# Basic spring question

1. Nov 1, 2007

### Goldenwind

A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.0 cm before reaching its new equilibrium length. The block is then pulled down slightly and released.

What is the frequency of oscillation?

I don't know how I'd get the frequency, so I just started with the things I knew.
F = ma = -kx
9.81m = -0.02k

...and then I got stuck.

Am I going in the right direction? If so, where next? If not, what should I try instead?

2. Nov 1, 2007

### Staff: Mentor

The frequency of oscillation of a mass on a spring is a function of the mass and the spring constant. Either look up the formula or derive it yourself by setting up the differential equation. (Look it up!)

Even though you don't have the mass or the spring constant, you are given enough information to make use of the formula (once you find it).
Good. You will definitely need this result.

3. Nov 1, 2007

### Goldenwind

Just the answer I was looking for - a point in the right direction, but not giving too much away. Thanks :)

Edit: Found the equation, got stuck again, got unstuck, and now happily finished the question. Thanks again ^^;

Last edited: Nov 1, 2007
4. Apr 23, 2008

### sleepy126zzz

can you give me a little more help with this? i still don't get it.

5. Apr 24, 2008

### Staff: Mentor

Where exactly are you stuck? Have you figured out the spring constant?

6. Apr 24, 2008

### sleepy126zzz

no, because i don't know what the force is. how do i find that?

7. Apr 24, 2008

### Staff: Mentor

It's just the weight of the block. Assume the block has mass "m". (You're not given the actual mass, but you don't need it.)

8. Apr 24, 2008

### sleepy126zzz

ok thanks! it's due at 11 so i'm gonna try to figure it out.