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Paul9
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Hi all! I'm having a bit of trouble writing the differential equation that governs this mechanical system of two springs and a hydraulic damper in series. Since there is no mass present I believe the resulting DE will be a first order equation of the form Ax'(t) + x(t) = f(t), where f(t) is the forcing function. However, I can not figure out how to manipulate the equations from the FBDs into this form. Once I have the DE I do not anticipate any trouble solving it. Any and all help you can provide to point me in the right direction is greatly appreciated.
Find the transfer function, Xo(s)/Xi(s), for the mechanical system in the diagram below. The displacements xi and xo are measured from their respective equilibrium positions.
Obtain the displacement, xo(t), when the input xi(t) is a step displacement of magnitude xi occurring at t=0.
Components in series require the force to be constant throughout the chain. Let's call this force "P".
The FBD at Junction I we will call Eq. 1:
P = k1*(xi-y)
FBD at Junction II we will call Eq. 2:
k1*(xi-y) = k2*(y - xo)
FBD at Junction III we will call Eq. 3:
k2*(y - xo) = b1*x'o
Since we are interested in the input and output, (xi and xo), my first thought is to eliminate the variable y by solving for it in Eq. 2 and substituting into Eq. 3.
y =[itex]\frac{k_{1}x_{i}+k_{2}x_{o}}{k_{1}+k_{2}}[/itex]
Subbing into Eq. 3:
[itex]k_{2}(\frac{k_{1}x_{i}+k_{2}x_{o}}{k_{1}+k_{2}}-x_{o}) = b*x'_{o}[/itex]
Dividing by [itex]k_{2}[/itex] and rearranging we get:
[itex]\frac{b}{k_{2}}x'_{o} + x_{o} = \frac{k_{1}x_{i}+k_{2}x_{o}}{k_{1}+k_{2}}[/itex]
Is this correct?
Again, thanks for any guidance you can give me :)
Homework Statement
Find the transfer function, Xo(s)/Xi(s), for the mechanical system in the diagram below. The displacements xi and xo are measured from their respective equilibrium positions.
Obtain the displacement, xo(t), when the input xi(t) is a step displacement of magnitude xi occurring at t=0.
Homework Equations
Components in series require the force to be constant throughout the chain. Let's call this force "P".
The FBD at Junction I we will call Eq. 1:
P = k1*(xi-y)
FBD at Junction II we will call Eq. 2:
k1*(xi-y) = k2*(y - xo)
FBD at Junction III we will call Eq. 3:
k2*(y - xo) = b1*x'o
The Attempt at a Solution
Since we are interested in the input and output, (xi and xo), my first thought is to eliminate the variable y by solving for it in Eq. 2 and substituting into Eq. 3.
y =[itex]\frac{k_{1}x_{i}+k_{2}x_{o}}{k_{1}+k_{2}}[/itex]
Subbing into Eq. 3:
[itex]k_{2}(\frac{k_{1}x_{i}+k_{2}x_{o}}{k_{1}+k_{2}}-x_{o}) = b*x'_{o}[/itex]
Dividing by [itex]k_{2}[/itex] and rearranging we get:
[itex]\frac{b}{k_{2}}x'_{o} + x_{o} = \frac{k_{1}x_{i}+k_{2}x_{o}}{k_{1}+k_{2}}[/itex]
Is this correct?
Again, thanks for any guidance you can give me :)
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