Hooke's law and elastic energyAQA A-Level Further Maths: Revision notes

A light elastic string (or spring) of natural length $l$ is stretched to length $l+x$, where $x$ is the extension. Hooke's law says the tension is proportional to the extension: $T=\frac{\lambda x}{l}\quad\text{or}\quad T=kx,\quad k=\frac{\lambda}{l}.$ $\lambda$ is the modulus of elasticity (in newtons) and $k$ is the stiffness (in N m⁻¹). The tension is the same at every point of a light string, and a string exerts no force when slack (length less than $l$). Example: $l=1.2$ m, $\lambda=36$ N, stretched to 1.5 m. The extension is $x=0.3$ m, so $T=\frac{36\times0.3}{1.2}=9$ N.

The energy stored in a stretched string is the area under the tension–extension line: $EPE=\frac12Tx=\frac{\lambda x^2}{2l}=\frac12kx^2.$ Example: $\lambda=36$ N, $l=1.2$ m, $x=0.3$ m gives $\frac{36\times0.09}{2.4}=1.35$ J. The EPE depends on the square of the extension, so doubling the stretch stores four times the energy. The energy stored when stretching from extension $x_1$ to $x_2$ is $\frac{\lambda}{2l}(x_2^2-x_1^2)$, not $\frac{\lambda}{2l}(x_2-x_1)^2$.

A particle hanging in equilibrium on a vertical string has tension equal to its weight: $\frac{\lambda x}{l}=mg$. This gives the equilibrium extension $x=\frac{mgl}{\lambda}$, or the modulus if $x$ is known. Example: a 2 kg particle hangs from a string with $l=0.5$ m and equilibrium extension $0.25$ m. Then $19.6=\frac{\lambda(0.25)}{0.5}$ and $\lambda=39.2$ N. The energy stored is $\frac{39.2(0.25)^2}{2(0.5)}=2.45$ J.

With a smooth surface or no resistance, mechanical energy is conserved, with three forms: $KE+GPE+EPE=\text{constant}.$ Use this when the string goes from stretched to natural length (all EPE becomes KE on a smooth horizontal surface) or when a particle falls on a vertical string. Example: a 0.3 kg truck released at extension 0.4 m on a string with $l=0.6$ m and $\lambda=24$ N has $3.2$ J of EPE, so its speed at natural length is $\sqrt{\frac{2\times3.2}{0.3}}=4.62\text{ m s}^{-1}$. Vertical example: a 2 kg particle pulled down to extension 0.6 m ($\lambda=39.2$ N, $l=0.5$ m) rises 0.6 m as the string returns to natural length, so $\frac12(2)v^2=14.112-2(9.8)(0.6)$ and $v=1.53\text{ m s}^{-1}$.

At the lowest point the particle is momentarily at rest, so KE is zero. A particle released from rest at the fixed end of a string (natural length $l$) falls $l+x$ in total, so $mg(l+x)=\frac{\lambda x^2}{2l},$ which gives a quadratic in $x$. Reject the negative root. The maximum speed occurs when the resultant force is zero, that is at the equilibrium extension $x=\frac{mgl}{\lambda}$; before it the particle speeds up (weight exceeds tension), after it the particle slows down. Substitute that extension in the energy equation to find $v_{\max}$. Example ($m=1.5$, $l=2$, $\lambda=49$): $14.7(2+x)=12.25x^2$ gives $x=2.26$ m. Equilibrium extension $0.6$ m, so $\frac12(1.5)v^2=14.7(2.6)-12.25(0.36)$ and $v_{\max}=6.71\text{ m s}^{-1}$.