Time period of a simple pendulum at the centre of earth 
So I thought of this hypothetical experiment and wondered whether it
be possible for it to even happen at center of earth even though
gravity is zero.

So considering a pendulum with length much much less than the radius of earth and center at earth center's can you find the time period of the pendulum (or is it even defined) ?
 A: This is quite an interesting question: let's consider the planet to be a solid sphere of constant density, and of some total mass $M$ and radius $R$. It's an exercise in undergraduate physics (using Gauss's law for Gravitation, for example) to show that the force exerted on a small mass $m$ at some distance $r$ from the origin is just: $$\mathbf{F} = - \frac{GM m r }{R^3} \hat{\mathbf{r}}.$$
You'll notice that the force is independent of the angles $\theta$ and $\varphi$, because of the symmetry of the problem. In other words, all that matters is how far away you are from the centre, your orientation is unimportant.
More interesting is the fact that the force is directly proportional to the distance from the centre, and in the opposite direction! If you wrote out the differential equation for this system, it would just be: $$\mathbf{a} = -\frac{GM}{R^3} r \,\,\hat{\mathbf{r}}.$$
This is just the differential equation for harmonic motion about the origin (i.e. the centre of the Earth)! By comparing it to the standard harmonic motion differential equation, you should be able to see that the angular frequency of oscillations $\omega$ is given by $$\omega = \sqrt{\frac{GM}{R^3}},$$ and so the time period is just $$T = \frac{2\pi}{\omega} = 2 \pi \sqrt{\frac{R^3}{GM}}.$$
For the Earth, this translates to a time period of $$T = 84.6 \text{ min}.$$
Curiously, this is also the time period of an infinite pendulum, admittedly for slightly different -- though related -- reasons.
What is particularly curious about the problem is the following: as you know, on the surface of the Earth, the Pendulum only approximates harmonic motion in the small-angle approximation. If the angle is too large, the force is no longer proportional to displacement. Interestingly, however, inside a solid sphere of constant density, the pendulum is truly harmonic! Which means you could move it by quite a large angle, and it would still take exactly $84.6 \text{ min}$ to oscillate!
It should also be clear that this time marks an upper bound on the amount of time it would take to complete a round-trip between any two points on the planet if you dug a tunnel between them and "fell" through it!
A: The answer depends on a detail that you have omitted: is the region of the Earth "within" the pendulum bob filled with mass, with a small planar slit cut to allow the pendulum to move friction-free, or is the pendulum in an empty pocket?
The answer for the first case has already been posted. The gravitational force on the bob acts towards the Earth's center with magnitude proportional to the distance from the center. The result is a nice harmonic motion with a period of roughly 84 minutes.
In the second case, though, no force acts on the bob. It is inside a uniform shell of mass, where the gravitational field is zero (as you can quickly show with Gauss's Law for gravity). The bob simply sits at rest.
A: The period can take a wide range of values.
If there is no mass "under" the pendulum, there is no gravitational force acting on it. In this case, the centripetal force acting on the bob will be determined entirely by the speed with which the pendulum is moving, which will determine the tension in the string. This is basically equivalent to swinging a tennis ball on a string over you head in a zero-G environment. You could pull the string lightly, imparting low centripetal acceleration and yield a long period, or pull the string very hard to impart a larger acceleration and a shorter period. If you don't give the pendulum a push to start, it'll never move at all, and if you do push it, the the period will be determined entirely by the initial velocity of the bob.
If there is mass under pendulum, the period can still take a wide range of values. You could effectively put the pendulum bob in orbit around the planetary core, and have the pendulum bob trace a circle without needing any string at all - the centripetal acceleration in this case is provided entirely by gravity, and will yield a period of about 85 minutes (see @Philip's answer). If the pendulum moved slower than that, it would fall into the core and stop oscillating. You can, however, make the pendulum circle the core even faster by imparting a larger initial velocity, in which case the centripetal acceleration is provided by the combination of gravity and tension in the string (this is basically how a space elevator works).
The period of the pendulum is upper-bounded by the amount of mass underneath the pendulum, and lower-bounded by tensile strength of the string. If the string is infinitely strong, you can make the pendulum orbit the core as fast as you want, equivalent to whipping a tennis ball on a string around your head as fast as possible.
The key difference from a typical pendulum is the fact that a typical pendulum always starts with an intial velocity of zero. In this case, if the pendulum starts with zero velocity, it will either fall into the core (if there is mass below), or not move at all (if there is no mass below) - you must select an initial velocity for the bob, which will be the only thing that determines the period of the pendulum. Since the initial velocity can take a wide range of values, so too can the period.
