Mass-spring system

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  • #1
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Main Question or Discussion Point

Vibration Free

Please, are correct?

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

Where frequency is

[tex]w = \sqrt{\frac{k}{m}}[/tex]

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

The characteristic equation is:

[tex]r^2 + w^2 = 0[/tex]
[tex]r = +or- iw[/tex] where [tex]i^2 = -1[/tex]

Then

[tex]x(t) = C_1e^{iwt} + C_2e^{-iwt}[/tex]

Calculating I can get
[tex]x(t) = a1cos(wt) + a2sin(wt)[/tex]


Now, I need to do to get the following equation. how do I find?
[tex]x(t) = Acos(wt - \delta)[/tex] (I think this is the equation we need to get the free vibration)
 

Answers and Replies

  • #2
162
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Ya, it's correct, and

[tex]
x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)
[/tex]

if you define

[tex]
A \equiv a_1^2 + a_2^2
[/tex]
and
[tex]
\delta \equiv arctan(a_1/a_2)
[/tex]

This is the equation of free vibration of an undamped single degree of freedom dynamic system with linear elastic stiffness.
 
  • #3
54
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Ya, it's correct, and

[tex]
x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)
[/tex]

if you define

[tex]
A \equiv a_1^2 + a_2^2
[/tex]
and
[tex]
\delta \equiv arctan(a_1/a_2)
[/tex]

This is the equation of free vibration of an undamped single degree of freedom dynamic system with linear elastic stiffness.
Why if [tex]A \equiv a_1^2 + a_2^2[/tex] and [tex]\delta \equiv arctan(a_1/a_2)[/tex] we have:
[tex]x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)[/tex]

What the calculations involved?
 
  • #4
144
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Use the multiple angle formula for cos, cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
 
  • #5
54
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Use the multiple angle formula for cos, cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
With
[tex]cos(a-b)=cos(a)cos(b)+sin(a)sin(b)[/tex]
[tex]x(t) = a_1cos(wt) + a_2sin(wt)[/tex]

I do not understand how to find [tex]A cos(wt - \delta)[/tex]
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
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Have you tried rather than just staring at the formulas?

"cos(a-b)= cos(a)cos(b)+ sin(a)sin(b)" with "[itex]\omega t- \delta[/itex]" instead of a- b gives you [itex]Acos(\omega t-\delta)= Acos(\omega t)cos(\delta)+ A sin(\omega t)sin(\delta)[/itex].

In order to have that equal to [itex]a_1cos(\omega t)+ a_2sin(\omega t)[/itex], you must have [itex]a_1= A cos(\delta)[/itex] and [itex]a_2= A sin(\delta)[/itex].

Dividing the first equation by the second gives
[tex]\frac{A sin(\delta)}{A cos(\delta)}= tan(\delta)= \frac{a_2}{a_1}[/tex]
so [itex]\delta= tan^{-1}(a_2/a-1)[/itex].

Squaring and summing the two equations gives
[tex]a_1^2+ a_2^2= A^2 cos^2(\delta)= A^2[/tex]
so [itex]A= \sqrt{a_1^2+ a_2^2}[/itex].

BobbyBear didn't quite have those equations in his post.
 
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  • #7
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Yes, thank you.

The general solution is
[tex]x(t) = Acos(wt - \delta)[/tex]

I find A with the initial conditions ?

[tex]x(0) = Acos(- \delta) = X_o[/tex]

[tex]x'(0) = -Awsin(- \delta) = V_o[/tex]

I'll be A in the function [tex] \delta [/tex] ???
 
  • #8
162
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Oopsie, I guess I tried to do it too quickly. HallsofIvy's relationships are the correct ones :)

And yes, on applying the initial conditions you obtain a 2 by 2 system (the one you wrote out), which you solve for A (the amplitude) and the angle [tex] \delta [/tex]
 
  • #9
54
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Oopsie, I guess I tried to do it too quickly. HallsofIvy's relationships are the correct ones :)

And yes, on applying the initial conditions you obtain a 2 by 2 system (the one you wrote out), which you solve for A (the amplitude) and the angle [tex] \delta [/tex]
I could not find A

[tex]X(0) = Acos( - \delta) = X_o[/tex]
[tex]X'(0) = -Awsin(- \delta) = V_o[/tex]

[tex]\delta = tan^{-1} \frac{V_o}{X_o w}[/tex]

And A ???
 
  • #10
162
0
Well, once you have [tex]
\delta
[/tex]

you can go to any of the two equations and get A, eg

[tex] A= X_0/cos(\delta) = \frac{X_0}{cos(arctan(V_0/X_0w))} = \frac{X_0}{X_0w/\sqrt{X_0^2w^2+V_0^2}} = \frac{\sqrt{X_0^2w^2+V_0^2}}{w}
[/tex]
 
  • #11
54
0
Well, once you have [tex]
\delta
[/tex]

you can go to any of the two equations and get A, eg

[tex] A= X_0/cos(\delta) = \frac{X_0}{cos(arctan(V_0/X_0w))} = \frac{X_0}{X_0w/\sqrt{X_0^2w^2+V_0^2}} = \frac{\sqrt{X_0^2w^2+V_0^2}}{w}
[/tex]
Sorry, I do not understand your calculation.

Acos(- \delta ) = X_o
 
  • #12
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[tex] A cos(\delta) = X_0 \rightarrow A = X_0 / cos(\delta) [/tex]

and

[tex] \delta = arctan(V_0/X_0w)

[/tex]

so

[tex] cos(\delta) = cos(arctan(V_0/X_0w))

[/tex]

yes?

(By the way, cos(-a) = cos(a) that's why I didn't bother with the minus sign)
 
  • #13
54
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[tex] A cos(\delta) = X_0 \rightarrow A = X_0 / cos(\delta) [/tex]

and

[tex] \delta = arctan(V_0/X_0w)

[/tex]

so

[tex] cos(\delta) = cos(arctan(V_0/X_0w))

[/tex]

yes?

(By the way, cos(-a) = cos(a) that's why I didn't bother with the minus sign)
Yes. Thank you
 
  • #14
54
0
I got a doubt in the beginning.
After obtaining the characteristic equation
[tex]r = \pm iw[/tex] with [tex]i^2 = -1[/tex]

How do we know [tex]x(t) = C_1e^{iwt} + C_2e^{-iwt}[/tex] ???
 
  • #15
162
0
I got a doubt in the beginning.
After obtaining the characteristic equation
[tex]r = \pm iw[/tex] with [tex]i^2 = -1[/tex]

How do we know [tex]x(t) = C_1e^{iwt} + C_2e^{-iwt}[/tex] ???
Because the theory of differential equations tells us that the general solution of a homogeneous linear ordinary differential equation with constant coefficients is given by a linear combination of exponentials of the form: [tex]
e^{rt}
[/tex]

with r being the roots of the characteristic polynomial, which in this case are [tex]\pm iw[/tex]
 
  • #16
54
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Because the theory of differential equations tells us that the general solution of a homogeneous linear ordinary differential equation with constant coefficients is given by a linear combination of exponentials of the form: [tex]
e^{rt}
[/tex]

with r being the roots of the characteristic polynomial, which in this case are [tex]\pm iw[/tex]
Thank you very much. You know about damped in my other topic ?
 
  • #17
54
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Someone has a graphic example of the system to vibration free ?
 
  • #18
54
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To plot the graph using Maple 12, which values I use for [tex]k[/tex], [tex]x_o[/tex] and [tex]v_o[/tex]?
 

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