Maximum separation between blocks on spring

  • Thread starter henry3369
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  • #1
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Homework Statement


Two blocks, each of mass m, are connected on a frictionless horizontal table by a spring of force constant k and equilibrium length L.

Find the maximum and minimum separation between the two blocks in terms of their maximum speed, vmax, relative to the table. (The two blocks always move in opposite directions as they oscillate back and forth about a fixed position.)

Homework Equations


Conservation of energy

The Attempt at a Solution


Ki + Ui = Kf + Uf
Ki = Uf
(1/2)mvmax2 = (1/2)kx2
Solve for x and the (1/2) cancel.
So:
x = sqrt(mvmax/k)
This gives the amount the string has stretched, so the distance between the blocks would be x + L which would result in:
Max Seperation = sqrt(mvmax/k) + L.
When I type in the solution I get a message that his is incorrect because the maximum separation does not depend on vmax.
 

Answers and Replies

  • #2
BvU
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Re-check the step (1/2)mvmax2 = (1/2)kx2, So: x = sqrt(mvmax/k)

For example with a dimension check.

And: what exactly is vmax if there are two blocks ? Is the centre of the spring fixed to the table ? Don't you have two half springs that way ?
 
  • #3
haruspex
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(1/2)mvmax2 = (1/2)kx2
In addition to BvU's comments, there are two blocks to store KE but only one spring to store PE.
 

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