The force of gravity will simply move the equilibrium point, being a costant force. So if $l_0$ is the initial lenght of the spring, the applying the orizontal component of the force $Mg\sin \theta$, you'll get the new lenght in equilibrium that is (<PERSON>'s Law) $$ \vec{F} = -k\vec{(x - x_0)} \Rightarrow x = F/k + x_0$$ $$x_1 = Mg\sin \theta / k + x_0$$ Since gravity is costant over time, any displacement applied to the string will act as if the initial lenght of the string is $x_1$, not $x_0$. So the amplitude of harmonic oscillation won't change, but you must consider the amplitude relative to $x_1$, not $x_0$. 766
You have to put the $0$ in your ruler on $x_1$, rather than $x_0$.
You can prove this statement solving the differential equation: $$F = ma = m\frac{d^2x}{dt^2} = -k(x - x_0) + Mg\sin \theta $$ $$ m\frac{d^2x}{dt^2} +kx = kx_0 + Mg\sin \theta $$ there is no $x$ in the $Mg\sin \theta$ so you can simply consider $$kx_0 + Mg\sin \theta$$ as a costant term $kx_1$ so you'll get $$ \frac{d^2x}{dt^2} + \frac{k}{m}x = \frac{k}{m}x_1 $$ the amplitude is given solving the differential equation http://en.wikipedia.org/wiki/Ordinary_differential_equation it doesn't depend on $x_0$ or $x_1$.
Considering the problem using conservation of energy, you'll get: Let's say that $h_0$ is the height relative to $x_0$ so $h = x\cos \theta $ energy at $x_0$ will be $$ mgh_0 + \frac{1}{2}m\upsilon^2 + \frac{1}{2}k(x_0 - x_0)^2$$ energy at a displacement $\Delta x$ from $x_0$ $$mgh_0 + mg\Delta h + \frac{1}{2}m\upsilon^2 + \frac{1}{2}k(\Delta x)^2$$ every force is conservative so $$ mgh_0 + \frac{1}{2}m\upsilon_0^2 + \frac{1}{2}k(x_0 - x_0)^2 = mgh_0 + mg\Delta h + \frac{1}{2}m\upsilon_1^2 + \frac{1}{2}k(\Delta x)^2$$ if the cart is release with $\upsilon_1 = 0$ $$ \frac{1}{2}m\upsilon_0^2 = mg\Delta x \cos \theta + \frac{1}{2}k(\Delta x)^2$$ the speed at $x_0$ depends only from the inital displacement $\Delta x$ and not from $x_0$ or $x_1$ absolutely.
Theoretically the amplitude won't change, but since there are other factors as friction, measures aren't perfectly matching to theory. The inclined plane present less friction rather than an horizontal plane due to the vertical component of the weight of the cart. So taking friction into account total energy will decrease while the cart is moving so, it will decrease the amplitude of motion.
Hope this help! 526