# Gravitational force in simple harmonic motion

1. Sep 7, 2014

### Linaran

1. The problem statement, all variables and given/known data
I am trying to derive the formula for simple harmonic motion of a mass hanging on a spring.

I understand the derivation for the situation when the mass and the spring are on an horizontal table. Then I go about deriving the same formula for the situation when a mass is hanging on a spring and I can't seem to get rid of the gravitational force components (F=m*g).

2. Relevant equations
The picture describes my problem very well.
http://prntscr.com/4koogs
I can't get rid of the m*g!!

3. The attempt at a solution
I'm very stuck.
I know you guys aren't supposed to hand out answers but this is a matter of understanding the fundamental I would really like a good explanation.

2. Sep 7, 2014

### BvU

The mg makes that y0 is not the same as the position of the end of the spring when there is no mass hanging from it...
With the spring on the table that's different.

3. Sep 7, 2014

### Staff: Mentor

You can choose "y" in a different way. You used the rest length of the spring as y=0, in this frame your formula is correct. If you want to reproduce the horizontal system with the formula, you'll have to pick a different position for y=0.

4. Sep 7, 2014

### Linaran

A professor on youtube set his y0 this way http://prntscr.com/4kqerq

But, if I understand correctly, the equilibrium point should be the point where the acceleration of the mass is 0?
How can I simply change the equilibrium point to my liking?

EDIT: if someone wants reference on which video was that.

EDIT2: Is the professors reference suppose to be a floor or something? I just can't see why should there be 0 acceleration.

Last edited by a moderator: Sep 25, 2014
5. Sep 8, 2014

### BvU

OK, I understand the circumstances now.

The blackboard shot shows a spring without any weight attached at left. The line y=0 is indeed chosen at the point where force from spring and force from gravity cancel each other: the point where the acceleration is zero. Not the forces, but the sum of the forces (the vector sum: one is down and equal in magnitude to the other, which is up). Note that the height of this y=0 depends on m (which isn't the case when the spring is horizontal on a table).

Middle drawing shows a weight attached, and we are led to assume this is a snapshot at the moment immediately after the hand (or other support) under the weight was taken away: speed is zero, no equilibrium at all, spring not stretched yet, so acceleration g downwards. Professor leaves it to us to guess if y(0) = y0 or not. A second look tells us not (y(0) = yinitial so to his credit: he works out the more general case).

The one on the right is then a generic drawing of an intermediate point, for which the differential equation will probably be set up in the following (sorry, don't have time to watch the whole thing).

When I now look at your notes in post #1, I see more or less the same, only you start at the lower end.

In your left picture you write y0 - equilibrium and I am pretty sure that you don't mean a minus sign but a $\equiv$, i.e. $y = 0 \Leftrightarrow$ equilibrium.

Your middle picture leads me to look at it as the point of maximum stretching at time 0, at the moment the weight is let go: y(0) = A, spring is stretched by A+y0 and $\dot y(0) = 0$.

You work it out and worry about the constant. But on the way, you step from $\vec F_S = k\left ( y_0 - y(t) \right)\hat\imath = -k \; y(t) \hat\imath$.
a) It should be $-y_0$
b) If you leave the $-y_0$ in, your problems are over, since you already know that $-k\; y_0 + mg = 0$.

(So I could have skipped the first seven paragraphs altogether: you understand things quite well but go a little too fast. Just like almost all of us )

Unstuck now ?

6. Sep 8, 2014

### BvU

If you have -- in another situation, because here the constant is now gone! -- a differential equation like$$\ddot y = -ky + C$$there is a simple way to get rid of the C by indeed shifting the reference point: define $y'= y + C/k$ and since ${d\over dt}\left (C/k\right ) = 0$ the differential equation for y' becomes $$\ddot y'= -ky = -k\left( y'- C/k\right) + C = -ky'$$

7. Sep 8, 2014

### Staff: Mentor

You cannot change the equilibrium point in space (this has a physical meaning), but you can change the definition of the coordinate system (this has not).

8. Sep 8, 2014

### BvU

Did we loose you, Linaran ? Or can some more clarification, different wording, or whatever else, shed more light on this issue? It is quite relevant, so it's good to understand it thoroughly. From your picture in post #1, I am quite optimistic in that respect.

9. Sep 8, 2014

### Linaran

I am sorry I had a physics exam today (not written but the other one -- "A+" english) and the whole day was a circus but I am glad to report I passed the exam.

I will be reading your instructions very slow and detailed BvU I just need some rest :)

10. Sep 9, 2014

### BvU

Congrats and take it easy!

11. Sep 9, 2014

### Linaran

I reviewed you instructions and I understand them they were very thorough, thanks for help :)