Harmonic oscillator negative mass coefficient My question is rather general. I am an engineer and I am studying a problem that can be approximated as a linear oscillator with frequency dependent coefficients. For some values of the frequency I find a negative mass coefficient. 
 I am baffled with the interpretation of this system. What is the solution in the case of negative mass coefficient? Is it bounded? 
Do you have any reference to textbooks I could have a look at?
Any help is appreciated 
Thanks in advance
 A: The mathematical description of the motion of this system can be found in a good  text of differential equations and is quite easy. 
Here I would like to point out a way to construct a physical model of the system you consider.
A physical example of this situation is  just the motion due to the inertial centrifugal force affecting every mass in a rotating reference  frame with fixed rotation axis, say $z$, and constant angular velocity $\Omega$ with respect to an inertial reference frame.  The force is radially directed: $$m\Omega^2 r {\bf e}_r \tag{1}$$ where $r$ is the distance from the rotation axis and ${\bf e}_r$ is the unit radial vector exiting from $z$.
To have an idea of the effect you sould remove the other inertial force, Coriolis'  one. A practical and very effective way is to consider a point of mass $m$ constranied to move on a frictionless rod radially emanating from the rotational axis $z$. Coriolis' force is cancelled by the reactive  force of the rod (normal to it) and the net motion of the point is only due to the centrifugal force since the rod is frictionless. This acts as a spring (with zero length at rest) but with negative constant $-m\Omega^2$ (the force of a radial  spring would be  $-kr {\bf e}_r$ so that (1) has the wrong sign $k= -m\Omega^2$).
In view of $$F= ma$$ (written in the rotating frame and referring to the radial direction only), changing sign to both sides, this is equivalent to a point with negative mass $-m$ under the action of a standard spring with positive constant.
$$(-m) \frac{d^2r}{dt^2} = -m\Omega^2 r\:.$$
The motion (not the the solution of the differential equation) is quite evident to every child... In particular the motion is never bounded.
