Given data
*The given length of the rod is l = 2.00 m
*The given diameter of rod is d = 1.0 mm = 1.0 × 10^-3 m
*The elastic modulus of brass is E = 9.10 × 10^10 N/m^2
The formula for the spring constant of the rod is given as
[tex]\begin{gathered} k=\frac{Ea}{l} \\ =\frac{E\times(\frac{\pi}{4}d^2)}{l} \end{gathered}[/tex]*Here a is the cross-sectional area of the rod
Substitute the known values in the above expression as
[tex]\begin{gathered} k=\frac{9.10\times10^{10}\times(\frac{3.14}{4}\times(1.0\times10^{-3})^2)}{2.00} \\ =3.57\times10^4\text{ N/m} \end{gathered}[/tex]Hence, the spring constant of the rod is k = 3.57 × 10^4 N/m
