Answer
$f = 2.68~Hz$
Work Step by Step
In part (a), we found that the spring constant of each spring is $~~k = 1.29\times 10^5~N/m$
We can find the mass on each spring:
$m = \frac{(1450~kg)+(5)(73.0~kg)}{4} = 453.75~kg$
We can find the oscillation frequency:
$f = \frac{1}{2\pi}~\sqrt{\frac{k}{m}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{1.29\times 10^5~N/m}{453.75~kg}}$
$f = 2.68~Hz$
