91Ó°ÊÓ

Figure for given system. The values of the spring constants and masses are:

$$\begin{gathered} {\mathrm{k}}_1=\frac{1600\mathrm{N}}{\mathrm{m}},{\mathrm{k}}_2=\frac{1500\mathrm{N}}{\mathrm{m}},{\mathrm{k}}_3=\frac{1400\mathrm{N}}{\mathrm{m}}\mathrm{and}{\mathrm{k}}_4=\frac{1200\mathrm{N}}{\mathrm{m}} \\ {\mathrm{m}}_1=800\mathrm{kg},{\mathrm{m}}_2=500\mathrm{kg},{\mathrm{m}}_3=400\mathrm{kg},{\mathrm{m}}_4=200\mathrm{kg} \end{gathered}$$

We can use the formula of frequency which is related to mass and spring constant. To maximize the amplitude in the second system both systems should have equal frequencies. Using this relation, we can check which mass and spring constant pair that satisfies the relation.

Formula:

The frequency of the oscillation in SHM, $f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k}{m}}$ (i)

We know for maximum amplitude of oscillation in system 2; it should have frequency equal to frequency of system 1.

$f_1=f_2$

From equation (i), we can get the frequency equation as follows:

$$\begin{gathered} \frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k_1}{m_1}}=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k_2}{m_2}} \\ \sqrt{\frac{k_1}{m_1}}=\sqrt{\frac{k_2}{m_2}} \\ \frac{k_1}{m_1}=\frac{k_2}{m_2} \end{gathered}$$

We can check the above equation is satisfied for the given masses and spring constants, that is

These values satisfy the above equation.

Hence, the pair of spring constant and mass with the values for first system spring constant is 1500N/m and mass is, while for second system spring constant is 1200N/m and mass is 400kg.