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tedwillis
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Hi all, I was wondering is you could help me with this springs question. We've only done springs hanging from a fixed support above being stretched, but now I've got a question where the spirng is being compressed.
So, here's some basic info about the question
So, the first question is to proove that the equation of motion for the ball is...
Here is a rough diagram of what I believe to be happening from the description of the problem:
So, with Hooke's law
[itex]T=mg[/itex]
[itex]ks=mg[/itex] (equation 1)
[itex]k=mg/s[/itex]
[itex]k=(5*9.8)/(-0.392)[/itex]
[itex]k=-125[/itex]
With Newton's law of motion (F=ma):
[itex]m*y''(t)=mg-T-R[/itex]
[itex]m*y''(t)=mg-k(s+y(t))-by'(t)[/itex]
[itex]m*y''(t)=mg-ks-ky(t)-by'(t)[/itex], from equation 1: ks-mg=0 so
[itex]m*y''(t)+ky(t)+by'(t)=0[/itex], So subbing in k=-125, m=5
[itex]5*y''(t)-125y(t)+by'(t)=0[/itex]
[itex]y''(t)+b/5y'(t)-25y(t)=0[/itex]
This is almost what I need, but for some reason I have -25y(t) instead of +25y(t). Can anyone see where I've gone wrong? The only thing I can think of is that k (and thus s) should be positive, but I am unsure why as it is in compression. We have been taught that stretching (extension) of the spring make "s" a postive number, so one would think that compression of the spring would make give "s" a negative value. Is k always > 0 by definition?
Homework Statement
So, here's some basic info about the question
- Obviously, acceleration due to gravity is 9.8 m/s^2
- The downward direction is considered the positive direction
- A spring is placed on a fixed support standing vertically
- The natural length of the spring is 5 metres
- An iron ball weighing 5kg is attached to the top of it
- This mass compresses the spring by 0.392m
- The ball is pulled to a distance of 0.4m above its equalibirum position then released
- Damping force is proportional to the instantaneous velocity of the ball
- y(t)=The displacement of the ball below the equalibrium position at t seconds after the release
Homework Equations
So, the first question is to proove that the equation of motion for the ball is...
- [itex]0=y''(t)+(b/5)y'(t)+25y(t)[/itex], where b kg/s is the damping constant
The Attempt at a Solution
Here is a rough diagram of what I believe to be happening from the description of the problem:
So, with Hooke's law
[itex]T=mg[/itex]
[itex]ks=mg[/itex] (equation 1)
[itex]k=mg/s[/itex]
[itex]k=(5*9.8)/(-0.392)[/itex]
[itex]k=-125[/itex]
With Newton's law of motion (F=ma):
[itex]m*y''(t)=mg-T-R[/itex]
[itex]m*y''(t)=mg-k(s+y(t))-by'(t)[/itex]
[itex]m*y''(t)=mg-ks-ky(t)-by'(t)[/itex], from equation 1: ks-mg=0 so
[itex]m*y''(t)+ky(t)+by'(t)=0[/itex], So subbing in k=-125, m=5
[itex]5*y''(t)-125y(t)+by'(t)=0[/itex]
[itex]y''(t)+b/5y'(t)-25y(t)=0[/itex]
This is almost what I need, but for some reason I have -25y(t) instead of +25y(t). Can anyone see where I've gone wrong? The only thing I can think of is that k (and thus s) should be positive, but I am unsure why as it is in compression. We have been taught that stretching (extension) of the spring make "s" a postive number, so one would think that compression of the spring would make give "s" a negative value. Is k always > 0 by definition?
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