# Deriving an expression for hte maximum displacement of a mass connected to a spring

1. Sep 28, 2009

### dozappp

1. The problem statement, all variables and given/known data
A block of mass 1.2kg, resting on a horizontal surface of coefficient of kinetic friction of .15, is attached to a linear spring of force constant 12.0 N/m. When the spring is unstretched the mass is at a position x=0. THe mass is pulled to the right streching the spring a distance x=0.8 m, and relased from rest at t=0. The mass accelerates the the left and momentarily comes to rest at point B.

a) Derive an expression for the speed of the mass as a function of x.
b)Calculate the maximum speed of the mass during this time. At what position x does this maximum speed occur?
c)Calculate the distance the spring is compressed whne the mass stops momentarily at point B.
d)Derive an expression for the maximum displacement of the mass at the end of each half oscillation as a function of hte number of half oscillations, n.

2. Relevant equations
Conservation of energy is what i used for a, b, and c. No idea how to do d.

3. The attempt at a solution

a)
.5*k*.8^2-.5*k*x^2-m*g*.15*(.8-x)=.5*m*v^2
k=12, m=1.2, g =9.8
sqrt(2(6*.8^2-6*x^2-1.764(.8-x))))=v
which simplifies to
v=Sqrt[-10x^2+2.94x+4.04]

b)
v=Sqrt[-10x^2+2.94x+4.04]
dv/dx = (-Sqrt[10]*(x-.147))/Sqrt[-x^2+.294+.404]
dv/dx = 0
x=.147
V(.147) = 2.06 m/s
therefore
2.06 m/s @ x=.147

c)
V=0
Sqrt[-10x^2+2.94x+4.04] = 0
x = -.505

.505 m compressed

d)

OK, I think I got the previous parts correct, but this part seriously breaks my brain. I simply do not know where to start. I tried to make a function like

1/2*k*X(n)^2-m*.15*g*(X(n)-X(n+1))=1/2*k*X(n+1)

but i dont know how to solve it. Many thanks in advance.

2. Sep 28, 2009

### dozappp

Re: Deriving an expression for hte maximum displacement of a mass connected to a spri

ok i redid my equasion to 1/2*k*((x(n))^2-mg(-1^n(x(n))-(-1^(n+1))((x(n+1))=1/2*k*x(n+1)^2