2 mass spring system

1. Nov 23, 2007

-RooneY-

Two blocks of masses m1 and m2 are connected by a spring of spring constant k.now forces F1 and F2 start acting on the two blocks in a manner to stretch the spring.Find the max.elongation of the spring.

What I could figure out was to conserve the energy. Energy stored in spring when elongation is x would be 1/2kx^2. Work done by F1 would be F1x and by F2 would be F2x. So,equating them, we would get x=2(F1 + F2)/k.

Thanks

2. Nov 24, 2007

-RooneY-

Anyone ?
The system will have some acceleration too. So, there must be some other catch in the problem.

3. Nov 24, 2007

azatkgz

Hmmmm....

Because the forces are not equal,the whole system starts to move.

$$F_1-kx=m_1a$$

$$kx-F_2=m_2a$$

4. Nov 24, 2007

-RooneY-

Yes..but that isnt really helping in find the max elongation. The elongation will increase and after reaching the max value, will start to decrease again.

5. Nov 24, 2007

azatkgz

$$F_1-F_2=(m_1+m_2)a$$

$$F_1-T_1=m_1a$$

$$T_2-F_2=m_2a$$

energy stored in the spring equals to

$$E=T_1x_1+T_2x_2$$

6. Nov 24, 2007

-RooneY-

Isnt T1=T2=kx ?

7. Nov 24, 2007

azatkgz

No,I just confused they are not egual.Spring is oscillating,this makes them to be inequal.

8. Nov 24, 2007

-RooneY-

Hey thanks

Got the answer as 2(f1M1 +f2M2)/k(M1+M2)

Thanks again

9. Nov 25, 2007

azatkgz

Must be

$$\frac{1}{2}k(x_1+x_2)^2=\frac{(F_1m_2+F_2m_1)}{m_1+m_2}(x_1+x_2)$$

so $$(x_1+x_2)=0$$ minimum elongation and

$$(x_1+x_2)=2\frac{F_1m_2+F_2m_2}{k(m_1+m_2)}$$ maximum.

10. Nov 25, 2007

-RooneY-

Hey wait....but wouldnt energy stored in spring be $$F_1x_1 + F_2x_2$$ since $$F_1 , F_2$$ are the external forces on the spring system ?

11. Nov 25, 2007

azatkgz

$$T_1x_1+T_2x_2=(F_1-m_1a)x_1+(F_2+m_2a)x_2$$

$$(F_1-F_2)d=\frac{1}{2}kx^2+\frac{1}{2}(m_1+m_2)v^2$$

if energy stored in the spring were $$F_1x_1 + F_2x_2$$,it means that spring is not moving.