## College Physics (4th Edition)

The midpoint moves a distance of $0.80~mm$
$Y = \frac{F/A}{\Delta L/L}$ $Y$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L$ is the original length We can find an expression for the change in length of the thinner wire: $\Delta L_1 = \frac{F~L}{Y~A_1}$ Since the thicker wire has twice the radius, the area $A_2$ of the thicker wire is $4 A_1$. We can find an expression for the change in length of the thicker wire: $\Delta L_2 = \frac{F~L}{Y~A_2}$ $\Delta L_2 = \frac{F~L}{Y~(4A_1)}$ $\Delta L_2 = \frac{1}{4}\times \frac{F~L}{Y~A_1}$ $\Delta L_2 = \frac{1}{4}\times \Delta L_1$ The total change in length of both wires is $\frac{5}{4}~\Delta L_1$. We can find $\Delta L_1$: $\frac{5}{4}~\Delta L_1 = 1.0~~mm$ $\Delta L_1 = 0.80~~mm$ Therefore, the midpoint moves a distance of $0.80~mm$.