3.4 Hooke's law

For natural length $l$, current length $L$ and extension $x=L-l$, a Hookean elastic element has force magnitudeF=\frac{\lambda x}{l}=kx,where $\lambda$ is the modulus of elasticity in newtons and $k=\lambda/l$ is stiffness in N m$^{-1}$. The model applies within the stated elastic range.

An elastic string has $l=0.80\text{ m}$ and $\lambda=100\text{ N}$. At length $0.92\text{ m}$,x=0.92-0.80=0.12\text{ m},\qquad T=\frac{100(0.12)}{0.80}=15\text{ N}. $$$$

For a Hookean string or spring of natural length $l$, modulus $\lambda$ and extension or compression magnitude $x$, the stored elastic potential energy isE_{\text{elastic}}=\frac{\lambda x^2}{2l}=\frac12Fx.$$Proof of this formula is not required.$$

Between extensions $x_1$ and $x_2$, use the endpoint difference\Delta E_{\text{elastic}}=\frac{\lambda}{2l}(x_2^2-x_1^2).$$For several active elastic elements, calculate and add one energy term for each element in each state.$$

For $l=0.80\text{ m}$, $\lambda=100\text{ N}$ and $x=0.12\text{ m}$,E_{\text{elastic}}=\frac{100(0.12)^2}{2(0.80)}=0.90\text{ J}.At natural length, $x=0$ and the stored elastic energy is zero.

$$At the first turning point the speed is again zero, but the particle need not be in equilibrium. Loss of gravitational potential energy equals gain in elastic energy:$$mgx\sin\alpha=\frac{\lambda x^2}{2l}.Besides the initial root $x=0$, the turning extension isx=\frac{2mgl\sin\alpha}{\lambda}.The speed is greatest earlier, where tangential acceleration is zero and $\lambda x/l=mg\sin\alpha$.