Are motion detectors set at 60 seconds or 1 minute??

[Iwonderaboutthings]

Are motion detectors set at 60 seconds or 1 minute??

This information comes from From:

"Properties of Periodic Motion." From the Phsics Classroom:

http://www.physicsclassroom.com/class/waves/Lesson-0/Properties-of-Periodic-Motion

It talks about a mass on a spring bouncing up and down a fixed position, and says that a detector is used to measure the periodic motion of this mass on a spring.

What confuses me however are the figures.

It states that a detector measures this at 0.60 seconds in the vertical Y direction, it also states this:

The small deviation from 2.3 s in the third cycle can be accounted for by the lack of precision in the reading of the graph.

Is this small error due to 0.60 seconds as 60 seconds = 1 minute?? Or is it the detector's unpredictable method of calculation?

Lets not forget that 0.60 seconds also has many conversions and exponents.

0.60 meter = 6e-16 petameter

0.60 meter = 3.712756412e+34 Planck length

THERE IS A VERY IMPORTANT REASON I ASK.

From what I gather this experiment seems much like a sinusoidal wave, and from what I also understand these sinusoidal waves have both electro and magnetic properties in the Y and X directions, but also the Z direction from what I gather still not sure though because of the issue with time in general.

With this:

Does science ever take into accountability that the mass on this spring must also be colliding with 2 invisible barriers, not counting the mass nor the spring itself?

The 2 invisible barriers in this case are the masse's minimum and maximum locations, everything else from their becomes inversely proportional.

Again:

The small deviation from 2.3 s in the third cycle??

It seems this calculation error is trying to resolve 1.

I am willing to accept my thinking may be incorrect and would rather be guided by professionals, but it looks like this may be the case of these invisible barriers in this simple mass on a spring example, and of the greater issues with singularities in general.

The way I see this is that nothing at-least as a physical object never really moves then due to displacements??

We also rely on a standard unit of measure as the second but how can this be correct as infrequent and frequent motions of any seem to not make any sense when it comes to all the above questions?

Again:

Does science ever take into accountability that the mass must also be colliding with 2 invisible barriers?

The more I learn about simple science the more and more it does not make sense due to the issues of time and space...I read that link many times, and still time seems to be undefined, why use it then>>> ? Why do we need time?????? It seems to be impossible.

Edited April 20, 2014 by Iwonderaboutthings

[swansont]

60 seconds is a minute, by definition.

 

0.6 seconds is less than 1 second.

 

But 0.6 seconds is not mentioned in the link, AFAICT. All mentions of 0.6 refer to the displacement of 0.6 m

 

Other than both being periodic phenomena, the mass on a spring has nothing to do with electromagnetic oscillations

[Iwonderaboutthings]

60 seconds is a minute, by definition.

 

0.6 seconds is less than 1 second.

 

But 0.6 seconds is not mentioned in the link, AFAICT. All mentions of 0.6 refer to the displacement of 0.6 m

 

Other than both being periodic phenomena, the mass on a spring has nothing to do with electromagnetic oscillations

The mass on the spring has nothing to do with electromagnetic oscillations??

 

 

Is this because of special relativity and the phenomena of space time, light photons and other quantum mechanical effects?

 

You see this is what I feel confuses many scientist both undergrads and professionals...

Our concept of what is what seems to be very very miss-understood or just not explained correctly, what is it then, is this physics? science? What is the point of the mass and spring example then?

 

I was under the presumption that it describes periodic motion as is the popular case...

 

I think really their is a whole new branch of science out there that maybe we are not told about perhaps??

 

At my level, the basics seem not to make sense anymore with the new current information on science, this is weird..

 

I think I am going to go back and examine every inch of the basics for now on..

Edited April 22, 2014 by Iwonderaboutthings

[swansont]

The mass on the spring has nothing to do with electromagnetic oscillations??

 

 

Is this because of special relativity and the phenomena of space time, light photons and other quantum mechanical effects?

No, it's because a mass on a spring is not the same thing as the oscillation of an electric or magnetic field. There's no reason to invoke relativity or the Planck length. There are no "electromagnetic properties" or quantum mechanical effects that have any relevance to the problem being discussed. It's a mass on a spring. Masses on springs existed and were being analyzed long before people knew of relativity or quantum mechanics.

 

You see this is what I feel confuses many scientist both undergrads and professionals...

Our concept of what is what seems to be very very miss-understood or just not explained correctly, what is it then, is this physics? science? What is the point of the mass and spring example then?

You're reading way too much into the problem. The point of the example is to understand how to solve a problem with an oscillator. That allows you to apply the concept to other systems, since there are several phenomena that oscillate.

 

I was under the presumption that it describes periodic motion as is the popular case...

 

I think really their is a whole new branch of science out there that maybe we are not told about perhaps??

 

At my level, the basics seem not to make sense anymore with the new current information on science, this is weird..

 

I think I am going to go back and examine every inch of the basics for now on..

Studying the basics is very important.

[Iwonderaboutthings]

No, it's because a mass on a spring is not the same thing as the oscillation of an electric or magnetic field. There's no reason to invoke relativity or the Planck length. There are no "electromagnetic properties" or quantum mechanical effects that have any relevance to the problem being discussed. It's a mass on a spring. Masses on springs existed and were being analyzed long before people knew of relativity or quantum mechanics.

 

You're reading way too much into the problem. The point of the example is to understand how to solve a problem with an oscillator. That allows you to apply the concept to other systems, since there are several phenomena that oscillate.

 

Studying the basics is very important.

