## College Physics (4th Edition)

If only the 45-kg person were present, the frequency would be $2.16~Hz$
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 $2.16~Hz$