# Mass m suspended by two springs in series

1. Oct 27, 2012

### idir93

Hello guys i'm desperately trying to understand the solution of this assignement .
What would be the differential equation og this Simple Harmonic Motion when you have a mass m suspended to 2 springs in series ?

2. Oct 27, 2012

### Puneeth423

[kx - mg = m (dv/dt) = m(d sqaure x/dt)]

3. Oct 27, 2012

### haruspex

If the springs are considered massless, what do you think the relationship would be between the tensions in the two springs?

4. Oct 27, 2012

### idir93

yes the springs are massless, i got this answer from a textbook : the tension is uniform in the upper spring and has a magnitude of kx, hence the tension in the lower spring also has a magnitude of kx
the displacement of M is 2x.
the diff eq is :
M2(dx²)/(dt) + kx = 0

Why 2 in the first argument and not in the second since we have two springs.

5. Oct 27, 2012

### Staff: Mentor

See if you can figure out what the effective spring constant is for the two springs in series.

6. Oct 27, 2012

### idir93

that's the problem my friend :( it's the reverse of series resistors on a circuit.
K (equivalent) = (k1k2)/(k1 + k2)
But i don't know how to get to it. maybe by moving backward

7. Oct 27, 2012

### Staff: Mentor

Compare the force stretching the springs to the overall amount of stretch. How does the overall system stretch compare to that of each spring?

8. Oct 27, 2012

### idir93

How ?

9. Oct 27, 2012

### Staff: Mentor

Imagine two springs in series, with spring constants k1 and k2. A force F stretches the two springs, thus F = k1x1 = k2x2.

For the system as a whole, you have:
F = k'(xtotal)

See if you can solve for k'.

10. Oct 27, 2012

### idir93

how can i get rid of x1 and x2 ?

11. Oct 27, 2012

### Staff: Mentor

Express them in terms of F and k.

12. Oct 27, 2012

### idir93

Thanks i did it with your big help, but i still don't understand this case when k1=k2 and x1=x2 because in my problem the two springs have the same k and since they are massless they'll be stretchend with the same x making a total of 2x in the displacement.
Why in the first argument of the diff eq we have M2(dx²)/(dt) and in the 2nd we just have kx not 2kx ?

13. Oct 27, 2012

### idir93

It's okay now i totally got it thank you very much for your help :) please tell me which degree do you have ?

14. Oct 27, 2012

### Staff: Mentor

When k1 = k2, k' = k/2.
d2x/dt2 + k'x = 0

d2x/dt2 + (k/2)x = 0

Multiply both sides by 2!

Edit: Looks like you figured it out while I was typing this.

15. Oct 27, 2012

Thanks again