# Springs Question - Second Order O.D.E

1. Oct 19, 2012

### tedwillis

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.

1. The problem statement, all variables and given/known data
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

2. Relevant equations
So, the first question is to proove that the equation of motion for the ball is...
• $0=y''(t)+(b/5)y'(t)+25y(t)$, where b kg/s is the damping constant

3. 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
$T=mg$
$ks=mg$ (equation 1)
$k=mg/s$
$k=(5*9.8)/(-0.392)$
$k=-125$

With Newton's law of motion (F=ma):
$m*y''(t)=mg-T-R$
$m*y''(t)=mg-k(s+y(t))-by'(t)$
$m*y''(t)=mg-ks-ky(t)-by'(t)$, from equation 1: ks-mg=0 so
$m*y''(t)+ky(t)+by'(t)=0$, So subbing in k=-125, m=5
$5*y''(t)-125y(t)+by'(t)=0$
$y''(t)+b/5y'(t)-25y(t)=0$

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?

Last edited: Oct 19, 2012
2. Oct 19, 2012

### HallsofIvy

Staff Emeritus
You have s- y when you should have y- s. Suppose y is very large. Then s- y will be negative so -k(s- y) will be positive and y''= -k(s- y) would be positive meaning that y will get larger. It should be the other way around- the spring should move y back toward the center. You should have either y''= k(s- y) or y''= -k(y- s).

3. Oct 19, 2012

### tedwillis

Woops. That was a typo actually.

It should say $my''(t)=mg-k(s+y(t))-by'(t)$ instead, and this is the path that the following calculations take.

In this sense, I am saying that the restoring force of the spring’s tension (T) is equal to the spring constant*(contraction (or extension)+y(t) (the distance below the equalibrium position)). Either I am not understanding you or the subject matter correctly or you have found the wrong error, simply a badly placed typo. Sorry on my behalf.

I'll fix the error up in the OP.

Last edited: Oct 19, 2012
4. Oct 21, 2012

### nishcha7

m*y'' = mg - T - R

my'' = mg - k(s-y) - By'

We write -k(s+y) because the spring is stretched but in this case spring is compressing that is why we use -k(s-y)
Hope it makes sense.
Cheers