Number of integration constants

  • I
  • Thread starter LagrangeEuler
  • Start date
  • #1
LagrangeEuler
708
20
If we have system of 3 ordinary differential equation in mechanics and we have two initial condition ##\vec{r}(t=0)=0## and ##\vec{v}(t=0)=\vec{v}_0 \vec{i}##. If we somehow get
[tex]\frac{d^2v_x}{dt^2}=-\omega^2v_x[/tex]
then [tex]v_x(t)=A\sin(\omega t)+B\cos(\omega t)[/tex]
Two integration constants and one initial condition for velocity. What to do? Should we put that one constant is equal to zero? So ##A=0##, ##B=v_0##?
 

Answers and Replies

  • #2
anuttarasammyak
Gold Member
1,943
1,012
From your solution of velocity,
cooridinate:
[tex]x=A\frac{1-\cos \omega t}{\omega}+B \frac{\sin \omega t}{\omega}[/tex]
accerelation:
[tex]a=A\omega \cos \omega t - B\omega \sin \omega t[/tex]
So
[tex]A= \frac{a_0}{\omega}[/tex]
[tex]B= v_0[/tex]
 
  • #3
LagrangeEuler
708
20
Yes, I agree. But suppose that I have only two initial conditions in the beginning. I do not know acceleration. When I integrate ##v_x(t)## I will get
[tex]x(t)=C\cos(\omega t)+D\sin (\omega t)+E[/tex]
so 3 integration constants. And because ##x(t)## satisfy some differential equation of second order, general solution should have only 2 integration constants. And I can put ##D=0##. Is it good reasoning?
 
Last edited:
  • #4
anuttarasammyak
Gold Member
1,943
1,012
My formula of x in #2 which includes initial coordinate condition is different from yours. Please criticize it.
 
  • #5
LagrangeEuler
708
20
I made a typo. From ##v_x(t)=A\sin \omega t+B \cos \omega t## you can write
##\frac{dx}{dt}=A\sin \omega t+B \cos \omega t##
by integrating
##x(t)=-\frac{A}{\omega}\cos \omega t+\frac{B}{\omega}\sin \omega t+E##.
Of course I can say that ##\frac{A}{\omega}=C## and ##\frac{B}{\omega}=D##
Initial condition is ##x(0)=0##. From that
##0=-\frac{A}{\omega}+E##.
I will get the same result as you.
 
Last edited:
  • #6
LagrangeEuler
708
20
From your solution of velocity,
cooridinate:
[tex]x=A\frac{1-\cos \omega t}{\omega}+B \frac{\sin \omega t}{\omega}[/tex]
accerelation:
[tex]a=A\omega \cos \omega t - B\omega \sin \omega t[/tex]
So
[tex]A= \frac{a_0}{\omega}[/tex]
[tex]B= v_0[/tex]
I do not think also that this is a general solution. Because by putting ##t=0## first term is always zero, regardless of value of ##A##.
 
  • #7
anuttarasammyak
Gold Member
1,943
1,012
by integrating
sin and cos are to be exchanged.

I agree with your observation that initial coordinate condition does not decide A nor B. As said in #2 initial velocity and initial acceleration decide them.

Popular oscillation equation for x has two constnts to decide the motion, initial cooridinate and initial velocity. Your oscillation equation for v, which has three times derivative of coordinates thus is not Newton's equation of motion, has two constants to decide the motion, initial velocity and initial acceleration. All one rank up in time derivative.
 
Last edited:

Suggested for: Number of integration constants

Replies
3
Views
125
Replies
15
Views
542
Replies
1
Views
326
Replies
18
Views
2K
Replies
4
Views
1K
Top