=-2.25in
=-8
Part of the system is shown schematically
at right. The two springs have force constants 
and the resistive force is 
(mechanism not shown). If the driven end is
fixed and the mass is at equilibrium, the two
springs will be under equal tension 
. If the
mass is now displaced a distance x to the right
from equilibrium, the lefthand spring will exert
a force on the car equal to
and the righthand spring will exert a force equal to
.
The net force on the car is equal to 
.
Now if the driven end is displace to the right
by a distance 
and the car is displaced a
distance x as before where, say 
, then
the force of the lefthand spring on the car is
and the force of the righthand spring is, as before,
. Now the net force on the car is
.
If we now include the damping force -Rv, the total force is now
using Newton's 2nd Law. Setting 
, 
, and
, we have
If 
is caused to vary sinusoidally by means of the drive unit
according to
where 
is the driving amplitude and 
is the
frequency of the driver, then the term 
becomes
where 
is
the amplitude of the driving force. This gives
which is solved in great detail in many mechanics books. For example, see the appropriate sections of Stephenson's Mechanics, 2nd Ed., which are attached to these sheets.