Wavefunction, its oscillating behaviour when $E > V(x)$ In Berkeley physics, for quantumphysics
It has been mentioned that -
The behavior of the wavefunction is oscillatory in the region where $E - V(x) >0$.
Mathematically it is correct - 
What is its physical interpretation? 
Is it correct to say that particle is oscillating? Or what reference can be made from the fact that behaviour is oscillatory? 
Intuitively one can ask why it will be oscillatory, just from taking energy into consideration.
 A: 
Is it correct to say that particle is oscillating ? 

It is not the particle that is oscillating,but the wavefunction is sinusoidal, due to the nature of wave equations. By the postulates of quantum mechanics, the meaning of the wavefunction is, when complex conjugate squared, to give the probability of finding a particle at (x,y,z,t). It is the probability that has oscilatory behavior, not the particle, particularly elementary particles which are point particles.
A: I can name two intuitive reasons. Firstly if the oposite is true, $E-V<0$, then the solution becomes a decaying exponential. So in this classicaly forbidden region the chance of finding the particle drops of very quickly.
Secondly the term $E-V$ is the kinetic energy operator: it tells you how quickly a particle moves (under some circumstances). For a free particle with solution $e^{ikx}$ the kinetic energy is given by $\hbar^2k^2/2m$. The Schrödinger equation becomes
$$\frac \partial{\partial t}\psi(x,t)=i\left(\frac {\hbar k^2}{2m}\right)\psi(x,t)$$
This means that, at a certain point, the phase of the wavefunction rotates and the larger $k$ the faster this phase rotates. This fast rotation of the phase translates in faster movement. A nice example is a wavepacket in the harmonic potential $V(x)=\frac 1 2 m\omega^2 x^2$. In this gif below you see that when the spatial frequency is large (large $k$, lots of colours) the wavepacket moves the fastest and stops moving when the kinetic energy is the smallest. For eigenfunctions this comparison breaks down because the solutions aren't actually moving but you should still keep in mind: large $E-V$ means the phase rotates fast.

Source of the gif:
https://physics.stackexchange.com/a/285557/93729
A: Note that while the free particle wave $e^{i k x}$ is oscillatory, it has a constant amplitude $|e^{i k x}| = 1$ for all $x$. So the probability of finding the particle is the same everywhere in space. The physical interpretation comes when you have e.g. interference between particles, where the phase of the oscillation can lead to destructive and constructive interference.
By the way, I am assuming that you mean $E > V_0$, where $V_0$ is a constant. The wave function in the $E > V(x)$ part of e.g. a harmonic oscillator is not oscillating in the same way.
