Difference between "Periodic motion" and "Oscillating Motion" So far I know one of them is a special case of the other: The Oscillating motion being the special case of Periodic motion. But I don't know the precise "Kinematical definition" of each one. I mean when you have an "Equation of motion" for a particle, how will you determine it's either a "Periodic motion" or an "Oscillating motion"? If some periodic functions appear in an equation of motion, can we call it a "Periodic motion"? If so how can we recognize it from "Oscillating motion"?
 A: In what follows I will only talk about the motion in one spacial dimension. Let me use the following definitions.


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*Periodic motion - the one in which all the processes repeat after the period $T$. 

*Oscillating motion - the one in which we have turning points.
Clearly, the second is a subclass of the first. Now let me explain in the language of the potentials. Again, in (classical!) mechanics we only have two options:
a) Motion in a bounded periodic potential with the energy above the absolute maximum, e.g. $V(x) = \sin(x)$ with $E>1$.
b)  Motion with turning points. There are two possibilities when this can happen, but they are indistinguishable in (classical!) mechanics.
Namely:
b1) Motion in a potential which has an absolute maximum, with the energy below this maximum (it's also useful to require for the energy not to coincide with any of the maxima in order to avoid infinite periods). Example: $V(x)=\sin (x)$ with $-1<E<1$
b2) Motion in potential of type $V(x)\to\infty$ as $|x|\to \infty$ with an arbitrary high energy. Exmaple: $V(x) = x^{2n}$, $E>0$.
Using our classification, we can say that both a) and b) cases belong to the $1$ class, while it is only $b$ that belongs to $2$.
OK, you may wonder now if these two cases are physically different. Of course, they are. It can be best understood when one thinks of such quantities as momentum or the (abbreviated) action.
Let's introduce the following definitions. Under the assumption that the system is closed, we can write its Hamiltonian as
$$H = \dfrac{p^2}{2m} + V(x) = E$$
where $E$ is the value of the Hamiltonian function - the energy of the system. Since the energy is conserved, we can solve the equation above for the momentum:
$$p(x) = \sqrt{2 m (E - V(x))}$$
In the case b), two interesting things happen to the momentum at the turning points: 1) It vanishes 2) It changes the sign. Neither of those happen in the case a).
Now let's talk about the classical abbreviated action, defined as the integral of the momentum over the period:
$$S = \int_{\mathcal{C}} p(x) \operatorname{d}x$$
The contour $\mathcal{C}$ denotes the part of the $x$-axis between the turning points, traveled there and back. The integral is nonzero dues to the fact that the momentum has changed its sign. If you are familiar with the complex analysis, this can be formulated in a slightly different way. We can complexify the position $x$ and consider the integral above as an integral along the closed contour in the complex $x$-plane. This integral is nonzero due to the presence of the singularity inside it. This singularity is the branch cut of the momentum - it does not let you to shrink the contour.

A good way of thinking about the motion in the periodic potential above the absolute maximum is thinking of the motion of the circle. It helps to imagine a finite interval of space, and a particle which by the end of the period $T$ goes from one to another end of it.
Now, you can ask yourself what is the Riemannian surface of the momentum in this case, and why is the integral nonzero - indeed, we have no singularities! In this case it is the topology who is responsible for the fact that the integration contour cannot be shrunk. Namely, the Riemannian surface is now either an infinite strip in the complex plane, or a parallelogram (depending on the analytic properties of the potential).
These ideas have very deep and interesting consequences:


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*Studying the global properties of the Riemannian surfaces allows to calculate certain things in a very cute and beautiful way (like calculating the classical action via employing the Picard-Fuchs equation for the Riemannian surface of the momentum)

*It turns out that exactly same ideas can be applied to the corresponding quantum-mechanical problems due to the fact that the quantum analogue of the momentum lives on the same Riemannian surface.
A: From a physical stand point they are very similar. When we consider the orbit of a planet around a star, we consider its period. Hence, orbits are periodic in nature. We don't refer to the oscillations of the planet.
In contrast, say we have a spring with a mass attached, and set it out of equilibrium. We don't typically talk about the periodic nature of the system, but rather its oscillatory behavior (the oscillation of the system). This is not to say that the system does not exhibit periodic motion, as the system will oscillate with some period.
Having an equation of motion, we are able to discern different types of oscillatory behavior such as forced, damped and coupled oscillators. These are all cases of oscillation being a 'special case' of periodic motion, as they contain elements of periodic motion (sin, cos, etc)
A: From what I understand, periodic motion from the physical point of view is quite general in the sense that any type of motion that repeats itself after some period of time would be term as periodic.
Where in the case of oscillation, the time period of the periodic motion is quite large, like the motion of the pendulum upto a scale of about seconds and above.
Which I find the periods having small periods concerning to vibrations.
In representing with a graph, the periodic motion would be repeating curves at some constant intervals, the intervals being time.
The curves which are comparable to our senses would be called oscillation and with very high frequency would be called vibration, where both are periodic motions in general.
