What are the equations modelling a vertical spring system with two masses?Modeling a vertical spring system with one mass is a pretty common problem. I looked around online and found some horizontal spring systems with two masses, but no examples of a vertical one.I'm curious, how would you set up equations modeling a vertical spring system like this: − − − − ∧ ∨ ∧ ∨ ( m 1 ) ∧ ∨ ∧ ∨ ( m 2 )Where the first spring has constant k 1 , and the second k 2 . I'll let y 1 ( t ) be the position of the top mass away from its equilibrium, and y 2 ( t ) the position of the bottom mass away from its equilibrium. I choose down to be the positive direction.For the bottom mass, there is an upward force of k 2 y 2 , and a downward gravitational force of m 2 g. So one equation should be m 2 y 2 ″ = − k 2 y 2 + m 2 gFor the top mass, the first spring pulls up with force − k 1 y 1 and a downward gravitational force of m 1 g. I'm not sure how to account for the forces of the second spring and second mass acting of the first mass. Is the equation something like m 1 y 1 ″ = − k 1 y 1 + m 1 g + something?I'm just curious how you would correctly set up the equations for this system.

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What are the equations modelling a vertical spring system with two masses?Modeling a vertical spring system with one mass is a pretty common problem. I looked around online and found some horizontal spring systems with two masses, but no examples of a vertical one.I&#039;m curious, how would you set up equations modeling a vertical spring system like this:  −  −  −  −    ∧    ∨    ∧    ∨    (      m    1    )    ∧    ∨    ∧    ∨    (      m    2    )Where the first spring has constant       k    1  , and the second       k    2  . I&#039;ll let       y    1    (  t  ) be the position of the top mass away from its equilibrium, and       y    2    (  t  ) the position of the bottom mass away from its equilibrium. I choose down to be the positive direction.For the bottom mass, there is an upward force of       k    2        y    2  , and a downward gravitational force of       m    2    g. So one equation should be      m    2        y    2    ″    =  −      k    2        y    2    +      m    2    gFor the top mass, the first spring pulls up with force   −      k    1        y    1   and a downward gravitational force of       m    1    g. I&#039;m not sure how to account for the forces of the second spring and second mass acting of the first mass. Is the equation something like      m    1        y    1    ″    =  −      k    1        y    1    +      m    1    g  +   something?I&#039;m just curious how you would correctly set up the equations for this system.

accionsanjoseC9L · Accepted Answer

Step 1Since it seems you are not incorporating a damping term, I&#039;ll proceed under that assumption.Let       k    i   denote the spring constant for the spring above mass       m    i  ,   i  =  1  ,  2. Then by Newton&#039;s Second Law   F  =  m  a and Hooke&#039;s Law which says that the restorative force of the y units beyond its equilibrium position, i.e.       F    restorative    =  −  k  y, we get                              m          1                          y          1          ″                                    =                                                            −                                  k                  1                                                  y                  1                                            ⏞                                            Hooke&#039;s Law from above                                                                              −                                  k                  2                                (                                  y                  1                                −                                  y                  2                                )                            ⏟                                            Hooke&#039;s Law from below                          ,                                      m          2                          y          2          ″                                    =                                                            −                                  k                  2                                (                                  y                  2                                −                                  y                  1                                )                            ⏟                                            Hooke&#039;s Law from above                              which is a linear coupled system in       y    1   and       y    2  .Step 2To see how Hooke&#039;s Law leads to these terms, first displace mass       m    1   by       y    1   units (taking &quot;down&quot; as the positive direction). Then Hooke&#039;s Law says spring 1 has a restoring force of   −      k    1        y    1  . Now pull mass       m    2   down       y    2   units. Since the total displacement of mass       m    2   from its equilibrium position is       y    2    −      y    1  , this explains the Hooke&#039;s Law term   −      k    2    (      y    2    −      y    1    ) in the second equation. Finally, to see where the second Hooke&#039;s Law term in the first equation comes from, just note that that will be the same force we just found for the second mass,   −      k    2    (      y    2    −      y    1    ), but when it acts on the first mass, it will be in the opposite direction, so it will be   −      k    2    (      y    1    −      y    2    ) which we can write as   +      k    2    (      y    2    −      y    1    ) if you prefer.

What are the equations modelling a vertical spring system with two masses? Modeling a vertical spring system with one mass is a pretty common problem. I looked around online and found some horizontal spring systems with two masses, but no examples of a vertical one.

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