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Simple question about integration in Mechanics

So I'm working out a trajectory.

I have Fˉ=mx¨=−λx˙\bar{F}=m\ddot{x}=-\lambda \dot{x}

This leaves me with the equation mx¨+λx˙=0m\ddot{x}+\lambda \dot{x}=0

The associated characteristic equation is mr2+λr=0mr^2+\lambda r=0

Hence r=−λmr=-\frac{\lambda}{m}

I know (from my answers) that the solution I am looking for is x˙=x˙(0)e−λmt\dot{x}=\dot{x}(0)e^{-\frac{\lambda}{m}t}

May someone please be kind enough to explain how I would derive that?
(edited 11 years ago)
Reply 1
Avatar for atsruser
atsruser
11
Original post by RamocitoMorales
So I'm working out a trajectory.

I have Fˉ=mx¨=−λx˙\bar{F}=m\ddot{x}=-\lambda \dot{x}

This leaves me with the equation mx¨+λx˙=0m\ddot{x}+\lambda \dot{x}=0

The associated characteristic equation is mr2+λr=0mr^2+\lambda r=0

Hence r=−λmr=-\frac{\lambda}{m}

I know (from my answers) that the solution I am looking for is x˙=x˙(0)e−λmt\dot{x}=\dot{x}(0)e^{-\frac{\lambda}{m}t}

May someone please be kind enough to explain how I would derive that?


You don't need to use the characteristic equation method. Just write:

Fˉ=mx¨=−λx˙⇒mdx˙dt=−λx˙\bar{F}=m\ddot{x}=-\lambda \dot{x} \Rightarrow m \displaystyle \frac{d \dot{x}}{dt} = -\lambda \dot{x}

and it's separable.
Original post by atsruser
You don't need to use the characteristic equation method. Just write:

Fˉ=mx¨=−λx˙⇒mdx˙dt=−λx˙\bar{F}=m\ddot{x}=-\lambda \dot{x} \Rightarrow m \displaystyle \frac{d \dot{x}}{dt} = -\lambda \dot{x}

and it's separable.


I get ∫1x dx˙=−λmt\int \frac{1}{x}\,d\dot{x}=-\frac{\lambda}{m}t

How would I integrate the LHS? :colondollar:
Reply 3
Avatar for atsruser
atsruser
11
Original post by RamocitoMorales
I get ∫1x dx˙=−λmt\int \frac{1}{x}\,d\dot{x}=-\frac{\lambda}{m}t

How would I integrate the LHS? :colondollar:


You wrote the integral incorrectly, with a missing dot (and constant). It should be:

∫1x˙ dx˙=−λmt+C\int \frac{1}{\dot{x}}\,d\dot{x}=-\frac{\lambda}{m}t + C

So you get a log form.
Original post by RamocitoMorales
I get ∫1x dx˙=−λmt\int \frac{1}{x}\,d\dot{x}=-\frac{\lambda}{m}t

How would I integrate the LHS? :colondollar:

Use the substitution x = e^u
Original post by atsruser
You wrote the integral incorrectly, with a missing dot (and constant). It should be:

∫1x˙ dx˙=−λmt+C\int \frac{1}{\dot{x}}\,d\dot{x}=-\frac{\lambda}{m}t + C

So you get a log form.


Oops, well that makes more sense now. I got

x˙=e−λmt\dot{x}=e^{-\frac{\lambda}{m}t}. The constant CC isn't there because I integrated between t=0t=0 and t=tt=t.

Also, how does the x˙(0)\dot{x}(0) come into it, as in the solution:

x˙=x˙(0)e−λmt\dot{x}=\dot{x}(0)e^{-\frac{\lambda}{m}t}


:hmmmm2:
Original post by RamocitoMorales
Oops, well that makes more sense now. I got

x˙=e−λmt\dot{x}=e^{-\frac{\lambda}{m}t}. The constant CC isn't there because I integrated between t=0t=0 and t=tt=t.

Also, how does the x˙(0)\dot{x}(0) come into it, as in the solution:



:hmmmm2:

You handled the t=0 limit incorrectly. What is e0e^0?
(edited 11 years ago)
Original post by Unkempt_One
You handled the t=0 limit incorrectly. What is e0e^0?


∫0t−λm dt=−λmt+λm⋅0=−λmt\displaystyle \int_{0}^{t} -\frac{\lambda}{m}\,dt=-\frac{\lambda}{m}t+\frac{\lambda}{m}\cdot 0=-\frac{\lambda}{m}t

Or ∫0t−λm dt=−λm∫0t dt=−λm(t−0)=−λmt\displaystyle \int_{0}^{t} -\frac{\lambda}{m}\,dt=-\frac{\lambda}{m}\int_{0}^{t}\,dt=-\frac{\lambda}{m}(t-0)=-\frac{\lambda}{m}t
(edited 11 years ago)
Original post by RamocitoMorales
∫0t−λm dt=−λmt+λm⋅0=−λmt\displaystyle \int_{0}^{t} -\frac{\lambda}{m}\,dt=-\frac{\lambda}{m}t+\frac{\lambda}{m}\cdot 0=-\frac{\lambda}{m}t

Or ∫0t−λm dt=−λm∫0t dt=−λm(t−0)=−λmt\displaystyle \int_{0}^{t} -\frac{\lambda}{m}\,dt=-\frac{\lambda}{m}\int_{0}^{t}\,dt=-\frac{\lambda}{m}(t-0)=-\frac{\lambda}{m}t

I mean on the left hand equation. Sorry, e0e^0 doesn't come into it. If you're integrating the RHS by t=0 and t, you're integrating the LHS between the velocities at those times: x˙(0)\dot{x}(0) and x˙(t)\dot{x}(t).
(edited 11 years ago)
Original post by Unkempt_One
I mean on the left hand equation. Sorry, e0e^0 doesn't come into it. If you're integrating the RHS by t=0 and t, you're integrating the LHS between the velocities at those times: x˙(0)\dot{x}(0) and x˙(t)\dot{x}(t).


Ah, thank you. I get my answer this way.
Original post by RamocitoMorales
Ah, thank you. I get my answer this way.

No problem. Sorry for being unclear.

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