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 Aug 30th 2017, 09:09 AM #1 Junior Member   Join Date: Aug 2017 Posts: 8 mass-spring prpblem the problem: Suppose at time zero, the bob was drawn upward four units from the equilibrium position, let C=2, K=2, m=1 lbm, initial speed=2 unit/sec find an expression for body's position. and in the solution it says: y''+6y'+5y=0 my question is: from where does the numbers (6) and (5) that is in te above DE came from ? 2. Relevant equations spring equation: my''+cy'+ky=f(t) consider f(t) =0 3. The attempt at a solution stopped at first because of the first step Last edited by aows61; Aug 30th 2017 at 01:31 PM. Reason: correction
 Aug 30th 2017, 12:25 PM #2 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 505 Have you got some kind of friction in the problem? A spring ODE doesn't usually have a velocity term unless some kind of dampening is involved. aows61 likes this. __________________ Burn those raisin muffins. Burn 'em all I say.
Aug 30th 2017, 12:28 PM   #3
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spring-mass

 Originally Posted by kiwiheretic Have you got some kind of friction in the problem? A spring ODE doesn't usually have a velocity term unless some kind of dampening is involved.
this is all the information that i have about the problem:
and here is a link of you want to check:
>>>https://i.imgur.com/xjqAOUY.jpg

Aug 30th 2017, 01:18 PM   #4
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 Originally Posted by aows61 this is all the information that i have about the problem: and here is a link of you want to check: >>>https://i.imgur.com/xjqAOUY.jpg
Can't see your link, why can't you attach it here?

y' is the velocity and you say that at t=0 the velocity = +2.

This is a differential equation you can solve for the boundary condition.

But what I don't know is what your sign convention is.

Which way is + and which way is -?

 Aug 30th 2017, 01:29 PM #5 Junior Member   Join Date: Aug 2017 Posts: 8 here is the attachment, just press on this link>>>>> https://imgur.com/xjqAOUY
 Aug 30th 2017, 01:30 PM #6 Junior Member   Join Date: Aug 2017 Posts: 8 for the sign convention: upward is negative and downward is positive
 Aug 30th 2017, 01:32 PM #7 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 505 I think he meant like this (reattached) Attached Thumbnails   aows61 likes this. __________________ Burn those raisin muffins. Burn 'em all I say.
 Aug 30th 2017, 01:33 PM #8 Junior Member   Join Date: Aug 2017 Posts: 8 THanks indeed @KIWIHERETIC , appreciate your help,
 Aug 30th 2017, 01:46 PM #9 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,261 I think the issue is either (a) the answer is incorrect in that it doesn't agree with the values of c and k that are given, or (b) there are units at play here that have not been properly defined. For example, they say m = 1 lb_m, so the first term of the DE must be divided by 32.2 lbm-ft/lbf-s^2 in order to get units of force in lbf for a given acceleration in ft/s^2. Unfortunately we are not told what the units of c and k are. If we assume that c is in lbf/(ft/s) and k is in lbf/ft, then the DE would be: $\displaystyle \frac 1 {32.2} y'' + 2 y' +2y = 0$ But the problem states that the initial displacement is "4 units," not 4 ft, and initial velocity is "2 units per second," not 2 ft per second, so I'm guessing that somewhere in the original problem statement they define what they mean by a distance of 1 "unit." They should also be clear about what the units are for both C and K. aows61 likes this. Last edited by ChipB; Aug 30th 2017 at 02:14 PM.
Aug 30th 2017, 01:54 PM   #10
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thanks

 Originally Posted by ChipB I think the issue is either (a) the answer is incorrect in that it doesn't agree with the values of c and k that are given, or (b) there are units at play here that have not been properly defined. For example, they say m = 1 lb_m, so the first term of the DE must be divided by 32.2 lbm-ft/lbf-s^2 in order to get units of force in lbf for a given acceleration in ft/s^2. Unfortunately we are not told what the units of c and k are. If we assume that c is in lbf/(ft/s) and k is in lbf/ft, then the DE would be: $\displaystyle \frac 1 {32.2} y'' + 2 y' +2y = 0$ But the problem states that the initial displacement is "4 units," not 4 ft, and initial velocity is "2 units per second," not 2 ft per second, so I'm guessing that somewhere in the original problem statement there were units defined for m, c and k that we don't know about.
thanks indeed for your contributions,
i also guess that there is some mistake within the solution.
even though there is some unit change it will not produce any 5 or 6, right ?
i solved it and the roots of the characteristic equation were complex (r1=-1-i, and r2=-1+i)

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