- #1
MMS
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Homework Statement
Given: Two masses M between 3 springs of constant K. equilibrium point for the left mass is x=a, for the right one x=b. We move the right mass to point x=b+A
1. What is the return force?
2. What is the work required to create the initial condition?
3. Can we write the enrgy of the system as: E=(1/2)*M*((w1)^2)*(A/2)^2+(1/2)*M*((w2)^2)*(A/2)^2 ?
4. Is there a time where Ψleft=A
The attempt at a solution
I'm currently stuck on 1 and it's pretty much confusing me but here is what I thought of:
The net force on each mass is the return force (for each).
For the left mass:
Well the left spring isn't affected at all since there is no displacement in it's position, therefore the only thing that this mass feels is K*A to the right.
Therefore, the return force is: F=K*A(x^^), x^^=x(hat)
For the right mass:
It feels to the right a force of -KA and to the left a force of KA. Summing those up with the right directions gives us a return force of F=-2K*A (x^^), x^^=x(hat)
All this, while taking into consideration that there is no displacement in the left mass.
To be honest I'm not quite sure if what I defined as the return force is actually the return force so I'd love some help with that as well.Any help is much appreciated. Thank you.
Given: Two masses M between 3 springs of constant K. equilibrium point for the left mass is x=a, for the right one x=b. We move the right mass to point x=b+A
1. What is the return force?
2. What is the work required to create the initial condition?
3. Can we write the enrgy of the system as: E=(1/2)*M*((w1)^2)*(A/2)^2+(1/2)*M*((w2)^2)*(A/2)^2 ?
4. Is there a time where Ψleft=A
I'm currently stuck on 1 and it's pretty much confusing me but here is what I thought of:
The net force on each mass is the return force (for each).
For the left mass:
Well the left spring isn't affected at all since there is no displacement in it's position, therefore the only thing that this mass feels is K*A to the right.
Therefore, the return force is: F=K*A(x^^), x^^=x(hat)
For the right mass:
It feels to the right a force of -KA and to the left a force of KA. Summing those up with the right directions gives us a return force of F=-2K*A (x^^), x^^=x(hat)
All this, while taking into consideration that there is no displacement in the left mass.
To be honest I'm not quite sure if what I defined as the return force is actually the return force so I'd love some help with that as well.Any help is much appreciated. Thank you.