Mass-spring system damped

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Damped vibration

[tex] m \frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + kx = 0 [/tex]

Characteristic equation is

[tex]mr^2 + \gamma r + k = 0[/tex]

[tex]r_1 = \frac{- \gamma + \sqrt{( \gamma )^2 - 4mk}}{2m}[/tex]
[tex]r_2 = \frac{- \gamma - \sqrt{( \gamma )^2 - 4mk}}{2m}[/tex]

In overdamped
[tex]( \gamma )^2 - 4mk > 0[/tex]

What I need to calculate to find the general solution:
[tex] x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}[/tex] ???
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi viciado123! Welcome to PF! :wink:

You need two "initial" conditions, to find the two constants C1 and C2.

Usually, they'll be the values of x at two particular times, or the value of x and dx/dt, or of dx/dt and d2x/dt2, at one particular time. :smile:
 
  • #3
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Hi viciado123! Welcome to PF! :wink:

You need two "initial" conditions, to find the two constants C1 and C2.

Usually, they'll be the values of x at two particular times, or the value of x and dx/dt, or of dx/dt and d2x/dt2, at one particular time. :smile:
I do not know how to find the equation [tex]x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}[/tex]

In the books we have to assume [tex]x(t) = e^{rt}[/tex] Why assume this ?

[tex]x_1(t) = C_1e^{r_1t}[/tex] and [tex]x_2(t) = C_2e^{r_2t}[/tex]
 
  • #4
tiny-tim
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It's because you can write the characteristic equation as (D - r1)(D - r2)x = 0 (where D is short for d/dt),

so the solutions are the solutions to (D - r1)x = 0 and (D - r2)x = 0,

which are the same as dx/dt = r1x and dx/dt = r2x,

or x = C1er1t and x = C2er2t. :smile:
 
  • #5
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It's because you can write the characteristic equation as (D - r1)(D - r2)x = 0 (where D is short for d/dt),

so the solutions are the solutions to (D - r1)x = 0 and (D - r2)x = 0,

which are the same as dx/dt = r1x and dx/dt = r2x,

or x = C1er1t and x = C2er2t. :smile:
Thanks. Is:
[tex]\int \frac{dx}{dt} = \int r_1xdt[/tex]
[tex]ln(x) = r_1t[/tex]
[tex]x(t) = e^{r_1t}[/tex]

I correct ?
 
  • #6
HallsofIvy
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I do not know how to find the equation [tex]x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}[/tex]

In the books we have to assume [tex]x(t) = e^{rt}[/tex] Why assume this ?

[tex]x_1(t) = C_1e^{r_1t}[/tex] and [tex]x_2(t) = C_2e^{r_2t}[/tex]
Be careful here. Textbooks suggest you assume a solution of that form (they don't say you must) because it leads to correct solutions, not necessarily because the solution "must" be of that form. You should quickly learn that you can also get polynomial solutions and trig functions as well as exponentials as solutions and, in fact, products of those kinds of things.

Also you need the crucial theoretical fact that the set of all solutions to a nth order linear differential equation forms an nth dimensional "vector space". In particular, that means that if you have two independent solutions, f1 and f2, of a second order linear differential equation, then any solution can be written as a linear combination of the two solutions: y(x)= C1f1(x)+ C2f2(x) for any solution, y(x), of the equation.

As for your original question, knowing that [itex]e^{r_1t}[/itex] and [itex]e^{r_2t}[/itex] are solutions to the differential equation (and knowing that exponentials with different coefficienta in the exponent are independent) tells you that any solution, y(x), can be written in the form [itex]C_1e^{r_1t}+ C_2e^{r_2t}[/itex].

It is also helpful to observe that, for complex r1 and r2, as you have here, [itex]e^{a+ bi}= e^ae^{bi}= e^a(cos(b)+ i sin(b))[/itex] so that your "exponential solutions" can be written as a combination of exponential and trigonometric solutions.
 
  • #7
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Be careful here. Textbooks suggest you assume a solution of that form (they don't say you must) because it leads to correct solutions, not necessarily because the solution "must" be of that form. You should quickly learn that you can also get polynomial solutions and trig functions as well as exponentials as solutions and, in fact, products of those kinds of things.

Also you need the crucial theoretical fact that the set of all solutions to a nth order linear differential equation forms an nth dimensional "vector space". In particular, that means that if you have two independent solutions, f1 and f2, of a second order linear differential equation, then any solution can be written as a linear combination of the two solutions: y(x)= C1f1(x)+ C2f2(x) for any solution, y(x), of the equation.

As for your original question, knowing that [itex]e^{r_1t}[/itex] and [itex]e^{r_2t}[/itex] are solutions to the differential equation (and knowing that exponentials with different coefficienta in the exponent are independent) tells you that any solution, y(x), can be written in the form [itex]C_1e^{r_1t}+ C_2e^{r_2t}[/itex].

It is also helpful to observe that, for complex r1 and r2, as you have here, [itex]e^{a+ bi}= e^ae^{bi}= e^a(cos(b)+ i sin(b))[/itex] so that your "exponential solutions" can be written as a combination of exponential and trigonometric solutions.
Thanks. I find [tex]C_1[/tex] and [tex]C_2[/tex] with initial conditions ?
The initial conditions is:
[tex]x(0) = X_o[/tex] Initial Position ?
[tex]x'(0) = V_o[/tex] Initial velocity?
It is ?
 
