/l wherever a tension appears. The unknown is usually the extension, so the equation is linear and solves at once.

With two strings, each has its own λ, l and x, and the extensions are linked by the geometry of the arrangement. The commonest arrangement is a particle held between two strings on a horizontal line, where the two stretched lengths add to a fixed total.

A mass hanging in equilibrium: the tension in the stretched string balances the weight1 m0.2 m2 kgT2g98x / 1 = 19.6x = 0.2 m
FIG. 2A mass hanging in equilibrium, with the tension in the stretched string balancing its weight.

GUIDED PRACTICE

A hanging mass

A particle of mass 2 kg hangs in equilibrium at the end of an elastic string of natural length 1 m and modulus 98 N, attached to a fixed ceiling. Find the extension and the total length, taking g = 9.8 m/s².

Show the working

In equilibrium the tension equals the weight: T = 2 × 9.8 = 19.6 N.

Hooke's law gives 98x/1 = 19.6, so x = 0.2 m.

The string is therefore 1.2 m long.

Doubling the mass would double the extension, since the relationship is linear right up to the elastic limit.

ASSESSMENT FOCUS

  • Write down l, λ and x separately before substituting. Mixing up the length and the extension is the standard error.
  • For a string, check whether it is taut. A slack string has zero tension and drops out of the equation.
  • For a spring, allow for a thrust as well as a tension, since a spring can push.
  • Modulus is a force in newtons, so a modulus quoted in metres is a misread question.

CHECK YOURSELF

An elastic spring of natural length 0.8 m has modulus 40 N. Find the thrust when it is compressed to a length of 0.6 m.

Show a hint

The compression plays the part of x.

Show the answer

The compression is 0.2 m, so the thrust is 40 × 0.2/0.8 = 10 N, pushing the ends apart.

Hooke's law is T = λx/l, where x is the extension, l the natural length and λ the modulus in newtons.

A string pulls only and has zero tension when slack. A spring also pushes, with the compression in place of the extension.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the hooke's law and elastic strings questions page.

CHECK YOUR PROGRESS

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  • Find a tension, an extension or a modulus given the other two.
  • Solve equilibrium problems involving one or more elastic strings.

Open the full revision checklist to see every objective in the course in one place.