Find s(t) Given Initial Conditions and F(s)

AI Thread Summary
To find the position function s(t) of an object under a periodic force F(s), one must first translate the force into acceleration a(s) and relate it to velocity and position. The integration of the equation v dv = a(s) ds leads to expressions for velocity and position, incorporating constants that depend on initial conditions. By applying specific initial conditions, such as s(0) = 0 and v(0) = 1, the constants can be determined, simplifying the solution to s(t) = sin(t). Definite integrals can be taken on both sides of the equation, but the choice of intervals must be consistent to maintain equality. Understanding these relationships and initial conditions is crucial for deriving a complete solution without arbitrary constants.
DSoul
Messages
4
Reaction score
0
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.
 
Physics news on Phys.org


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)
 
Last edited:


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?
 


DSoul said:
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.
DSoul said:
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).
 


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 let's say [s1;0] for the left side and [v(s1);v(0)] for the other side?
 


DSoul said:
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 let's 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?
 


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.
 


DSoul said:
(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
 
Last edited:

Similar threads

Replies
13
Views
1K
Replies
4
Views
2K
Replies
8
Views
4K
Replies
13
Views
2K
Replies
17
Views
3K
Replies
59
Views
4K
Replies
6
Views
335
Back
Top