Help with springs and potential energy problem

[GeoMike]

Ok, this seems simple, and it probably is, but I'm just not seeing why I'm getting two non-equivalent solutions...

Homework Statement

"A massless platform is supported by 4 equal springs (same length, same spring constant, etc.)
A mass m is placed on the center of the platform causing an equal compression of x in each spring. What is the spring constant k of a single spring in terms of the mass m, the acceleration of gravity g, and the compression of the spring x?

Homework Equations

F_spring = -kx
F_grav = mg
U_spring = 1/2(kx^2)
U_grav = mgx

The Attempt at a Solution

What's throwing me off is that there seem to be two valid ways to solve this, each resulting in a different value of k...

If I use the fact that the force acting on each spring is 1/4 the force of gravity acting on the mass I get:

kx=mg/4
So... k = (mg)/(4x)

But, then isn't the elastic potential energy stored in each spring 1/4 the gravitational potential energy in the mass when it is first placed on the platform (before any compression)? If so, I get:

1/2(kx^2) = mgx/4
So... k = (mg)/(2x)

The back of the book gives (mg)/(4x). Which I understand, the problem is that the second answer seems valid to me, but obviously the second one is wrong. I just can't see where is my error in reasoning. Help?

Thank you!
-GM-

 

[PhanthomJay]

Ok, this seems simple, and it probably is, but I'm just not seeing why I'm getting two non-equivalent solutions...

Homework Statement

"A massless platform is supported by 4 equal springs (same length, same spring constant, etc.)
A mass m is placed on the center of the platform causing an equal compression of x in each spring. What is the spring constant k of a single spring in terms of the mass m, the acceleration of gravity g, and the compression of the spring x?

Homework Equations

F_spring = -kx
F_grav = mg
U_spring = 1/2(kx^2)
U_grav = mgx

The Attempt at a Solution

What's throwing me off is that there seem to be two valid ways to solve this, each resulting in a different value of k...

If I use the fact that the force acting on each spring is 1/4 the force of gravity acting on the mass I get:

kx=mg/4
So... k = (mg)/(4x)

But, then isn't the elastic potential energy stored in each spring 1/4 the gravitational potential energy in the mass when it is first placed on the platform (before any compression)? If so, I get:

1/2(kx^2) = mgx/4
So... k = (mg)/(2x)

The back of the book gives (mg)/(4x). Which I understand, the problem is that the second answer seems valid to me, but obviously the second one is wrong. I just can't see where is my error in reasoning. Help?

Thank you!
-GM-

That is a very sharp observation on your part. The error in your reasoning is that mechanical energy is not conserved, since you must apply a non conservative force to slowly lower the beam to its new rest postion; otherwise it would osscilate back and forth about its final rest position, extending the spring twice as much in so doing.

 

[GeoMike]

That is a very sharp observation on your part. The error in your reasoning is that mechanical energy is not conserved, since you must apply a non conservative force to slowly lower the beam to its new rest postion; otherwise it would osscilate back and forth about its final rest position, extending the spring twice as much in so doing.

So basically I forgot to account for the fact that if the mechanical energy is conserved (i.e. the mass is simply released, not slowly lowered) then an oscillation about x occurs requiring an additional term to account for the fact that there is a non-zero velocity as the mass passes through x (it has kinetic energy as well).

Right?
-GM-

 

[Doc Al]

Right!