# Mass-spring system

## Main Question or Discussion Point

Vibration Free

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

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)

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Ya, it's correct, and

$$x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)$$

if you define

$$A \equiv a_1^2 + a_2^2$$
and
$$\delta \equiv arctan(a_1/a_2)$$

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

Ya, it's correct, and

$$x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)$$

if you define

$$A \equiv a_1^2 + a_2^2$$
and
$$\delta \equiv arctan(a_1/a_2)$$

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?

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

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
Homework Helper
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.

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Yes, thank you.

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

I find A with the initial conditions ?

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

$$x'(0) = -Awsin(- \delta) = V_o$$

I'll be A in the function $$\delta$$ ???

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 $$\delta$$

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

$$X(0) = Acos( - \delta) = X_o$$
$$X'(0) = -Awsin(- \delta) = V_o$$

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

And A ???

Well, once you have $$\delta$$

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

$$A= X_0/cos(\delta) = \frac{X_0}{cos(arctan(V_0/X_0w))} = \frac{X_0}{X_0w/\sqrt{X_0^2w^2+V_0^2}} = \frac{\sqrt{X_0^2w^2+V_0^2}}{w}$$

Well, once you have $$\delta$$

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

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

Acos(- \delta ) = X_o

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

and

$$\delta = arctan(V_0/X_0w)$$

so

$$cos(\delta) = cos(arctan(V_0/X_0w))$$

yes?

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

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

and

$$\delta = arctan(V_0/X_0w)$$

so

$$cos(\delta) = cos(arctan(V_0/X_0w))$$

yes?

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

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

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

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: $$e^{rt}$$

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

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: $$e^{rt}$$

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 ?

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

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