# Kinematics (rectilnear motion) spring.

1. Dec 25, 2013

### whynot314

the 14 in. spring is compressed to an 8 in length, where it is released from rest and accelerates black A. the acceleration has an initial value of 400ft/sec^2 and then decreases linearly with the x-movement of the black, reaching zero when the spring regains its original 14 in. length. calculate the time for the black to go to 3 in.

just wondering If I am on the right track. I believe that is the right function for acceleration of a displacement. I am unsure about initial velocity though. Yes I made the adjustment to a=-240x+3360
[URL=http://s1341.photobucket.com/user/nebula-314/media/20131225_150014_zpsac8609aa.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20131225_150014_zpsac8609aa.jpg[/URL][/PLAIN]

Last edited: Dec 25, 2013
2. Dec 25, 2013

### haruspex

How do you get x = -6in. for the compressed length?

3. Dec 26, 2013

### whynot314

yea idk I took x to be in the positive direction to the right, thus giving me a negative X.

I want to use the equation of acceleration as a function of position a=f(s). and.

$\int^{V}_{V0}$vdv=$\int^{S}_{s0}$f(s)ds

to derive me velocity as function of position then, use that to further drive the equations for time.

I am just having trouble with the initial acceleration function.

4. Dec 26, 2013

### haruspex

The spring has a relaxed length of 14 in. When it is compressed to being only 8 in. long, how much has it been compressed by?
Your diagrams on the RHS of the OP are correct, but the graph on the left is wrong. I think you have been inconsistent in what you mean by 'x'. In the lower diagram you have x as the change in length, which starts at 0 and becomes -6. But in the graph you have x changing from 14 to -6.

5. Dec 26, 2013

### whynot314

Ok my new strat, a= $\frac{-2400x}{7}$+4800
(general integration setup
$\int^{V}_{0}$vdv=$\int^{X}_{0}$ ($\frac{-2400x}{7}$+4800)dx

gives me V= $\sqrt{\frac{-2400x^2}{7}+9600x }$

then, velocity as a function of position.

$\int^{t}_{0}$dt=$\int^{3}_{0}$$\frac{dx}{ \sqrt{\frac{-2400x^2}{7}+9600x }}$

put that into wolfram
gives me .0360 sec, book says .0370 sec

Last edited: Dec 26, 2013
6. Dec 26, 2013

### whynot314

This time I used the position of 8 in as x=0, which allows me to generally set up an integration for velocity where initial position is zero. does this look correct?

7. Dec 26, 2013

### whynot314

got it guys finally!!!, Yes you were right haruspex I was having HUGE difficulty as to what was being meant by "X". Thanks for the help haruspex.

[URL=http://s1341.photobucket.com/user/nebula-314/media/20131226_161422_zps90413be0.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20131226_161422_zps90413be0.jpg[/URL][/PLAIN]

8. Dec 26, 2013

### haruspex

I confirm the answer. My method is rather different.
We know it will be SHM with an amplitude of 6 in.
Just plug the known facts in to find omega, then find the time at x = -3in.:
$x = -6 \cos(\omega t)$
$\ddot x = 6 \omega^2 \cos(\omega t)$
At t = 0:
$4800 = 6 \omega^2$, $\omega = 20\sqrt 2$
After 3 in. of expansion:
$x = 3 = -6 \cos(\omega t_3)$
$\omega t_3 = \pi/3$

9. Dec 26, 2013

### whynot314

Yea I saw those equations else where, but I wanted to use the method the book was giving me.