# Homework Help: How to find the spring constant?

1. Aug 26, 2009

### jrodgers

1. The problem statement, all variables and given/known data
When a mass m is placed on the end of a spring in the vertical position, it stretches the spring a distance d. Find the spring constant k.

2. Relevant equations
Force of a spring = Fs = -ky
Elastic Potential Energy = Us= (1/2)ky^2
Gravitational Potential Energy = Ug = mgy

3. The attempt at a solution
Method 1 - Conservation of total mechanical energy.
When the spring is not streched, the mass has only gravitational potential energy Ug=mgd. After the spring drops a distance d, the mass has only elastic potential energy. Hence
mgd = (1/2)kd^2
or
k = 2mg/d
Method 2 - Newton's second law.
When the mass has dropped a distance d, it has streched the string a distance d. Hence the sum of the forces in the vertical position gives
Fs -Fg=0
or
kd-mg=0
or
k=mg/d

My question is, why do these two methods give different values?

Thank you.

2. Aug 26, 2009

### Staff: Mentor

Are you attaching the mass to the unstretched spring and then dropping it? Or are you gently lowering the mass to its equilibrium position?

Your method #1 assumes that you let it drop and that "d" is the distance to the lowest point of its oscillation. Note that that lowest point is not the equilibrium position.

Your method #2 assumes that you just lowered the mass gently, placing it at its equilibrium position.

Last edited: Aug 26, 2009
3. Aug 26, 2009

### jrodgers

A. Yes I see, the equilibrium point for the spring is at height d where the spring has no elastic potential energy. At height d, the potential energy for the mass is mgd. At the bottom (height 0, the equilibrium position of the system), the system has no potential energy, only the elastic energy of the spring.

B. But these are both conservative forces. The path taken between initial and final times should not matter?

C. Since Method 2 is just a static analysis, a mass in equilibrium; I don't see how the history of how the system was assembled can make a difference? Draw the free-body-diagram that currently exists, and then use Newton's second law.

D. I'm supposing in Method 1 that the spring is so stiff that it gently lowers the mass to its final reference position, making the path followed by Method 1 and Method 2 identical.

4. Aug 26, 2009

### Staff: Mentor

Not once you've attached the mass. The equilibrium point will be where mg = kx.
The bottom position (in method #1) is not the equilibrium position.

In method #2 you assume it is at rest at the equilibrium point. That means a non-conservative force was introduced (your hand, perhaps) that lowered the spring gently.

You can do a force analysis at any point and draw a free body diagram. But the mass will only be in equilibrium at one point.

That won't work. If you gently lower the mass until it's at equilibrium, the spring will stretch by x = mg/k. If you drop the mass, it will oscillate about the equilibrium point. At its lowest point, the spring will have stretched by an amount equal to 2mg/k.

5. Aug 26, 2009

### jrodgers

Q1. So you can't have an over damped vertical spring-mass system that just comes straight to rest?

Q2. So what you are saying is that both methods give the correct equation it's just that the d will be different in each case?

6. Aug 26, 2009

### Staff: Mentor

Why can't you? (But that requires a damping force.)

Either method can be used to find the spring constant; yes, the d will be different in each case.

7. Aug 26, 2009