# Find out my mistake!

Suppose an object of mass 'm' is attached to a vertical spring of spring constant 'k' and slowly lowered to the equilibrium position. We have to find the extension in the spring.

I have two approaches-
1) At the equilibrium position,
force due to tension in the spring and weight of the object balance each other
i.e. kx = mg (x = extension produced in the spring)
or x = mg/k

2) Consider the object-spring-earth system.
When the block is lowered by a distance x, its potential energy decreases by mgx (taking the equilibrium position as reference level).
The elastic energy of the spring increases by 1/2 kx2.
Energy is conserved as no other external force on the system does work.
so mgx = 1/2 kx2
or x = 2mg/k

I'm curious to find out which one is correct.
Please provide convincing answers to both approaches.

## Answers and Replies

Filip Larsen
Gold Member
It may help to think about what "slowly lowered" really means.

'Slowly lowered' means there is no increase in kinetic energy of the block as it descends. I did not take any kinetic energy expression in my conservation of energy equation.

Borek
Mentor
mgx = 1/2 kx2

Is this an equilibrium position?

Doc Al
Mentor
Energy is conserved as no other external force on the system does work.
Then how did you 'slowly lower' the mass to the equilibrium point? The key is that mechanical energy is not conserved and that there is an external force on the system.

Abdul,imagine adding the mass to the spring whilst holding it at its unstreched length and then suddenly releasing the mass.What do you see happening next and how does the system eventually settle down?How does this differ from the event described in your question?

Is this an equilibrium position?

Sorry if my sentence was misleading. By equilibrium position I meant that the position of unstretched length of the spring is taken as 0 potential energy level.

Then how did you 'slowly lower' the mass to the equilibrium point? The key is that mechanical energy is not conserved and that there is an external force on the system.

I got your point. But how do we find out the magnitude of that external force which we are providing to slowly lower it down?

Abdul,imagine adding the mass to the spring whilst holding it at its unstreched length and then suddenly releasing the mass.What do you see happening next and how does the system eventually settle down?How does this differ from the event described in your question?

I think the spring will be stretched more in this case ( amplitude of oscillation will be more ) and the system settles down at the same equilibrium position as that in the situation given in my question.

As Doc Al pointed out, I think the only difference is that we are not providing any external force to the block in suddenly releasing the mass.

Borek
Mentor
Sorry if my sentence was misleading. By equilibrium position I meant that the position of unstretched length of the spring is taken as 0 potential energy level.

I was not referring to the initial, I was asking if the position calculated using your second approach is the equilibrium position - that is, whether the mass will stay there for ever.

You have calculated something, and you have calculated it correctly, question is, whether you have calculated what you think you did.

Doc Al
Mentor
I got your point. But how do we find out the magnitude of that external force which we are providing to slowly lower it down?
The external force required will be whatever is needed to give zero net force. For example, when you are just beginning to lower the mass the spring force is zero. So the external force must equal the weight of the mass. By the time you've reached the equilibrium position, the external force needed is zero, since the spring force will equal the weight of the mass.

If you calculate the work done by that external force (which will be negative work), you'll find that it accounts for the 'missing' mechanical energy.

I was not referring to the initial, I was asking if the position calculated using your second approach is the equilibrium position - that is, whether the mass will stay there for ever.

You have calculated something, and you have calculated it correctly, question is, whether you have calculated what you think you did.

Yes, that position is equilibrium position.
I did not calculate it correctly. The answer in my second approach is wrong!

The external force required will be whatever is needed to give zero net force. For example, when you are just beginning to lower the mass the spring force is zero. So the external force must equal the weight of the mass. By the time you've reached the equilibrium position, the external force needed is zero, since the spring force will equal the weight of the mass.

If you calculate the work done by that external force (which will be negative work), you'll find that it accounts for the 'missing' mechanical energy.

In that case the work done by the external force equals -mgx + 1/2kx2. After substituting this extra term in L.H.S. of my energy conservation equation, I found that it satisfies the equation. How do we find out the extension using energy conservation then?

