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.

 

[rwisz]

As a scientist, it is important to understand the principles and assumptions behind the methods used to solve a problem. In this case, both methods are correct but they are based on different assumptions and principles.

The first method, using forces in equilibrium, assumes that the spring is not stretched beyond its natural length and therefore the force exerted by the spring is directly proportional to the displacement of the mass. This is known as Hooke's Law. In this case, the spring is considered to be ideal and there is no energy loss or gain.

The second method, using energy bookkeeping, takes into account the potential energy stored in the spring due to its deformation. This method assumes that the spring is not ideal and there is some energy loss or gain as the mass is attached to it. This is a more accurate representation of a real-world scenario.

Therefore, both methods are correct but they are based on different assumptions and principles. It is important to understand these principles in order to determine which method is more appropriate for a given situation. In this case, the second method may be more accurate as it takes into account the energy loss in the spring, but the first method can still provide a good approximation in certain situations.