Mass m suspended by two springs in series

AI Thread Summary
The discussion centers on deriving the differential equation for a mass suspended by two massless springs in series, focusing on the principles of Simple Harmonic Motion. The effective spring constant for the two springs is calculated as k' = k/2 when both springs have the same spring constant k. The tension in each spring is uniform and equal to kx, leading to a total displacement of 2x for the mass. The correct differential equation is expressed as d²x/dt² + (k/2)x = 0, indicating the relationship between the mass and the spring system. The participants collaboratively clarify the concepts and calculations involved in the problem.
idir93
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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 ?
 
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idir93 said:
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 ?

[kx - mg = m (dv/dt) = m(d sqaure x/dt)]
 
idir93 said:
What would be the differential equation og this Simple Harmonic Motion when you have a mass m suspended to 2 springs in series ?
If the springs are considered massless, what do you think the relationship would be between the tensions in the two springs?
 
haruspex said:
If the springs are considered massless, what do you think the relationship would be between the tensions in the two springs?

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

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
 
idir93 said:
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
Compare the force stretching the springs to the overall amount of stretch. How does the overall system stretch compare to that of each spring?
 
How ?
 
idir93 said:
How ?
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
how can i get rid of x1 and x2 ?
 
  • #11
idir93 said:
how can i get rid of x1 and x2 ?
Express them in terms of F and k.
 
  • #12
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
It's okay now i totally got it thank you very much for your help :) please tell me which degree do you have ?
 
  • #14
idir93 said:
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.
When k1 = k2, k' = k/2.
Why in the first argument of the diff eq we have M2(dx²)/(dt) and in the 2nd we just have kx not 2kx ?
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
Thanks again
 
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