$w=mg$

$4=32m$

$m=\frac{1}{8}$

Substituting the constant values into the differential equation results in the following.

$\frac{d^{2}x}{d{t}^2}\frac{B}{m}\frac{dx}{dt}\frac{k}{m}x=0$

$\frac{d^{2}x}{d{t}^2}(8-\sqrt[\phantom{2}]{16})x=0$

Step $2$

We have found that the following differential equation defines the motion of this spring.

$\frac{d^{2}x}{d{t}^2}8\frac{dx}{dt}16x=0$

Therefore, the auxiliary equation is as follows.

$m^2=0$

Solving for $m,$ the root of the auxiliary equation is $m_1=$ with multiplicity $2.$