Potential energy of two masses connected by 3 identical strings

  • Thread starter mccoy1
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  • #1
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Hi fellows,
'Two particles of mass m are connected by three identical springs of relaxed length l and constrained to move longitudinally. What is the potential energy when the two masses are displaced from their equilibrium positions?

What I know: let k = force constant of the strings, x1 and x2 be displacements of the masses, so:
P.E (V) = (k/2)x1^2+(k/2)x2^2 + (I know there is going to be a third term here but I don't know what it's)..
The book says that that third term is (k/2)[x2-x1]^2....how did they derive that? Any insight would be appreciated.
Thanks you all.
 

Answers and Replies

  • #2
what do you think is the compression of the middle spring,
1/2(kx^2),right
where x is the cange in length of the middle spring ,that is (x^2 = (x1-x2)^2)
 
  • #3
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what do you think is the compression of the middle spring,
1/2(kx^2),right
where x is the cange in length of the middle spring ,that is (x^2 = (x1-x2)^2)

Thanks for the help. I still have problem though. If that's a compression of the middle spring, then the two masses are moving towards each other( hence compressing the middle spring). How about if the two particles are moving in the opposite directions?
 
  • #4
then it's the elognation, which will again lead to a rise in the potential energy of the spring
 
  • #5
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Great...so would it still be x2-x1 or x2+x1 this time?
 
  • #6
one of the terms x1, or x2 will get negative
we are measuring them from their eqilibrium positions
this means that the equation holds for both the cases

and yes ,were x1 and x2 magnitudes of displacement only ,what you have put forth
would hol beyond doubt.
 
  • #7
117
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Great. Thank you for your time and help.
 

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