Oscillations of elastic strings and springsEdexcel International A Level Further Maths: Revision notes

The tension in an elastic string or spring of natural length $l$ and modulus of elasticity $\lambda$ is $T=\frac{\lambda x}{l}$, where $x$ is the extension. A spring can also be compressed, producing a thrust $\frac{\lambda x}{l}$ with $x$ the compression. A string can only pull and is slack if it is shorter than its natural length. For oscillations along the line of the string or spring, resolve forces along the line and apply Newton's second law.

For a particle on a smooth horizontal table attached to a fixed point, the equilibrium position is where the spring has its natural length, so the extension $x$ is the displacement from the centre. Newton's second law gives $m\ddot x=-\frac{\lambda x}{l}$, so $\ddot x=-\frac{\lambda}{ml}x$. This is simple harmonic motion with $\omega^2=\frac{\lambda}{ml}$ and period $T=2\pi\sqrt{\frac{ml}{\lambda}}$. The amplitude is the greatest extension from equilibrium, and the maximum speed is $a\omega$.

For a particle hanging from a string or spring, first find the equilibrium position: $mg=\frac{\lambda e}{l}$, so $e=\frac{mgl}{\lambda}$. At a displacement $x$ below equilibrium the tension is $\frac{\lambda(e+x)}{l}$. Newton's second law, taking downwards as positive, is $mg-\frac{\lambda(e+x)}{l}=m\ddot x$. Since $mg=\frac{\lambda e}{l}$ the constant terms cancel, leaving $m\ddot x=-\frac{\lambda x}{l}$, so $\ddot x=-\frac{\lambda}{ml}x$: SHM about the equilibrium position with the same $\omega^2=\frac{\lambda}{ml}$ as the horizontal case.

The oscillation is SHM only while the string is taut, i.e. while the particle is not above the natural length position. Compare the amplitude $a$ with the equilibrium extension $e$. If $a\le e$ the string stays taut for the whole motion. If $a>e$ the string goes slack when $x=-e$ (a distance $e$ above equilibrium). At that point $v^2=\omega^2(a^2-e^2)$. After that the particle moves as a projectile under gravity until the string becomes taut again.

Energy is often quicker for speeds and greatest displacements: kinetic $+$ gravitational $+$ elastic $\frac{\lambda x^2}{2l}$ is conserved. At the lowest point $v=0$. If the string is slack at the highest point, the elastic energy there is zero. Example: $m=0.5$, $l=1.2$, $\lambda=24$, pulled down $0.6$ m below equilibrium ($e=0.245$): the lowest extension is $0.845$ m, the elastic energy is $\frac{24(0.845)^2}{2.4}=7.14$ J, and the highest point is $\frac{7.14}{0.5g}=1.46$ m above the lowest point.