M3.2 - Elastic strings and springs

An elastic element has natural length $l$, the length at which it is undeformed. If its current length is $L$, its extension is $x=L-l$. Under Hooke's law, the elastic-force magnitude is proportional to this extension.

Here $T$ is tension in newtons and $\lambda$ is the modulus of elasticity, also measured in newtons. The ratio $x/l$ is dimensionless. When a spring is described using stiffness $k$ instead, the equivalent form is $T=kx$, with $k=\lambda/l$ and unit $\mathrm{N\,m^{-1}}$.

First calculate extension from the actual geometry, not from the total length alone. Apply Hooke's law to find tension, then place that force on the free-body diagram and resolve it in the chosen direction. If two equal sections pull symmetrically, include both resolved tensions; do not replace them by one tension.

A particle of mass $1.5$ kg hangs from a vertical light elastic string with natural length $0.8$ m and modulus $40$ N. At an instant the string has length $1.1$ m, so $x=0.3$ m and$$T=\frac{40(0.3)}{0.8}=15\text{ N}.$$Taking upward as positive, the particle's acceleration at that instant is$$a=\frac{T-mg}{m}=\frac{15-1.5(9.8)}{1.5}=0.20\,\mathrm{m\,s^{-2}}.$$Hooke's law supplies the force; Newton's second law supplies the acceleration.

Stretching an elastic string or deforming a spring stores elastic potential energy (EPE). For a Hooke-law element, the force rises linearly from zero, so the stored energy is the area under the force-extension relation.

Use the extension $x=L-l$ at the particular state being considered. For an elastic string, $E_{\mathrm e}=0$ whenever the string is slack; for a spring, $x^2$ gives positive stored energy for either extension or compression. Elastic energy is measured in joules.

Choose initial and final states, calculate the elastic extension separately at each state, and use one consistent height datum. If only gravity and ideal elastic forces do work,$$K_i+U_{g,i}+E_{\mathrm e,i}=K_f+U_{g,f}+E_{\mathrm e,f}.$$Otherwise add the work of any other force with its correct sign. Energy equations compare states; they do not require the acceleration to be constant.

A $1$ kg particle is attached below a vertical elastic string of natural length $1$ m and modulus $20$ N. It is released from rest with extension $0.10$ m and falls $0.20$ m, so the final extension is $0.30$ m. The initial and final EPE values are $0.10$ J and $0.90$ J. Conservation of mechanical energy gives$$0+0.10+1(9.8)(0.20)=\frac12v^2+0.90,$$hence $\tfrac12v^2=1.16$ and $v=1.52\,\mathrm{m\,s^{-1}}$ to three significant figures.