Elastic strings, springs and stored energy

What you’ll learn

• Use Hooke’s law with extension and natural length.
• Combine elastic, gravitational and kinetic energy.

Understand the idea

In the ideal elastic model, force is proportional to extension. The modulus $\lambda$ and natural length $l$ give stiffness $\lambda/l$. Extension is the actual length minus the natural length, not the total length. A spring can act in compression; a string becomes slack instead of pushing when its extension would be negative.

Elastic energy is the area under the linear force–extension graph. Include it in the energy balance together with kinetic energy and gravitational potential energy. At vertical equilibrium, tension balances weight, but at a turning point the velocity is zero without necessarily having zero resultant force.

Key formulae

$$T=\frac{\lambda x}{l}\quad\text{(taut string)}$$

$$E_{\mathrm{elastic}}=\frac{\lambda x^2}{2l}$$

$$k=\frac\lambda l$$

$$\frac12mv^2+mgh+\frac{\lambda x^2}{2l}=\text{constant}\quad\text{(conservative model)}$$

Worked example

A $1\,\mathrm{kg}$ mass hangs at rest from an elastic string of natural length $2\,\mathrm m$ and modulus $98\,\mathrm N$. Find its extension. Take $g=9.8\,\mathrm{m\,s^{-2}}$.

1. Equilibrium gives $T=mg=9.8\,\mathrm N$.
2. Hooke's law gives $T=98x/2=49x$.
3. Solve $49x=9.8$.

Answer: $x=0.2\,\mathrm m$.