Two Springs Displaced Horizontally

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1. Two springs each of natural length a and spring constant C are connected at one end(see figure).Consider a two dimensional displacement given by (x,y).
(a)Write the potential energy as a function of x and y.
(b)Find the force vector for a given (x,y) pair.

2. [tex]\vec{F}[/tex]=-k[tex]\vec{x}[/tex](Hookes Law) [tex]\vec{x}[/tex] is the displacement along the spring from equilibrium point. Magnitude wise l=L-L_{0}
3. First I assumed for the sake of simplicity potential energy as [tex]U(l)=(1/2)l^2C[/tex]and summed them up since U is a scalar quantity. Then to find the force I differentiated U with respect to x and y respectively. But I am very curious if this solution is right cause a friend of mine has provided another solution which does not correspond with mine. I am including both of the solutions as attachments in jpeg format. By the way I need a vectorial solution for this thanks for your contribution in advance.
 
Attachments
  • Handwritten derivation of potential energy for two springs with displacements, showing the vector sums, the integral of force along each spring, and the result U_total = 2axC, followed by the total force vector F = -C(Y + X) = (1a+x+a+x)i + (y-3)j.
    03122010349.jpg
    21.6 KB · Views: 482
  • Handwritten derivation of potential energy and force for two springs of natural length a and constant C, with a diagram of two springs meeting at a point displaced by (x, y), and the expression for F = -grad U.
    03122010347.jpg
    13.9 KB · Views: 467
Septim said:
1. Two springs each of natural length a and spring constant C are connected at one end(see figure).Consider a two dimensional displacement given by (x,y).
(a)Write the potential energy as a function of x and y.
(b)Find the force vector for a given (x,y) pair.




2. [tex]\vec{F}[/tex]=-k[tex]\vec{x}[/tex](Hookes Law) [tex]\vec{x}[/tex] is the displacement along the spring from equilibrium point. Magnitude wise l=L-L_{0}



3. First I assumed for the sake of simplicity potential energy as [tex]U(l)=(1/2)l^2C[/tex]and summed them up since U is a scalar quantity. Then to find the force I differentiated U with respect to x and y respectively. But I am very curious if this solution is right cause a friend of mine has provided another solution which does not correspond with mine. I am including both of the solutions as attachments in jpeg format. By the way I need a vectorial solution for this thanks for your contribution in advance.

The attachments are pretty hard to read. Can you re-scan them? Why are they so dark?
 
They are taken with a cell phone unfortunately I do not have the oppurtunity to scan it but I will do my best to provide better photos.
 
I think these photos taken without a flash is better. Our main problem is potential energy and force is dependent on the y component of the displacement or not?
 
Attachments
  • Handwritten physics solution on lined paper: derivation of total potential energy U = 2ax C for a spring system, with a labeled diagram of two springs and a vector triangle diagram above.
    03122010350.jpg
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  • Handwritten derivation on lined paper: a coil spring sketch labelled a, a spring stretched between two points with displacement y, and equations for the potential U(l), lengths l_A and l_B, force F = -grad U, and F = -C(...) expressions with square roots of (a+x)^2 + y^2 and (a-x)^2 + y^2.
    03122010351.jpg
    19.6 KB · Views: 448