Creating New Logic’s Laws by Using Coding in Microsoft Excel

Law (1):

(p → q) I ((p → q) → r) I (((p → q) → r)) → s) I ((((p → q) → r)) → s) → t) I……

This Logic always “TRUE”

Examples:

Law (1) for three propositions (variables)

Law (1) for four propositions (variables)

Law (1) for five propositions (variables)

Law (2):

(p → q) & ((p → q) → r) & (((p → q) → r)) → s) & ((((p → q) → r)) → s) → t) &……

This logic, under below conditions always is “TRUE”:

·     -    p = False and other propositions (variables) are “TRUE”

·      -   p and q = False and other propositions (variables) are “TRUE”

·     -    All propositions (variables) are “TRUE”

Examples:

Law (2) for three propositions (variables)

Law (2) for four propositions (variables)

Law (2) for five propositions (variables)

To be continued .....

Analysis and Design Open Oscillatory Systems with Forced Harmonic Motion (3)

 Following the article of Analysis and Design Open Oscillatory Systems with Forced Harmonic Motion (2), the purpose of this article is to present a model for finding the characteristics of an oscillatory system with forced harmonic motion where the acceleration will be equal zero. In this case, there is an interesting point that is related to an important difference between SHM and an oscillatory system with forced harmonic motion.

If an object is oscillating under simple harmonic motion, its linear velocity will be zero at the highest and lowest points where we have maximum displacement which is named the amplitude (A).On the other hand; at the maximum level of the displacement (x = A), the acceleration has also its maximum magnitude while at the middle (x = 0), acceleration is zero due to stop at those points in order to change direction while velocity gains its maximum magnitude at the equilibrium point (x = 0). At the extreme ends (x = A), when we have the maximum force and kinetic energy, the acceleration has its maximum magnitude. Therefore, the maximum of acceleration magnitude in simple harmonic motion occurs at maximum displacement (A) and acceleration at the middle is zero when we have the displacement equal to zero just like the below diagrams: 

But there is a different story about the oscillatory systems with forced harmonic motion. In this case, at the some points where the displacement is not zero (x = a), we have the acceleration equal to zero. Please see below diagrams:

Now, the question is: What are the characteristics of an oscillatory system with forced harmonic motion where we have the acceleration equal to zero at the point of x = a?

Below model is able to answer above question:

The components of above model are as follows:

1. In right side on cells L6:M6, we have inputs including given data of “Fm”, and “t”.

2. In left side on cells H6:J7, we have other inputs including lower and upper ranges for independent variables of “m”, “b” and “ω″/ω” to reach the answers for the acceleration equal to zero which are the characteristics of driven oscillatory system. Here, there are lower and upper ranges which are changed by click on cell A2 and also this change will again go back by click on cell B2 (Go & Back).

3. On cells I9:I18, we have outputs which are the characteristics for driven oscillatory system responding to the acceleration equal to zero.

You can see below clip as the example for this model:

All researchers and individual people, who are interested in having this model, don’t hesitate to send their request to below addresses:

Analysis and Design Open Oscillatory Systems with Forced Harmonic Motion (2)

Following to articles of “Analysis and Design Open Oscillatory Systems with Forced Harmonic Motion (1)” (https://emfps.blogspot.com/2014/02/analysis-and-design-open-oscillatory.html) and “A Model to Solve a System of SHM Equations” (https://emfps.blogspot.com/2018/08/a-model-to-solve-system-of-shm-equations_23.html), the purpose of this article is, to find out the characteristics of unknown open oscillatory system which is under forced harmonic motion in the moment of close to resonance. In this case, three models have been designed. One model presents the results for two independent variables “m”, “b” and  another model gives us the results for three independent variables “m”, “b” and “ω''/ω” and the third model shows us the results for higher than a specific velocity.

Statement of the problem

In many situations of oscillatory motions, we have to control the point of the resonance to prevent a huge collapse of system. On the other hand, sometimes we need to know the width of the resonance curve and the characteristics of the system such as natural frequency to increase the velocity of the system by using a forced harmonic motion. In fact, the question is: What is the amount of the force and angular frequency for a forced harmonic motion where the amplitude and velocity of an oscillatory system will significantly increase? These two models are able to answer to this question.

Model (1): The results for two independent variables “mass” and “linear damping constant”

Please be informed that here we encounter with seven independent variables and we have to process all these variables simultaneously for solving the nonlinear equations. For first try, I breakdown the problem and start my analysis by using the most important independent variables which are “m” and “b” and also utilizing the method mentioned in article of “A Model to Solve a System of SHM Equations” (https://emfps.blogspot.com/2018/08/a-model-to-solve-system-of-shm-equations_23.html) accompanied by some tricks in Microsoft excel. I think this is the easiest way. This model says us what are maximum displacement and velocity for a defined range of “m” and “b” close to the point of the resonance.

