Answer
66 Hz
Work Step by Step
Let' s consider the equation $f=\frac{1}{2\pi}\sqrt {\frac{k}{m}}$ for the simple harmonic motion of a spring. Also, we know that for a spring $F=kx$. But for a solid rod $F=YA(\frac{\Delta L}{L_{0}})$. So we can write,
$F=(\frac{YA}{L_{0}})\Delta L$. We find that x is analogous to $\Delta L$ and k is analogous to $(\frac{YA}{L_{0}})$. Therefore we can write,
$f=\frac{1}{2\pi}\sqrt {\frac{k}{m}}=\frac{1}{2\pi}\sqrt {\frac{YA}{L_{0}m}}=\frac{1}{2\pi}\sqrt {\frac{Y(\pi r^{2})}{L_{0}m}}$ ; Let's plug known values into this equation.
$f=\frac{1}{2\pi}\sqrt {\frac{(1.1\times10^{11}N/m^{2})\pi(3\times10^{-3}m)^{2}}{(2\space m)(9\space kg)}}\approx66 Hz$
