Maximum separation between blocks on spring

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SUMMARY

The discussion centers on calculating the maximum separation between two blocks connected by a spring on a frictionless table. The derived formula for maximum separation is incorrect as it incorrectly incorporates the maximum speed, vmax. The correct approach involves recognizing that the maximum separation is determined solely by the spring constant k and the equilibrium length L, without dependence on vmax. The correct maximum separation formula is x + L, where x is derived from the spring's potential energy and kinetic energy conservation.

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  • Study the principles of conservation of energy in mechanical systems.
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henry3369
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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 separation = 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.
 
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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 ?
 
henry3369 said:
(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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