kinetic energy and conservative force field The kinetic energy of a particle is a periodic function in time. Does it imply that the particle is in a conservative force field and there are no dissipative forces acting on it at any instance of time ?
EDIT (in view of comment by Willie)
Please consider the question as "Is the force acting on the particle conservative ?" Also consider the kinetic energy to be periodic with non zero period.
 A: No.  I can take a ball and swing it back and forth periodically with my hand.  The motion is periodic, but the situation is not conservative - my body generates a lot of heat.
A simple mathematical example is a forced, damped harmonic oscillator.  It has a steady-state periodic solution that dissipates energy.
If you want to know whether a force field is conservative, take its curl.  Time-independent force fields (force is a function of position but not time) are conservative iff their curl is zero.
A: There is quite a lot wrong with this question. 


*

*What is a conservative force field? As usually defined, a conservative force field is a field that is defined everywhere in space that is the gradient of some potential energy (which may be time dependent); equivalently we ask that the curl of the force field is zero. This requires knowing the force everywhere in space. 

*Suppose we know the trajectory $\vec{\gamma}(t)$ of a particle. Then by taking the second time derivative we can compute the acceleration, and using Newton's second law we can compute the force. Okay, so this gives us knowledge of the (possibly time-dependent) force field $\vec{F}$ and the space-time coordinates $(t,\vec{\gamma}(t))$: we have that $\vec{F}(t,\vec{\gamma}(t)) = m \ddot{\gamma}(t)$. 


*

*But what does this tell us? Absolutely nothing! At any given time, we are only given knowledge at one single point of the force. The force field could be anything at other points in space. Given the value of a function at a point, you can extend it arbitrarily! You can extend it so that the force, at that instant in time, is conservative; or you can extend it so that the force is not conservative. You can extend it anywhich way you want and still have a force compatible with the motion of your particle. (Another way to look at it using Mark Eichenlaub's answer: a particle executing simple harmonic motion could either be in a harmonic potential (so is in a conservative force field), or it could equally well be in the steady state of a driven, damped harmonic oscillator, in which case you have dissipation. 

*Okay, you say, let us suppose then the force field is time independent. This way by specifying $\gamma(t)$ you can specify what the force would be along every point visited by the particle. Now you run into two problems: (a) what if the trajectory visits the same point twice, but exhibits different acceleration at those two different times? Then it is impossible to find a well-defined time-independent force field that generate this motion. (b) what if the trajectory only visits a single curve of points? Then you have the problem as before: if a function is prescribed along a one-dimensional subset of three-dimensional space, you can essentially extend the function anyway you want. You can extend the force-field to be conservative, or you can extend it to be non-conservative. 


*And then there is the additional problem that just giving the kinetic energy is insufficient to specify the velocity of the trajectory: the kinetic energy can allow you to determine the modulus $|\dot{\gamma}|(t)$ of the particle, but it cannot allow you to determine the direction the particle is moving in. So associated with a kinetic energy there are infinitely many trajectories that intersects itself infinitely many times, and there are also infinitely many trajectories which trace out a smooth embedded curve in 3 dimensional space.  

