How to find the spring constant?

[jrodgers]

Homework Statement

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.

Homework Equations

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

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.

 

[Doc Al]

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

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.

 

[jrodgers]

Thank you for replying.

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.

 

[Doc Al]

A. Yes I see, the equilibrium point for the spring is at height d where the spring has no elastic potential energy.

Not once you've attached the mass. The equilibrium point will be where mg = kx.

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.

The bottom position (in method #1) is not the equilibrium position.

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

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.

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.

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.

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.

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.

 

[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?

 

[Doc Al]

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

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

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?

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

 

[jrodgers]

Thank you so much, your feedback was most helpful.