Is Huygens-Fresnel principle applicable for waves other than light? In Wikipedia it was mentioned Luminous disturbance so I Did get confused that this principle only works for light waves and not for all of the Waves. Like some mechanical waves example wave on string.
Wiki text:

In 1678, Huygens [1] proposed that every point which a LUMINOUS DISTURBANCE reaches becomes a source of a spherical wave; the sum of these secondary waves determines the form of the wave at any subsequent time.

 A: The principle that every point on a wavefront can be thought of as an emitter of spherical (or, in 2D, circular) waves is applicable to any waves - see any introductory high school course on waves, where the demonstrations are typically done with water surface waves. Note that, as @ignacio pointed out, the construction is only exact in odd dimensions (in practice that means 3D), but even in 2D it is quite convincing:
Nice demonstration with water waves in 2D
And the link given by Ignacio:
Mathematical proof that Huygens construction is only valid in odd dimensions
A: Then smaller a slit, then clearer you can see, that behind the slit the dissipation of the water wave happens in a pure circular way (Huygens spherical waves). This seems clear if one realise, what happens if you put a pin into water. The pin displaces the water and a part of this displaced water due to the elasticity (which is the possibility of being displaced but not to get broken) does not goes into deepness; the water dissipates around the pin. Since there was water too, the displaced water accumulates, the water level gets higher and one get a wave around the pensil. Due to the displacement direction this wave moves away from the pin and - due to the elasticity of the water - behind the maximum amplitude you see a minimum amplitude and so on until the energy is dissipated to heat (what happens very fast, if you not move the pin up und down periodically to support the wave production).
The slim slit is nothing else than a half pin. A wider slit is like a oscillating rod instead of the pin. The dissipation process is always a circular process and seen best at the ends of the oscillator bar or the edges of a slit in an obstacle or the pin (which means, at every point of the pin surface).
You mentioned mechanical waves. If I understood right, you mean waves in elastic solid bodies. There is no principle difference between such waves and water waves. Waves phenomena is per definition energy transfer without material transfer. A first difference is in the amount if energy one need to see the waves. For metal bodies it is easier to hear the wave instead to see the wave. A second difference is, how fat the heat dissipation happens. In liquides it happens fast, in metal bodies it could happens very slow (slow damping process, best seen for artful designed shape and well composed material like in a bell). Inalastic bodies do not vibrate, but such bodies do not exist in reality.
A two-dimensional body like a rope has a variying tension along the rope during the vibrating motion. This lead to heat dissipation. To be precise, the rope is a three-dimensional body, which could be vibrated in a two-dimensional way. The tension is a three-dimensional process and the dissipation is threedimensional too; the rope section does variying.
Additional remark
For electromagnetic radiation there are different processes; behind a slit we have nearby the edges deflection but not dissipation processes. Water waves always dissipate in the medium and produce heat (increase the temperature of the medium). For EM radiation in vacuum this dissipation doesn't happens. Nearby edges there is a pure deflection (otherwise one will see a color shift in this part of the radiation; perhaps this will happens due to scattering processes, but this seems to be negligible). The deflection of EM radiation is not the same as the dissipation for water waves.
To make a clear statement, Huygens spherical waves lower the wave amplitudes, that is what we call dissipation. The deflection of EM radiation is different. It changes the direction of some part of the radiation behind edges (not taking in account scattering processes, this seems negligible too), but does not dissipate the energy into the vacuum. This is the reason, why the cant catch any light far enough left or right the deflection pattern. The intensity distribution behind an obstacle will end and the shadow is really a shadow. Remember water waves behind an obstacle, there is the wave really a spherical.
