Solve Mass-Spring System Vibration Free

[viciado123]

Vibration Free

Please, are correct?

m \frac{d^2x}{dt^2} + kx = 0

Where frequency is

w = \sqrt{\frac{k}{m}}

\frac{d^2x}{dt^2} + \frac{k}{m}x = 0

The characteristic equation is:

r^2 + w^2 = 0
r = +or- iw where i^2 = -1

Then

x(t) = C_1e^{iwt} + C_2e^{-iwt}

Calculating I can get
x(t) = a1cos(wt) + a2sin(wt)


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

 

[BobbyBear]

Ya, it's correct, and

<br /> x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)<br />

if you define

<br /> A \equiv a_1^2 + a_2^2<br />
and
<br /> \delta \equiv arctan(a_1/a_2)<br />

This is the equation of free vibration of an undamped single degree of freedom dynamic system with linear elastic stiffness.

 

[viciado123]

Ya, it's correct, and

<br /> x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)<br />

if you define

<br /> A \equiv a_1^2 + a_2^2<br />
and
<br /> \delta \equiv arctan(a_1/a_2)<br />

This is the equation of free vibration of an undamped single degree of freedom dynamic system with linear elastic stiffness.

Why if A \equiv a_1^2 + a_2^2 and \delta \equiv arctan(a_1/a_2) we have:
x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)

What the calculations involved?

 

[n1person]

Use the multiple angle formula for cos, cos(a-b)=cos(a)cos(b)+sin(a)sin(b)

 

[viciado123]

Use the multiple angle formula for cos, cos(a-b)=cos(a)cos(b)+sin(a)sin(b)

With
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
x(t) = a_1cos(wt) + a_2sin(wt)

I do not understand how to find A cos(wt - \delta)

 

[HallsofIvy]

Have you tried rather than just staring at the formulas?

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

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

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

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

BobbyBear didn't quite have those equations in his post.

 

[viciado123]

Yes, thank you.

The general solution is
x(t) = Acos(wt - \delta)

I find A with the initial conditions ?

x(0) = Acos(- \delta) = X_o

x&#039;(0) = -Awsin(- \delta) = V_o

I'll be A in the function \delta ?

 

[BobbyBear]

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 \delta

 

[viciado123]

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 \delta

I could not find A

X(0) = Acos( - \delta) = X_o
X&#039;(0) = -Awsin(- \delta) = V_o

\delta = tan^{-1} \frac{V_o}{X_o w}

And A ?

 

[BobbyBear]

Well, once you have <br /> \delta <br />

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

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}<br />

 

[viciado123]

Well, once you have <br /> \delta <br />

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

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}<br />

Sorry, I do not understand your calculation.

Acos(- \delta ) = X_o

 

[BobbyBear]

A cos(\delta) = X_0 \rightarrow A = X_0 / cos(\delta)

and

\delta = arctan(V_0/X_0w)<br /> <br />

so

cos(\delta) = cos(arctan(V_0/X_0w))<br /> <br />

yes?

(By the way, cos(-a) = cos(a) that's why I didn't bother with the minus sign)

 

[viciado123]

A cos(\delta) = X_0 \rightarrow A = X_0 / cos(\delta)

and

\delta = arctan(V_0/X_0w)<br /> <br />

so

cos(\delta) = cos(arctan(V_0/X_0w))<br /> <br />

yes?

(By the way, cos(-a) = cos(a) that's why I didn't bother with the minus sign)

Yes. Thank you

 

[viciado123]

I got a doubt in the beginning.
After obtaining the characteristic equation
r = \pm iw with i^2 = -1

How do we know x(t) = C_1e^{iwt} + C_2e^{-iwt} ?

 

[BobbyBear]

I got a doubt in the beginning.
After obtaining the characteristic equation
r = \pm iw with i^2 = -1

How do we know x(t) = C_1e^{iwt} + C_2e^{-iwt} ?

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: <br /> e^{rt}<br />

with r being the roots of the characteristic polynomial, which in this case are \pm iw

 

[viciado123]

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: <br /> e^{rt}<br />

with r being the roots of the characteristic polynomial, which in this case are \pm iw

Thank you very much. You know about damped in my other topic ?

 

[viciado123]

Someone has a graphic example of the system to vibration free ?

 

[viciado123]

To plot the graph using Maple 12, which values I use for k, x_o and v_o?