# Spring and Mass Oscillating on a Slope

1. Apr 16, 2009

1. The problem statement, all variables and given/known data

An elastic spring has modulus of elasticity $$\lambda$$ and natural length $$l_{0}$$. The spring is 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 forces). The body sticks permanently to the free end of the spring after first contact.

2. Relevant equations

$$\dot{x}=0$$ when the spring is compressed by $$mgl_{0}$$/$$\lambda$$

3. The attempt at a solution

I set the x axis going to the slope with the origin at $$l_{0}$$ and had $$mgsin\alpha$$ as the force going down the hill and $$x\lambda$$/$$l_{0}$$ as the force going up.

With this i said that $$m\ddot{x}$$=mgsin$$\alpha- x\lambda$$/$$l_{0}$$ and so found the equation of motion to be

x=Asinwt+Bcoswt+$$mgl_{0}\alpha$$/$$\lambda$$ where w$$^{2}$$=$$\lambda$$/$$ml_{0}$$

Then i find B by saying that x(0)=0 (setting t=0 to be when the mass meets the spring), but then when i try and find A by saying x(T)=$$mgl_{0}$$/$$\lambda$$, but i get stuck with A being in terms of tanwT, and when i try to find tanwT i find the value i get for coswT is impossible. Have i set up the situation correctly?

2. Apr 25, 2009

### dpal9

(I took out the latex from the post as it doesnt seem to be working)..

When you said:
Do you mean x=Asinwt+Bcoswt+(mgl_0*sin(alpha))/lambda?

Also, don't you find B by differentiating the above with respect to t, and then equating this to your initial condition of x'=0?

i.e... x'(t)= w(Acoswt - Bsinwt)

so x'(0) = 0 (your initial condition)
x'(0) = wA

so A=0 ?

3. Apr 26, 2009