# Chapter 10 - Simple Harmonic Motion and Elasticity - Problems - Page 277: 51

$5.2\times10^{-4}\,m$

#### Work Step by Step

Given/known: $F=890\,N$, $L_{0}=9.1\,m$, $A=\pi r^{2}=\pi (0.50\times10^{-2}m)^{2}$, $Y=2.0\times10^{11}\,N/m^{2}$ (Young's modulus of steel). Rearranging Equation 10.17, we have $\Delta L=\frac{FL_{0}}{AY}=\frac{890\,N\times9.1\,m}{\pi (0.50\times10^{-2}m)^{2}\times2.0\times10^{11}\,N/m^{2}}=5.2\times10^{-4}\,m$

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