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Mass Spring Damping system

  • Thread starter IState21
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  • #1
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Homework Statement


fjl3dl.png

That is the system.

B1 and B2 are the dampers
I am asked to obtain the differential equation model for the system. I haven't dealt with these in a while so I'm not sure where to start.

Homework Equations


Like I stated earlier i haven't messed with one of these in a while so I'm not really sure how to get started. Any help to get the wheels goin would be greatly appreciated.
 

Answers and Replies

  • #2
collinsmark
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I am asked to obtain the differential equation model for the system. I haven't dealt with these in a while so I'm not sure where to start.

Homework Equations


Like I stated earlier i haven't messed with one of these in a while so I'm not really sure how to get started. Any help to get the wheels goin would be greatly appreciated.
The direct approach is to treat each mass individually (at least at the beginning), and apply Newton's second law of motion,

[tex] m_n \underline{\ddot y_n} = \sum \vec F_n [/tex]

[Edit: In my notation above, the n subscript refers to the particular mass in question, and not any particular force acting on that mass. (There may be multiple forces associated with a given n.)]


You'll end up with a set of simultaneous differential equations (in this case, two simultaneous differential equations, one equation for each n).

(Hint: I see 3 forces acting on m1 and 5 forces acting on m2)

Once you have your set of simultaneous differential equations, you could express them together using linear algebra (if the equations are linear -- which they are in this case), or whatever representation your coursework calls for.
 
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  • #3
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For the first equation I get
[tex]m_{2}\frac{d^{2}y}{dt^{2}}=F_{1}-K_{1}y-B_{1}\frac{dy}{dt}[/tex]
does that look correct?
 
  • #4
collinsmark
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For the first equation I get
[tex]m_{2}\frac{d^{2}y}{dt^{2}}=F_{1}-K_{1}y-B_{1}\frac{dy}{dt}[/tex]
does that look correct?
No not quite.

Are you sure you mean m2 for this equation and not m1?

Also the force felt on m1 by the spring k1 depends on the difference between y1 and y2; not just y1 alone.

Similarly, the force felt on m1 by the dampener B1 is proportional the difference in velocities between m1 and m2.
 
  • #5
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No not quite.

Are you sure you mean m2 for this equation and not m1?

Also the force felt on m1 by the spring k1 depends on the difference between y1 and y2; not just y1 alone.

Similarly, the force felt on m1 by the dampener B1 is proportional the difference in velocities between m1 and m2.
You are right, I meant to put [tex]m_{2}[/tex]

So are you saying
[tex]m_{1}\frac{d^{2}y}{dt^{2}}=F_{1}-K_{1}(y_{1}-y_{2})-B_{1}(\frac{dy_{1}}{dt}-\frac{dy_{2}}{dt}))[/tex]
 
  • #6
collinsmark
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So are you saying
[tex]m_{1}\frac{d^{2} {\color{red}y}}{dt^{2}}=F_{1}-K_{1}(y_{1}-y_{2})-B_{1} \left( \frac{dy_{1}}{dt}-\frac{dy_{2}}{dt} \right)[/tex]
That looks quite a bit better. :approve:

(I modified the format of the parenthesis a little. Also red emphasis mine.)

Just make sure you label the y in the d2y/dt2 term. Remember, there's two ys, y1 and y2. You need to specify which one.
 
  • #7
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That looks quite a bit better. :approve:

(I modified the format of the parenthesis a little. Also red emphasis mine.)

Just make sure you label the y in the d2y/dt2 term. Remember, there's two ys, y1 and y2. You need to specify which one.
for m2 i got:

[tex]m_{2}\frac{d^{2}y_{2}}{dt^{2}}=F_{2}+K_{1}y_{1}+B_{1}\frac{dy_{1}}{dt}-k_{2}y_{2}-B_{2}\frac{dy_{2}}{dt}[/tex]

how does that look?
 
  • #8
collinsmark
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for m2 i got:

[tex]m_{2}\frac{d^{2}y_{2}}{dt^{2}}=F_{2}+ {\color{red}K_{1}y_{1}}+{\color{red}B_{1}\frac{dy_{1}}{dt}}-k_{2}y_{2}-B_{2}\frac{dy_{2}}{dt}[/tex]

how does that look?
Similar [but not identical to] before, the force from spring k1 depends on the difference in position of y1 and y2. Similarly, the force from the dampener B1 depends on the difference between the velocities of the two masses.
 
  • #9
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Similar [but not identical to] before, the force from spring k1 depends on the difference in position of y1 and y2. Similarly, the force from the dampener B1 depends on the difference between the velocities of the two masses.
You are right, i forgot to figure that in

[tex]m_{2}\frac{d^{2}y_{2}}{dt^{2}}=F_{2}+K_{1}(y_{1}-y_{2})+B_{1}(\frac{dy_{1}}{dt}-\frac{dy_{2}}{dt})-k_{2}y_{2}-B_{2}\frac{dy_{2}}{dt}[/tex]

correct?
 
  • #10
collinsmark
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You are right, i forgot to figure that in

[tex]m_{2}\frac{d^{2}y_{2}}{dt^{2}}=F_{2}+K_{1}(y_{1}-y_{2})+B_{1}(\frac{dy_{1}}{dt}-\frac{dy_{2}}{dt})-k_{2}y_{2}-B_{2}\frac{dy_{2}}{dt}[/tex]

correct?
'Looks good to me! :approve:
 

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