Answer
$a_{max}=3.9\times 10^{13}m/s^2=(4.0\times 10^{12})g$
Work Step by Step
To find the maximum acceleration of a mass-spring system, use the relation $$a_{max}=A\omega^2$$ Using the fact that $\omega=2\pi f$, the acceleration becomes $$a_{max}=A(2\pi f)^2=4\pi^2 A f^2$$ Substituting known values of $A=0.10\times 10^{-11}m$ and $f=10^{12}Hz$ yields a maximum acceleration of $$a_{max}=4\pi^2(0.10\times 10^{-11}m)(10^{12}Hz)^2=3.9\times 10^{13}m/s^2$$ To find this acceleration in terms of $g$, divide the acceleration by $g$ to get $$n=\frac{3.9\times 10^{13}m/s^2}{9.8m/s^2}=4.0\times 10^{12}$$ This means that the acceleration is $a=(4.0\times 10^{12})g$.
