How do I incorporate friction and mass when analyzing spring motion? Problem 12 of section B of this PDF file reads

Two springs with spring constant k1 and k2 are attached to a body of mass (m) in two different
  configurations in 2 cases (A and B) as shown.  

The mass rests on a surface with coefficient of
  friction between the mass and surface as $K$. 
  The mass is moved to the right, parallel to the axis of the springs by a distance $d$ (from natural
  lengths of the springs) and is released. The time taken till the body comes to rest in case A is:

The given answer is equal to B but I get it to be less than B.
$$T= 2\pi \sqrt{\frac{m}{k}} $$
But I'm not sure how to incorporate effect of friction and mass.
In case of A, there is a $$ frictional force F = Kmg $$
But how do I relate this to T ? 
I referred the pdf link above. I was trying to solve the question. The answer is 
$$ T_{a} = T_{b} $$
An alernate way that I just figured is given below. Please clarify -
$For A $ ,$$ -k_{1}d - k_{2}d - Kmg = m\omega_{a} ^2 x $$ $For B$,
$$ -k_{1}d - k_{2}d - Kmg = m\omega _{b}^2 x $$
$$ \omega_{a} = \omega_{b} $$
$$ T=\frac{2Pi}{\omega} $$
$$ T_a=T_b $$
I am not a physics student. So I wanted to understand this conceptually and get the answer from the relevant equations. Also is my approach to the problem correct ?
 A: You actually do not need to solve the equation at all in order to determine that, whatever the answer, it must be the same for A and B.  
The only difference between A and B is the direction of k1.  However if you look at $F=-kx$ is always a restoring force whose magnitude is proportional to the magnitude of the displacement from equilibrium.  So the mass always experiences a restoring force of the same magnitude regardless of whether k1 is being compressed or stretched.
So in both cases the forces are the same and therefore the time must be the same also.
A: Since both springs are connected effectively in parallel, the both cases will have a same equation of motion. The equation of motion: $ma = F$
$$ \tag{1}
 m \frac{d^2x}{dt^2} = -2kx - f_d.
$$
The variable $x$ is the displacement for the oscillator displacement from its equilibrium position, $m$ the mass, $f_d$ is the dragging or frictional force. It is not easy to model a constant static frictional force, because it is oscillation motion, the displacement $x$ changes sign regularly. It is more convinient to model a dragging force propotional to velocity, $f_d = - b v$, the velocity will take care the sign of the force. Therefore, lets adopt this dragging force model:
$$ \tag{2}
m \frac{d^2x}{dt^2} = -2kx -b v;
$$
where $b$ is the damping coefficient. Write Eq. (2) in standard form of a second order differential equation:
$$ \tag{3}
m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + 2kx= 0;
$$
Scale the equation by mass $m$, and define $\omega_0 = \sqrt{\frac{2k}{m}}$
, $\gamma = \frac{b}{2m}$. We then write the equation as:
$$ \tag{4}
\frac{d^2x}{dt^2} + 2\gamma \frac{dx}{dt} + \omega_0^2 x = 0;
$$
Assume that $\gamma \lt \omega_0$ (the underdamping regime). The solution for Eq. (4):
$$
  x(t) = e^{-\gamma t} \left\{ A \cos (\omega t) + B \sin(\omega t) \right\}.
$$
Where $\omega = \sqrt{\omega_0^2 - \gamma^2}$, A and B are two constnats to be determined by the initial condition: $x(0)$ and $v(0)$, the initial displacement and initial velocity.
