How Do You Calculate the Frequency of Oscillation for a Massless Spring?

[DizBelieF14]

Homework Statement

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 25 cm below yi.

(a) What is the frequency of the oscillation?

Homework Equations

SHM Equations: X=Acos(), etc.
omega = sqrt(k/m)
T = 2PI * sqrt(m/k)

The Attempt at a Solution

a = w^2 * A ... w = 6.26, so f = 1 -- wrong obviously

I'm relatively sure I need to find omega to get the frequency, but for some reason I simply can't get omega. Any help would be great, because I've been staring at it for about an hour. I'm sure I'm missing something really stupid.

 

[Doc Al]

Hint: Use the information given to find k.

 

[DizBelieF14]

The first thing I tried to do was find K, but for some reason cannot see what to do. (I'm usually relatively good with this stuff.)

Obviously you know that F=KX, and I thought that the F was the weight, but then you do not know the mass. If you could give me anything else, it would be appreciated, but understand that it might be difficult since it's a rather easy problem. Either way, thank you.

 

[Doc Al]

I just noticed that you are not given the mass either. But no matter, you have all the information you need.

Here's another hint: Find the equilibrium point of the oscillation. What's the net force on the mass at that point?

 

[DizBelieF14]

At equilibrium, the net force/acceleration is zero. I thought that was where you would say that mg=kx, but once again you don't have m or x, and I can't see multiple equations that could be used. I'm assuming since you don't need the mass, it will cancel out somewhere? Or do you not use it at all...

Sorry, I don't know why I can't get this, I did the other 29 problems in less time than I've spent on this. :shy:

 

[Doc Al]

At equilibrium, the net force/acceleration is zero.

Right.

I thought that was where you would say that mg=kx,

Right! This is what you need.

but once again you don't have m or x, and I can't see multiple equations that could be used.

You have all the information you need to find x. You don't need m or k--all you need is enough info to solve for the frequency (using the equations you have for omega or period). Look at those equations and see how m and k appear.

Sorry, I don't know why I can't get this, I did the other 29 problems in less time than I've spent on this.

This one is trying to be cute. You'll get it.

 

[DizBelieF14]

For some reason, it's just not working out. Maybe if I come back later I will get it. Thank you for all of your help though, it's much appreciated.

 

[Doc Al]

Since I don't know what you are trying, I can't offer any specific advice on why it doesn't work.

Did you figure out x yet?

 

[DizBelieF14]

I know mg=kx will somehow get me the answer, but I don't know why I can't see how. I realize that if you move the k over, you get mg/k, and m/k is in the period equation, but can't seem to get it. (also, the other way around for the omega equation).

I keep getting x in terms of f or T... x=1/(4f^2)

Sometimes, my brain just decides to not work...

 

[Mindscrape]

That question doesn't make much sense from my perspective. If the spring is at the rest length, which I assume is meant to mean the equilibrium point, then when it is released nothing will happen.

If it is supposed to be mean that the spring is at a rest position, then is brought to some distance below the equilibrium that would make sense. Maybe it was just worded awkwardly. Still, I don't see how there is enough information (unless Doc Al sees something I don't). I think you need either how far below the ceiling the equilibrium point is, or how long it takes for the spring to do something, or the mass.

 

[DizBelieF14]

I'm under the impression that the "rest position" is the equilibrium position for simply the spring, which is above the equilibrium position with the object attached. Still, I'm stumped...

EDIT: Wait, X is just .125, since it goes down .25 ... Yeah, wow...

 

[Doc Al]

That question doesn't make much sense from my perspective. If the spring is at the rest length, which I assume is meant to mean the equilibrium point, then when it is released nothing will happen.

If it weren't for gravity, then the unstretched position would be the equilibrium point. But with gravity acting, it is not. The equilibrium point is where the net force on the mass is zero.

EDIT: Wait, X is just .125, since it goes down .25 ... Yeah, wow...

Now you're cooking!

 

[Mindscrape]

The equilibrium point is when all the forces acting on the spring sum to zero, which includes gravity. I guess that is where my misinterpretation comes in, and now I am curious how you are viewing it.

The differential equation would still be:
$$mx'' + kx = 0$$

Gravity does not need to be explicit:
The reason behind gravity not explicitly appearing in the differential equation has to due with building the equation around the equilibrium point. When the spring and mass are at rest the sum of the forces must all equal zero, and since there is no velocity or acceleration at this point then ks=mg where s is the distance from the ceiling to the mass. The only time a restoring force will be applied is when the spring has been stretched away from this equilibrium point, but if the spring is lifted up from the equilibrium point the force of gravity on the mass will be equal to the restoring force that would occur if it were stretched because ks=mg. In other words, whether the spring is lifted up or stretched down the forces will be equal; thus, the equation does not need to explicitly use a term for the force of gravity because it is implicit in the restoring force. However, from my point of view, the problem would have to give how far the spring is stretching from the ceiling or the mass so that the spring constant could be determined.

 

[Doc Al]

The equilibrium point is when all the forces acting on the spring sum to zero, which includes gravity. I guess that is where my misinterpretation comes in, and now I am curious how you are viewing it.

The differential equation would still be:
$$mx'' + kx = 0$$

Gravity does not need to be explicit:
The reason behind gravity not explicitly appearing in the differential equation has to due with building the equation around the equilibrium point.

All good. x is now the displacement from the equilibrium point, not the unstretched position.

When the spring and mass are at rest the sum of the forces must all equal zero, and since there is no velocity or acceleration at this point then ks=mg where s is the distance from the ceiling to the mass.

When the mass is oscillating, the velocity is a maximum at the equilibrium point.

The only time a restoring force will be applied is when the spring has been stretched away from this equilibrium point, but if the spring is lifted up from the equilibrium point the force of gravity on the mass will be equal to the restoring force that would occur if it were stretched because ks=mg. In other words, whether the spring is lifted up or stretched down the forces will be equal; thus, the equation does not need to explicitly use a term for the force of gravity because it is implicit in the restoring force.

Yes, the restoring force equals the combined effect of the the spring force (measured from its unstretched position) and gravity. It turns out that the combined force equals "kx" where "x" is now measured from equilibrium.

However, from my point of view, the problem would have to give how far the spring is stretching from the ceiling or the mass so that the spring constant could be determined.

You are given the upper and lower bounds of the motion, which should enable you to find the equilibrium point. (Note that when the mass is released the spring is unstretched.) That's all you need to know to solve the problem.