If you have force depending on distance, then how to find distance depending on time.

  • Thread starter DSoul
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  • #1
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This problem has been bothering me for a while now, hope you can help me.

Let's say that the initial velocity of an object, with mass of m is 0 and the initial position is s0 and the force acting on the object is defined as F(s), how do i find the s(t), where t is time. If it's any help, then the F(s) should be periodic. I can also write the exact problem I'm working on, but a general solution would be nice.

Thank You in advance.
 

Answers and Replies

  • #2
rcgldr
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Generally you'll have to translate force into acceleration, this gives you a(s). You can start with a = dv/dt, multply by ds/ds, => a = (dv/ds) (ds/dt) = v dv/ds. This leads to v dv = a(s) ds, which will be the first integration step. You mentioned F(s) is periodic, so take the simple case a(s) = -s, this results in:
v dv = -s ds
1/2 v2 = - 1/2 s2 + c
v = sqrt(c - s2)
ds/dt = sqrt(c - s2)
ds/sqrt(c - s2) = dt
let c = d^2
ds/sqrt(d2 - s2) = dt
sin-1(s / |d|) = t + e (or - cos-1(s / |d|)= t + e)
s = |d| sin(t + e) (or s = -|d| cos(t + e)
 
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  • #3
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Thank you for your answer. Correct me of I'm wrong, but I understand that d and e are random constants. Isn't there a way to solve it so there wouldn't be any random constants in the answer?
 
  • #4
rcgldr
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Thank you for your answer. Correct me of I'm wrong, but I understand that d and e are random constants.
Yes, they are random constants.
Isn't there a way to solve it so there wouldn't be any random constants in the answer?
You need to supply enough initial conditions to solve for the constants, for example, if s(0) = 0 and v(0) = 1, then s(t) = sin(t).
 
  • #5
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Say i wanted to take definite integrals from both sides of the equation: a(s) ds = v dv, then what should be the intervals for both sides? Should they be equal, or lets say [s1;0] for the left side and [v(s1);v(0)] for the other side?
 
  • #6
rcgldr
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Say i wanted to take definite integrals from both sides of the equation: a(s) ds = v dv, then what should be the intervals for both sides? Should they be equal, or lets say [s1;0] for the left side and [v(s1);v(0)] for the other side?
I'm not sure, since this would restrict the equality to defined intervals which could affect the outcome, and you'd still need limits for the ds/sqrt(...) = dt definite integral.

For my example, knowing s(0) and v(0) was enough to solve the example problem. I'm not sure of the advantage of including a second state for s1 earlier on. How would you choose s1?
 
  • #7
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I know that v(s1) = 0 and i also know how to calculate v(0). But okay, as your way seems better, could you please tell me how you got to the point where s(t) = sin(t). As I'm quite new to all this i didn't really understand how you got rid of those constants. If you could do it step by step, that'd be wonderful. Again, thanks in advance.
 
  • #8
rcgldr
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(how to) get rid of those constants.
I used the derived equation for s(t) and it's derivatives, acceleration wasn't used:

s(t) = |d| sin(t + e)
v(t) = |d| cos(t + e)
a(t) = -|d| sin(t + e)

if s(0) = 0 then e = 0 or e = ± π (assuming |d| ≠ 0)
if v(0) = 1 then |d| = 1 and e = 0
 
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