Borek
Mentor
Yes, that position is equilibrium position.

And that's what you are missing. This is NOT an equilibrium position.

And that's what you are missing. This is NOT an equilibrium position.

Yes you are right. The equilibrium position in this case will be still lower.

Borek
Mentor
No, equilibrium position will be higher. You are calculating displacement as the mass drops down.

Doc Al
Mentor
In that case the work done by the external force equals -mgx + 1/2kx2. After substituting this extra term in L.H.S. of my energy conservation equation, I found that it satisfies the equation. How do we find out the extension using energy conservation then?
Not sure what you're looking for. The equilibrium position is defined by kx = mg, not by energy conservation. (Your second approach is wrong.)

Not sure what you're looking for. The equilibrium position is defined by kx = mg, not by energy conservation. (Your second approach is wrong.)

Yes, you cannot use energy conservation to get equilibrium position. The reason is simple. Imagine yourself positioning the mass hung from a spring. While you working to position the mass gently, you are applying external forces to make it settle down. So unless you account for these forces, the K.E wont balance. So yu must use the force balance to get the equilibrium position. How much energy would that external force over a distance translate to? Initially the spring wont provide any force and so you would have to support the entire weight. As you lower, you would need lesser force (linearly). So the average force would be (mg/2) over a distance of x. so work = mgx/2

So mgx - mgx/2 = kx^2/2

you get x = mg/k, exactly what you got with force balance!

Ok, Now what if you did not extract that energy of the falling mass? ie you just dropped the mass.

It would overshoot the equlibrium position! By how much? by twice the "X" we found above. But when it is at 2X position(momentary rest), the forces dont balance !!
So a net force up exists and so it goes up to get to "X". But again it overshoots!!!. It is now executing an oscillation!!

Another way to look at this problem:
===========================
When you gently allowed to settle down, you always made sure that the netforce is zero (no acceleration). With increase in spring force with distance you supported less weight by that amount. So it settled down at x = mg/k

When you dropped it suddenly there was a net force that is mg and inertial forces that did not become zero until X=2x. At X, inertial force vanished and only mg and KX remained. This is an imbalance and hence bounced up to catch up with position "x". But it overshoots

The velocity changed from 0 to maximum at "x" and then back to 0 at X=2x
The accelaration changed from max "g" to 0 at "x" and "-g" at X=2x

Last edited:
Yes, you cannot use energy conservation to get equilibrium position. The reason is simple. Imagine yourself positioning the mass hung from a spring. While you working to position the mass gently, you are applying external forces to make it settle down. So unless you account for these forces, the K.E wont balance. So yu must use the force balance to get the equilibrium position. How much energy would that external force over a distance translate to? Initially the spring wont provide any force and so you would have to support the entire weight. As you lower, you would need lesser force (linearly). So the average force would be (mg/2) over a distance of x. so work = mgx/2

So mgx - mgx/2 = kx^2/2

you get x = mg/k, exactly what you got with force balance!

Ok, Now what if you did not extract that energy of the falling mass? ie you just dropped the mass.

It would overshoot the equlibrium position! By how much? by twice the "X" we found above. But when it is at 2X position(momentary rest), the forces dont balance !!
So a net force up exists and so it goes up to get to "X". But again it overshoots!!!. It is now executing an oscillation!!

Another way to look at this problem:
===========================
When you gently allowed to settle down, you always made sure that the netforce is zero (no acceleration). With increase in spring force with distance you supported less weight by that amount. So it settled down at x = mg/k

When you dropped it suddenly there was a net force that is mg and inertial forces that did not become zero until X=2x. At X, inertial force vanished and only mg and KX remained. This is an imbalance and hence bounced up to catch up with position "x". But it overshoots

The velocity changed from 0 to maximum at "x" and then back to 0 at X=2x
The accelaration changed from max "g" to 0 at "x" and "-g" at X=2x

I did not read your post earlier.
Thanks for your explanation