# Spring/SHM question

1. Apr 13, 2006

### jimmy_neutron

Here is my question, hope someone can help.
Q: You attach a mass m on a spring with a spring constant k. You then lower the mass so it hangs on the spring in equilibrium.
a) If you pull the mass down a distance y, so its below the equilibrium position, then let go, the mass will bounce up and down. Draw the force diagram for the mass just after you let go.
b)At this instant, what is the net force on the mass? Is the force upwards or downwards? and is the force linear with distance y?
c)Will the periodic bouncing of the mass be SHM?

For a), the forces acting on the mass would be Fs which points upward, and mg which points downward. What am I missing?
For b) would the net force just be F= Fg + Fs ? The spring force is upwards and the force of gravity is downards, so the net force is 0?? I am not sure if it is linear.
For c) yes it will be SHM, because the net force is proportional to the displacement from the equilibrium point?

Thanks for the help!!!

2. Apr 13, 2006

### Hootenanny

Staff Emeritus
The force of a spring is given by $F = -kx$. Reconsider your answers to (a) and (b)

Regards
-Hoot

Last edited: Apr 13, 2006
3. Apr 13, 2006

### thunder

Hello hootenanny, sound slike you know your physics pretty well! If you have aminute, I am needing some hlp with a tough couple of problems dealing with Boyl's Law...I posted it in this topic. Can you help me? Thanks, hoot!

4. Apr 13, 2006

### jimmy_neutron

I know F= -kx, but I am not sure how that changes my answers to part a and b, can you explain in further detail
Thanks for the help.

5. Apr 13, 2006

### Hootenanny

Staff Emeritus
Well, why would the force be $Fs$ as you have stated? (mg is correct btw)

6. Apr 13, 2006

### jimmy_neutron

Well Fs = kd
the spring constant and the distance the mass is pulled, is a force acting on the mass right?

7. Apr 13, 2006

### Hootenanny

Staff Emeritus
Yes, so you should write -kx, not Fs.

That goes for part (b) also. Take note of the direction of the force.