Answer
If only the 45-kg person were present, the frequency would be $2.16~Hz$
Work Step by Step
We can find the spring constant $k$:
$\omega = 2\pi~f$
$\sqrt{\frac{k}{m}} = 2\pi~f$
$k = (2\pi~f)^2~m$
$k = (2\pi)^2~(2.00~Hz)^2~(1020~kg+45~kg+52~kg+67~kg+61~kg)$
$k = 196,600~N/m$
We can find the frequency if only the 45-kg person were present:
$f = \frac{\omega}{2\pi}$
$f = \frac{1}{2\pi}~\sqrt{\frac{k}{m}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{196,600~N/m}{1020~kg+45~kg}}$
$f = 2.16~Hz$
If only the 45-kg person were present, the frequency would be $2.16~Hz$
