Spring and Mass Oscillating on a Slope

[adg203]

Homework Statement

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.

Homework Equations

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

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?

 

[dpal9]

(I took out the latex from the post as it doesn't seem to be working)..

When you said:

found the equation of motion to be

x=Asinwt+Bcoswt+(mgl_0*alpha)/lambda

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 ?

 

[adg203]

Hi,

Yeah i did mean sin(alpha). I agree that if you let x'(0)=0 then A=0, however we would then need a second equation to find B and we do not know what x(0) is. We are only told that the mass is at the top of the hill, and we are given no detail as to where the top of the hill is.