# Understanding spring mass system

1. Apr 19, 2005

### chandran

I am just learining basic dynamics from spring mass system.
there is a horizontal spring fixed to the wall and a mass is at the end.
When the mass is displaced by a force F to the right the spring stretches to a distance X. Now i draw the STATIC FBD. The
eqn is f-kx=net force on the mass. Now i draw the kinetic diagram and the equation i derive is

f-kx=ma where a is the acceleration of the mass m. So for this spring mass system the dynamic equation is
f-kx-ma=0.

Is what i have done is true and correct?

2. Apr 19, 2005

### dextercioby

Is the force acting continuously on the object...?Or it just stretches the string & lets it go to oscillate...?

That equation,vector or scalar,f-kx-ma=0 makes no sense...

Daniel.

Daniel.

3. Apr 19, 2005

### ramollari

Yes, it is correct as long the hand pulling the mass is not releasing it.

$$F - kx = 0$$

4. Apr 19, 2005

### Staff: Mentor

This is just a statement that the net force on the mass equals f - kx.
Now you've applied Newton's 2nd law: Net Force = ma. Note that rewriting it as "Net Force - ma = 0" adds nothing.

If the mass is in equilibrium, then f - kx = 0; if not, then it will accelerate.

5. Apr 23, 2005

### marlon

Here, you can learn more on basic dynamics. make sure that you are able to solve the given exercises...