Solving a Simple Mass-Spring Problem: Two Methods, Two Answers

[mnova]

Homework Statement

I work a simple problem in two different ways and get two different answers. I'd like to know which way is wrong, and why.

This is a simple mass-spring problem. The spring hangs from a ceiling, and a mass is attached to its end.

We know the spring constant, k, and the mass, m, The problem is to find x, how much the spring has stretched from its natural length by the mass.

The Attempt at a Solution

First way - Forces in equilibrium
The mass is acted upon by two forces, its weight downward, and the spring tension upward. Since the mass is not moving, the forces must cancel.
kx = mg. or x = mg/k (answer one)

Second way - energy bookkeeping
After the mass is attached to the end of the spring, it falls the distance x. It loses potential energy mgx. This energy must be stored in the spring. Since F = -kx = -dV/dx where V is the springs potential energy, then V = kx**2/2 . This must be equal to the potential energy lost by the mass.
kx**2/2 = mgx or x = 2mg/k (answer two)

Where are my analyses going astray?

 

[PhanthomJay]

You haven't gone astray, these are 2 different problems. For case 1, you slowly lower the mass with your hand applying an external force, until it rests in its equilibrium position, and you let go of your hand. The spring extension is mg/k. For case 2, you let the mass fall, thus extending the spring twice as much , per conservation of mechanical energy , with no other external forces applied.

 

[mnova]

PhanthomJay,

Am I following you correctly?

Let's call the amount the spring stretches in case #1: x1
and for case #2 (in which the mass falls): x2

Are you saying:
In case #2, the mass drops and overshoots x1, falling to x2 before bouncing back. It then oscillates about x1, finally damping to a stop after one-half the potential energy gained by the spring at x2 has been dissipated.

 

[PhanthomJay]

PhanthomJay,

Am I following you correctly?

Let's call the amount the spring stretches in case #1: x1
and for case #2 (in which the mass falls): x2

Are you saying:
In case #2, the mass drops and overshoots x1, falling to x2 before bouncing back.

yes, where x2 = 2x1, for an ideal spring.

It then oscillates about x1, finally damping to a stop

for an ideal spring, it will keep on oscillating, but for a damped spring, yes

after one-half the potential energy gained by the spring at x2 has been dissipated.

Since the PE is a function of x^2, the PE of the spring at it's at rest equilibrium position is less than 1/2 of its PE at its max point.