# Spring motion

1. Nov 30, 2013

### Psyguy22

1. The problem statement, all variables and given/known data
When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled down from this position and released. It undergoes 57 oscillations in 39.0 s. What was the distance d?

2. Relevant equations
F=-Kx
F=mg
T=2pi*sqrt(K/m)

3. The attempt at a solution
I started by putting the first two equations equal.
$-K x= m g$
Solving for x you get
$[1] (m g) /-K= x$
Then I solved for m/K in the third equation.
$[2] (2 π / T )^2=m / K$
Then I solved for osc/s which came to be 1.46 osc/s
Substituting {2} for {1} I came up with
$g*(- 2 π/ T)^2= x$ BUt i feel like this equation is wrong. Any help?

2. Nov 30, 2013

### xophergrunge

Do not set the first two equations equal to each other, they are not equal. The two forces acting on the mass, the weight and spring force, add to give the total force which will be zero when x=d.

3. Nov 30, 2013

### Staff: Mentor

This is incorrect. You have K and m reversed.

Here the problem is that minus sign. Set the magnitudes of the forces equal.

Fix those problems and you'll be OK.

4. Nov 30, 2013

### Psyguy22

Ok i fixed the two mistakes, but I'm still coming up with the wrong answer. I think I have something wrong with the peiod but am unsure how to fix it

5. Nov 30, 2013

### Staff: Mentor

The period is the time it takes for one oscillation. Above you solved for the osc/sec, which is the frequency, not the period.