Below figure as well as shows you the components of this model:

The components of above model are as follows:

1. In right side on cells K6:M6, we have inputs including given data of “ω″/ω”, “Fm”, and “t”.

2. In left side on cells H6:I7, we have other inputs including lower and upper ranges for independent variables of “m” and “b” to reach the answers which are the solution for driven oscillatory system close to the point of the resonance. Here, there are lower and upper ranges which are changed by click on cell A2 and also this change will again go back by click on cell B2 (Go & Back).

3. On cells I9:I17, we have outputs which are maximum displacement and velocity for driven oscillatory system close to the point of the resonance.

Note: On I9 and I10, we have maximum displacement and velocity for a defined range of “m” and “b”. Please do not consider them as the amplitude of harmonic motion. On cell I16 and I17, we have the amplitude of displacement (A) and the amplitude of velocity (Av).

You can see below clip as the examples for this model:

Model (2): The results for three independent variables “mass”, “linear damping constant” and “ω″/ω”

In this model, there are the defined ranges for “m”, “b” and “ω″/ω” which are our three independent variables and by changing the forced harmonic motion (Fm) and time (t), we are willing to know what are the characteristics of a unknown oscillatory system that give us the maximum displacement and velocity close to the point of the resonance.

Below figure as well as shows you the components of this model:

The components of above model are as follows:

1. In right side on cells L6:M6, we have inputs including given data of “Fm”, and “t”.

2. In left side on cells H6:J7, we have other inputs including lower and upper ranges for independent variables of “m”, “b” and “ω″/ω” to reach the answers which are the solution for driven oscillatory system close to the point of the resonance. Here, there are lower and upper ranges which are changed by click on cell A2 and also this change will again go back by click on cell B2 (Go & Back).

3. On cells J9:J17, we have outputs which are maximum displacement and velocity for driven oscillatory system close to the point of the resonance.

Note: On J9 and J10, we have maximum displacement and velocity for a defined range of “m”, “b” and “ω″/ω”. Please do not consider them as the amplitude of harmonic motion. On cell J16 and J17, we have the amplitude of displacement (A) and the amplitude of velocity (Av).

 You can see below clip as the examples for this model:

Model (3): The results of three independent variables “mass”, “linear damping constant” and “ω″/ω” for higher than a specific velocity

In this model, we want to design an unknown oscillatory system for a specific velocity. We can have all characteristics of the system for focus, lower and higher than a specific velocity. Here, I consider it for a higher than specific velocity. Since there are many answers, I have only fixed 15 answers for this model.

 Below figure as well as shows you the components of this model:

As you can see, all components of this model are the similar to model (2). Only I have added15 absolute value for velocities more than a specific velocity and also on cell M7, we have specific amount of velocity.

You can see below clip as the examples for this model:

Data Analysis and Conclusion

 One of the most important applications of above models is to create the value from data analysis. Please see below results:

Above table shows us that there are many answers for a constant velocity. But if you carefully focus on the results, you will find out an interesting property about driven oscillations and resonance.

What is this property?

If an oscillatory system has very low angular frequency (ω) and linear damping constant (b) in the conditions close to the point of the resonance and in zero time (t =0), surprisingly by increasing the mass, the displacement  (x) will go up and by decreasing the mass, the displacement will go down. In this situation, the amplitude for any changes on mass will stay the constant forever.

All researchers and individual people, who are interested in having these models, don’t hesitate to send their request to below addresses:

A Model to Solve a System of SHM Equations

Following to article of “A Model to Solve a System of Nonlinear Equations by Using Microsoft Excel plus VBA” (http://emfps.blogspot.com/2018/02/a-model-to-track-location-of-particle.html), the purpose of this article is, to present another example about solving a system of nonlinear equations for Simple Harmonic Motion (SHM).

Suppose we have two oscillatory motions which are in two perpendicular directions. We can write a system of equations for them as follows:

Assume the angular frequency for both SHMs is the same and also there is the difference between initial phases equal 90 degree. In this case, we will have below system of equations:

This model is able to solve above system of two equations for given data of “x”, “y”, “Ax” and “Ay”.

 There are two methods for finding “ω”, “t” and “φ”. 

Method (1):

The convert of two equations to one equation and solving only one equation for three independent variables in accordance with the method mentioned in article of “Solving a Nonlinear Equation with Many Independent Variables by Using Microsoft Excel plus VBA” (https://emfps.blogspot.com/2016/10/can-we-solve-nonlinear-equation-with.html) as follows:

Below figure as well as shows you the components of this model:

Let me explain you about the components of above model as follows:

1. In right side on cells L6:O6, we have inputs including given data of “x”, “y”, “Ax” and “Ay”.

2. In left side on cells H6:J7, we have other inputs including lower and upper ranges for independent variables of “ω”, “t” and “φ” to reach the answers which are the solution for system of two nonlinear equations. Here, there are lower and upper ranges which are changed by click on cell A2 and also this change will again go back by click on cell B2 (Go & Back).

