# Thread: Solve a problem using differential equations.

1. ## Solve a problem using differential equations.

I would be grateful if anybody could help me in this problem.
A body of mass m is moving around an equilibrium position. It is affected by a return force proportional to the distance to the equilibrium position, and a friction force (absorbing force)
proportional to velocity.

With the power equation using you can then set up the equation.

Set m = 0.10 kg and k = 2.6 N/m. Now determine the solution that satisfies the conditions
s (0) = 0.030 m and s’(0) = 0 m/s if b = 0 Ns/m (damping is neglected)

2. What ideas have you had so far?

3. I just think it is a second order diffrential equation.

4. You are correct. Does it have constant coefficients? Is it homogeneous?

5. Do you mean like m and b in the power equation?

6. I mean that your DE is

$m\ddot{s}=-ks-b\dot{s},$

and you're given values for $m,k$ as well as initial conditions. Are the coefficients of the terms of the DE constant, or do they vary?

Is the DE homogeneous?

The answers to these questions greatly affect the type of solutions you get, as well as the methods of solution.

By power equation, do you mean the series solution method? Or do you mean the characteristic equation?

7. I have no idea about that but I just know the final answer is s=0.030*cos (t * (26^(1/2))) but no idea about how the solution would be.

8. Hmm. I suggest you take a good long look at Chris L T521's Differential Equations tutorial, especially Post # 3 in the Second Order DE's portion. He's pretty much got it all there. I'm a little surprised that you've been presented with a DE like this, and you don't yet have the tools to solve them! Self-study is often the best, though, as I'm finding out.

9. m* ((d^2s)/(dt^2)) = - ks – b* (ds/dt)
So

m* s’’ = - ks – b* s’

There is a second order DE
m s’’ + b s’ + ks = 0
s’’ + b/m s’ + k/m s = 0

if we set m=0,10 and k=2,6 and b=0 then
s’’ + 0 s’ + (2,6/0,10) s = 0
s’’ + 0,26 s = 0
r^2 – 0. r – 0,26= 0
r^2 – 0,26 = 0
r1= + i (0,26)1/2 r2= - i (0,26)1/2

α= 0 β= ± (0,26)1/2

So
s= e 0t (C. cos((0,26)^(1/2) . t) + D. sin ((0,26)^(1/2) . t))
s= (C. cos((0,26)^(1/2) . t) + D. sin ((0,26)^(1/2) . t))
s’= (D. (0,26)^(1/2).cos ((0,26)^(1/2) . t) -
C. (0,26)^(1/2) .sin((0,26)^(1/2) . t))

1- if s(0) = 0,030 then
0,030 = (C. cos((0,26)^(1/2) . 0) + D. sin ((0,26)^(1/2) . 0))
= (C. cos(0) + D. sin ((0))
=C. 1 + 0
So C= 0,030
2- If s’(0)=0 then
s’= (D. (0,26)^(1/2).cos ((0,26)^(1/2) . 0) -
C. (0,26)^(1/2).sin((0,26)^(1/2) . 0))=0
D. (0,26)^(1/2) .1) + 0=0

So D=0

The equation will be s= 0,030.cos ((0,26)^(1/2).t)

10. 2,6/0,10 = 0,26?