Kinematics (rectilnear motion) spring.

In summary: Thanks for the confirmation.In summary, 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.
  • #1
whynot314
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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]
 
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  • #2
How do you get x = -6in. for the compressed length?
 
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  • #3
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.

[itex]\int^{V}_{V0}[/itex]vdv=[itex]\int^{S}_{s0}[/itex]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
whynot314 said:
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.

[itex]\int^{V}_{V0}[/itex]vdv=[itex]\int^{S}_{s0}[/itex]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.
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.
 
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  • #5
Ok my new strat, a= [itex]\frac{-2400x}{7}[/itex]+4800
(general integration setup
[itex]\int^{V}_{0}[/itex]vdv=[itex]\int^{X}_{0}[/itex] ([itex]\frac{-2400x}{7}[/itex]+4800)dx

gives me V= [itex]\sqrt{\frac{-2400x^2}{7}+9600x }[/itex]

then, velocity as a function of position.

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


put that into wolfram
gives me .0360 sec, book says .0370 sec
 
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  • #6
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?

20131226_150252_zps042894e0.jpg
 
  • #7
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
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 into 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##
 
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  • #9
Yea I saw those equations else where, but I wanted to use the method the book was giving me.
 

1. What is the definition of kinematics?

Kinematics is the study of the motion of objects without considering the forces that cause the motion.

2. How is kinematics related to rectilinear motion?

Rectilinear motion is a type of kinematics where an object moves in a straight line with constant velocity.

3. What is a spring in the context of kinematics?

A spring is a mechanical device that is used to store and release energy, typically by stretching or compressing.

4. How is spring constant related to kinematics?

Spring constant, also known as the spring stiffness, is a measure of how stiff a spring is and is used in kinematics equations to calculate the force exerted by a spring.

5. Can kinematics be used to analyze the motion of a spring?

Yes, kinematics can be used to analyze the motion of a spring by using equations such as Hooke's law to calculate the displacement, velocity, and acceleration of the spring.

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