Answer
$m=21.2Kg$
$608.67N/m$
Work Step by Step
We can calculate the required mass and stiffness of each of the springs as follows:
$K_{eq}=K_1+K_2=2K$
We know that
$\omega_n=\sqrt{\frac{K_{eq}}{m_{eq}}}=\sqrt{\frac{2K}{m}}$
As $T=\frac{2\pi}{\omega_n}$
$\implies T=2\pi \sqrt{\frac{m}{2K}}$
Now $\frac{T_1}{T_2}=\frac{\sqrt{\frac{m_1}{2K}}}{\sqrt{\frac{m_2}{2K}}}=\frac{m_1}{m_2}$
$\implies \frac{(0.83)^2}{(1.52)^2}=\frac{m}{m+50}$
This simplifies to:
$m=21.2Kg$
Now $T=2\pi\sqrt{\frac{m}{2K}}$
We plug in the known values to obtain:
$0.83=2\pi \sqrt{\frac{21.2}{2K}}$
This simplifies to:
$K=608.67N/m$
