Amplitude of motion for a cart on an inclined plane attached to a spring A very practical question.
So, let's imagine that we have a small cart attached to a spring lying on an inclined plane. If I move my cart by 10 cm (measured as if there were a ruler lying along the inclined plane) such that the spring is compressed, and then release it, will my cart oscillate in simple harmonic motion with an amplitude of 10cm measured along the plane? 
I thought about this for some time now. But, I think the answer is simply no. Because there will always be the impact of a component of the force of gravity. So, the cart should still move in SHM, but not with an amplitude of 10 cm, right? If I am wrong, please explain with detail.
 A: 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 (Hooke'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 $x_0$. 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! 