  • #8
tiny-tim
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Thanks. I find [tex]C_1[/tex] and [tex]C_2[/tex] with initial conditions ?
The initial conditions is:
[tex]x(0) = X_o[/tex] Initial Position ?
[tex]x'(0) = V_o[/tex] Initial velocity?
It is ?
Hint: x(0) = C1+ C2

x'(0) = … ? :smile:
Be careful here. Textbooks suggest you assume a solution of that form (they don't say you must) because it leads to correct solutions, not necessarily because the solution "must" be of that form. You should quickly learn that you can also get polynomial solutions and trig functions as well as exponentials as solutions and, in fact, products of those kinds of things.

Also you need the crucial theoretical fact that the set of all solutions to a nth order linear differential equation forms an nth dimensional "vector space".
No, in this case, the solutions must be of that form.

And you can prove it just by putting y = (D - r1)x, and solving for y, then solving for x.

The "textbook suggestions" (ie, mere intelligent guesswork :rolleyes:) really apply to "particular solutions", but not to "general solutions" such as this or any other "polynomial = 0" equation. :wink:
 
  • #9
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Hint: x(0) = C1+ C2

x'(0) = … ? :smile:


No, in this case, the solutions must be of that form.

And you can prove it just by putting y = (D - r1)x, and solving for y, then solving for x.

The "textbook suggestions" (ie, mere intelligent guesswork :rolleyes:) really apply to "particular solutions", but not to "general solutions" such as this or any other "polynomial = 0" equation. :wink:
Am I correct?

[tex]x(0) = C_1 + C_2 = x_o[/tex]
[tex]x'(0) = C_1r_1 + C_2r_2 = v_o[/tex]

I find:

[tex] C_1 = \frac{x_or_2 - v_o}{r_2 - r_1}[/tex]

[tex] C_2 = \frac{v_o - x_o r_1}{r_2 - r_1}[/tex]

Correct ?
 
  • #11
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Yup! :biggrin:
Thanks.

In case [tex] \gamma^2 -4mk < 0[/tex] is underdamped

[tex]r_1 = \frac{- \gamma}{2m} + ri[/tex] and [tex]r_1 = \frac{- \gamma}{2m} - ri[/tex]

By Euler's formula

[tex]x(t) = e^{-(\frac{\gamma}{2m})t} (C_1 cos(rt) + C_2 sin(rt))[/tex]

Are the same initial conditions?
What is r? How do I calculate r ?
 
  • #12
tiny-tim
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Are the same initial conditions?
What is r? How do I calculate r ?
Do you understand why [tex]r_1 = \frac{- \gamma}{2m} + ri[/tex] and [tex]r_2 = \frac{- \gamma}{2m} - ri[/tex] ?

If so, the value of r is obvious. :wink:

(and use x0 and v0 as before)
 
  • #13
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Do you understand why [tex]r_1 = \frac{- \gamma}{2m} + ri[/tex] and [tex]r_2 = \frac{- \gamma}{2m} - ri[/tex] ?

If so, the value of r is obvious. :wink:

(and use x0 and v0 as before)

r is the imaginary part ?

[tex]r = \frac{\sqrt{\gamma^2 -4mk}}{2m}[/tex]
 
  • #14
tiny-tim
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That's it! :smile:
 
  • #16
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Someone has a graphic example of the system to overdamped and underdamped ?
 
  • #17
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I tried to plot in matlab overdamped. But it appeared nothing
Code:
plot((.1*((-30-sqrt(30^2-100*(4*1.3)))*e^((-30+sqrt(30^2-100*(4*1.3)))*t/(2*1.3))/(2*1.3)-(-30+sqrt(30^2-100*(4*1.3)))*e^((-30-sqrt(30^2-100*(4*1.3)))*t/(2*1.3))/(2*1.3)))/(-sqrt(30^2-100*(4*1.3))/(1.3)), t = 0 .. 5)
 
  • #18
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In underdamped:
[tex]C_1 = x_o[/tex]

[tex]C_2 = \frac{v_o}{r}[/tex]

It is correct ?
 
  • #19
tiny-tim
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Yup! :biggrin:
 
  • #20
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Yup! :biggrin:
I find

[tex]x'(0) = \frac{- \gamma}{2m}(C_2r) = v_o[/tex]

[tex] C_2 = - \frac{2mv_o}{\gamma r}[/tex]

Where is my mistake?
 
  • #21
tiny-tim
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I find

[tex]x'(0) = \frac{- \gamma}{2m}(C_2r) = v_o[/tex]

[tex] C_2 = - \frac{2mv_o}{\gamma r}[/tex]

Where is my mistake?
ooh, sorry … that's what happens when I try to flip between two pages. :redface:

C2 isn't v0/r

because you have to differentiate the whole of e-(γ/2m)t(C1 etc), before putting t = 0,

so you get C2r plus an extra -(γ/2m)C1 = v0.
 
  • #22
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ooh, sorry … that's what happens when I try to flip between two pages. :redface:

C2 isn't v0/r

because you have to differentiate the whole of e-(γ/2m)t(C1 etc), before putting t = 0,

so you get C2r plus an extra -(γ/2m)C1 = v0.
Ok. Thanks
 

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