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We want to find the spring constant for the spring in each wheel. Its relationship to other known quantities is $k=m\omega^2$. Since each spring in the car is supporting a quarter of the total car weight, we have $m=M/4$. The angular frequency has the value

\begin{displaymath}\omega=2\pi f=18.85\;{\textrm s^{-1}},\end{displaymath}

which gives $k$ a value of 129 kN/m.

Next, we're going to add an amount of 5$\times$73.0 kg to the total weight. Then each spring supports 1/4 of this new total weight and, since the spring constant doesn't change when the weight changes, we have

\begin{displaymath}\omega'=\sqrt{\frac{k}{M'/4}}=16.84\; {\textrm s^{-1}}.\end{displaymath}

Dan Cross 2006-10-18