# Solve Mass-Spring System Vibration Free

In summary, the conversation discusses the concept of vibration free and the equation for free vibration, which is x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta). The values for A and \delta can be found by applying the initial conditions, x(0) and x'(0), to the equation. The characteristic equation is also discussed, along with the general solution for a homogeneous linear ordinary differential equation with constant coefficients. Finally, there is a question about plotting a graph using Maple 12 and the values of k, x_o, and v_o for a damped system.
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)

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.

BobbyBear said:
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)

n1person said:
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 $$A cos(wt - \delta)$$

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.

Last edited by a moderator:
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 $$\delta$$

BobbyBear said:
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

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

BobbyBear said:
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)

BobbyBear said:
$$A cos(\delta) = X_0 \rightarrow A = X_0 / cos(\delta)$$

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 $$\pm iw$$

BobbyBear said:
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 $$v_o$$?

## 1. What is a mass-spring system?

A mass-spring system is a physical system that consists of a mass attached to a spring. The mass undergoes simple harmonic motion when the spring is stretched or compressed.

## 2. How do you solve a mass-spring system for vibration free?

To solve a mass-spring system for vibration free, you need to find the natural frequency of the system and then adjust the parameters such as the mass and spring constant to match this frequency. This will result in a system that vibrates with minimal amplitude and without any external forces.

## 3. What is the natural frequency of a mass-spring system?

The natural frequency of a mass-spring system is the frequency at which the system will vibrate without any external forces or damping. It is determined by the mass of the object and the stiffness of the spring.

## 4. How does damping affect a mass-spring system?

Damping is a force that resists the motion of a mass-spring system. It can reduce the amplitude of vibrations and change the natural frequency of the system. In some cases, damping can also eliminate vibrations altogether.

## 5. What are some real-life applications of mass-spring systems?

Mass-spring systems can be found in various applications such as in mechanical watches, car suspensions, and earthquake-resistant buildings. They are also used in musical instruments such as pianos and string instruments to produce sound.

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