3. On cells H10:J24, we have outputs which are the answers to above system of two nonlinear equations.

4. On cell M7, we have Error which is the difference between equation of outputs (-y.Ax /x.Ay) and equation of inputs (Tan (ωt+φ))

5. On cells K10:K24, we have the solution of “x” by replacing the answers.

6. On cells M10:M24, we have the solution of “y” by replacing the answers.

7. On cells O10:O24, we have the difference between item 5 (“x”) and cell L6 which are the errors of our answers.

8. On cells P10:P24, we have the difference between item 6 (“y”) and cell M6 which are the errors of our answers.

You can see below clip as the examples for this model:

Method (2):

Using the method stated in article of “A Model to Solve a System of Nonlinear Equations by Using Microsoft Excel plus VBA” (http://emfps.blogspot.com/2018/02/a-model-to-track-location-of-particle.html

In this method, we directly solve a system of two nonlinear equations.  

Below figure as well as shows you the components of this model:

The explanations of the components are the same with method (1) except the error.

You can see below clip as the examples for this model:

All researchers and individual people, who are interested in having this model, don’t hesitate to send 

their request to below addresses:

Is Angular Frequency a Vector or Scalar Quantity?

Angular frequency is the changes of angular displacement to the changes of time:

Therefore, if we say that the angular frequency is a vector, then an angle should be also a vector. In all reference books in physics, you can find that only a very small angles can be considered as a vector. But I am willing to tell you that there are several another factors (except very small angles) which can be defined as impact factors such as Amplitude of an oscillatory motion in which high amplitude does not allow you to use the rules dominated on vectors to analyze an oscillatory system with combination of many harmonic motions.

The surprising news is: "The most important impact factors are not very small angles and amplitude but the direction of the motion. In some situations and conditions, you can even consider a very large angle as a vector where this reverses the concept defined in reference books in Physics."

If you do not truly apply the concept of vector for angular frequency, the result of your analysis and design for an oscillatory system with combination of many harmonic motions, will go in wrong way.

Workshop Series: The Big Data Science Analysis with Microsoft Excel plus VBA

"The Big Data Science Analysis with Microsoft Excel plus VBA"

 There are two big strategies for these workshop series as follows:

1. Solving the complicated problems by producing new methods and models

2.  Creating the value from the big data 

Here is PowerPoint movie about "The Big Data Science Analysis with Microsoft Excel plus VBA"

The Distances among The Particles in The Space (3)

You can preview the conjectures (1), (2), (3) and (4) on below links:

 http://www.emfps.org/2018/04/the-distances-among-particles-in-space.html

 http://www.emfps.org/2018/04/the-distances-among-particles-in-space-2.html

Conjecture (5): 

By using this theorem or conjecture, I am willing to show you that there is below figure among five points in the space:

Suppose we have point P1 (x, y, z) and an independent variable “t” where there is below functions between them:

If points Pm and Pn have below coordination:

Pm (f(x,y,z,t), f(x,y,z,t), f(x,y,z,t))

Pn (g(x,y,z,t), g(x,y,z,t), g(x,y,z,t))

Then, in according to conjecture (3), the distance of points Pm and Pn will be equal with the points P1, P2 and P3 stated in conjecture (3). (Please see conjecture (3) on above links).

It means:

d(Pm,P1)= d(Pm,P2) = d(Pm,P3) = d(Pn,P1) = d(Pn,P2) = d(Pn,P3) =R

Example (1):

Suppose we have below data:

P1 (-11, 3, 31)

t = 129

The results will be as follows:

Example (2):

Suppose we have below data:

P1 (-3.57, 9.4, -1.84)

t = 32.83

The results will be as follows:

The Mapping A System of five Particle for Given Gravity Potential Energy

Suppose we have five particles Pm, Pn, P1, P2 and P3 with the distances among them in infinity which have the same masses of “m”. If an external force brings all these particles in new location just like above figure, how can we map the location of these particles for a constant gravity potential 

energy?

The method of analysis is just like the steps stated in previous article 

(http://www.emfps.org/2018/04/the-distances-among-particles-in-space-2.html)

The only difference is to solve a system of three nonlinear equations instead of a system of two nonlinear equations in previous articles as follows:

Example:

To simplify above system of equations, I consider “t” as a constant number.

Assume we have:

r = 0.24 m

t = 3

Thus, we should solve below system of three equations:

I found 190 models that some of them are as follows:

For testing of these models, I use model (1) in above table and we can see the results as follows: