Dynamics problem - spring on a slope

[HmBe]

1. An elastic spring has modulus of elasticity $$\lambda$$ and natural length l₀. This spring is placed on the slope of a hill with an angle $$\alpha$$ to the horizontal such that one end of the spring is fixed at the foot of the hill and the other end can move freely along the slope. A body of mass m, starting from rest at the top of the hill, is moving down the hill (neglect friction). Knowing that the body sticks permanently to the free end of the spring after first contact, find its subsequent position x(t) for the case:


dx/dt=0 when the spring is compressed by mgl₀/$${\lambda}$$


2. F=ma
t=$$\lambda$$L/l₀ where L is the extension



3. I set my axis so that the bottom of the hill is x=0, with x increasing along the slope. I then rewrote L as l₀-x. Then I found the tension as a function of x, including a -mgsin$$\alpha$$. I set this equal to m*d²x/dt², and divided through by m, to get a ODE. However it is proving really tricky to solve, as I've got loads of constants and stuff I'm not quite sure what I'm meant to do with. I've tried using the initial conditions, but it just gets more and more messy.

This is my first post, sorry about the rubbish formatting. Hope this is in the right section. Let me know if I've done anything wrong. Cheers!

 

[HmBe]

Ok I tried a different approach, setting the x-axis so x=0 at l₀

L (the extension) = -x

T=-$$\lambda$$x/l₀

let w²=$$\lambda$$/l₀m

d²x/dt²+w²x=-gsin$$\alpha$$


Working through this I get x=g((sin$$\alpha$$-1)coswt-sin$$\alpha$$)/w²

Can anyone confirm this..?