# How to find max velocity in a spring-mass system?

Tags:
1. Jul 11, 2017

### Helly123

1. The problem statement, all variables and given/known data
Two masses connected with a spring with contant k. The string streched by l . Find the max velocity of mass m!
M2 ___spring___ M1

M2--stretched by x2--____spring____--x1--M1
l = x1+ x2
2. Relevant equations

F = k.l
Mass1.x1 = mass2.x2
(x= displacement?)
a=w^2 x
v = wx

Ep + Ek1 = Ep + Ek2
Ep = 1/2kx^2
Vmax at equilibrium = $\sqrt{\frac{A^2k}{m}}$

3. The attempt at a solution
kl = m.a
a = k.l/m
While l = x1 + x2
After i get a, i can find w, and can find v

My question is, why can't i use Vmax at equilibrium = $\sqrt{\frac{A^2k}{m}}$ ? The A = x1
m1x1 = m2x2
X1 = m2.(l-x1)/m1
X1=m2.l /(m1+m2)
Substitute x1 to A. But i get different answer.

2. Jul 11, 2017

### TSny

Is k the effective spring constant for m1's motion? Hint: Write Hooke's law in terms of x1 rather than L.

Last edited: Jul 11, 2017
3. Jul 12, 2017

### Helly123

Ah, i get it..
The Hooke's law = k.x1
F=m.a
m1.a = kx1
K= m1.a/x1
Then substitute k to equation.
So, when we use total constant the displacement is the total L. While we use only partly displacement the k constant different one?

4. Jul 12, 2017

### TSny

Yes. The force felt by each mass is F = kL where L is the total amount of stretch of the spring from equilibrium. Using some of your equations in your first post, you should be able to express L in terms of just x1. Then you can identify the effective spring constant k1 for mass m1.

5. Jul 12, 2017

### J Hann

If the system can move freely then isn't the total momentum always zero?

6. Jul 12, 2017

Yes.