On another mass and spring example there is a regression performed:

 

http://www.physicsclassroom.com/class/waves/Lesson-0/Motion-of-a-Mass-on-a-Spring

 

 

 

 

 

 

They use Hook's Law:

 

 

 

BUT! in this example on a spring hanging " with no mass connected to it"

 

It states: The spring hangs in a relaxed, un-stretched position.

 

 

Does the force of gravity 9.8 m/s get deduced????

Does 9.8 m/s have no effect on the spring in this un-stretched position?

 

 

The experiment is how much force would certain distances become if you " pulled" on this spring at random.

 

 

Here are the values of this regression analysis:

 

 

This is only one of them:

 

Amount of Stretch (m)=0.0199

Force on Spring (N) = 4.900

Mass (kg) = 0.500

Now this is the regression analysis:

slope = 0.00406 m/N

y-intercept = 3.43 x10^-5 (pert near close to 0.000)

regression constant = 0.999

Stretch = 0.00406•Force + 3.43x10^-5

In this part here is " Force" the gravity on earth's surface??? 9.8 m/s?

On another note here: It seems that the exponent 3.43x10^-5 should be treated as a distance as mass, is this the case?

If the force is 9.8 m/s then should the regression constant = 0.999 simply = 10 seconds?

Like I said its been some time since I have looked at the basics but it seems as if though this would be the case..But if so, it looks to complicate things in other areas of science much much smaller.

I think what I am seeing is that " mass or force" = distance, and time = electricity = frequency, it looks backward the way they are doing it or inversely proportional but proportional to what? 1??

This does not make any sense..

Edited April 23, 2014 by Iwonderaboutthings

[swansont]

On another mass and spring example there is a regression performed:

 

http://www.physicsclassroom.com/class/waves/Lesson-0/Motion-of-a-Mass-on-a-Spring

 

 

 

 

 

 

They use Hook's Law:

 

 

 

BUT! in this example on a spring hanging " with no mass connected to it"

 

It states: The spring hangs in a relaxed, un-stretched position.

 

 

Does the force of gravity 9.8 m/s get deduced????

Does 9.8 m/s have no effect on the spring in this un-stretched position?

Un-stretched means with no extra masses attached.

 

The experiment is how much force would certain distances become if you " pulled" on this spring at random.

 

 

Here are the values of this regression analysis:

 

 

This is only one of them:

 

Amount of Stretch (m)=0.0199

 

Force on Spring (N) = 4.900

 

Mass (kg) = 0.500

 

No, there's not just one. There's a table of nine different values! There's one graph, so there's one regression analysis.

 

 

 

Now this is the regression analysis:

 

 

slope = 0.00406 m/N

y-intercept = 3.43 x10^-5 (pert near close to 0.000)

regression constant = 0.999

 

 

Stretch = 0.00406•Force + 3.43x10^-5

 

 

In this part here is " Force" the gravity on earth's surface??? 9.8 m/s?

 

 

 

 

 

 

On another note here: It seems that the exponent 3.43x10^-5 should be treated as a distance as mass, is this the case?

Distance as mass? What?

 

3.43x10^-5 = 0.0000343

 

It's a distance, and essentially zero.

 

If the force is 9.8 m/s then should the regression constant = 0.999 simply = 10 seconds?

 

There is no time associated with the analysis. The regression constant of 0.999 means the straight line is a very good fit to the data.

 

Like I said its been some time since I have looked at the basics but it seems as if though this would be the case..But if so, it looks to complicate things in other areas of science much much smaller.

 

I think what I am seeing is that " mass or force" = distance, and time = electricity = frequency, it looks backward the way they are doing it or inversely proportional but proportional to what? 1??

 

This does not make any sense..

Hooke's law is that F=-kx. There's a restoring force proportional to the amount of stretch. No time, no frequency, no electricity at all in this experiment.

Edited April 23, 2014 by swansont
Playing hooke-e

[imatfaal]

[unwarranted pedantry]

 

it's Hooke with an E -> http://en.wikipedia.org/wiki/Robert_Hooke

 

[/unwarranted pedantry]

[swansont]

Yeah, I know. Typo, not caught by the spell-checker.

[imatfaal]

I wondered if it had actually been caused by an "auto-correct" feature.

[swansont]

Not for me. I turn that stuff off. I'd rather have something spelled wrong than type and come out with the wrong fusebox.

[Iwonderaboutthings]

Un-stretched means with no extra masses attached.

 

 

No, there's not just one. There's a table of nine different values! There's one graph, so there's one regression analysis.

 

 

 

Distance as mass? What?

 

3.43x10^-5 = 0.0000343

 

It's a distance, and essentially zero.

 

 

There is no time associated with the analysis. The regression constant of 0.999 means the straight line is a very good fit to the data.

 

Hooke's law is that F=-kx. There's a restoring force proportional to the amount of stretch. No time, no frequency, no electricity at all in this experiment.

 

 

No time, no frequency, no electricity at all in this experiment.

 

 

Then what was the point of this experiment and Hook's Law??

 

Everyone makes such a big deal about how famous he became after this law..

 

 

 

 

I assume the restoring force would be g 9.8 m/s on the earth's surface then?

 

http://en.wikipedia.org/wiki/Restoring_force

 

 

What does g on the earth's havto do do with this "No time, no frequency, no electricity at all in this experiment"

 

Does this mean earth's g force on the surface of earth is null = 0????

 

I heard it said before that earth " appears to be stationary" in lectures dealing with Holographic Universes

And String Theory.

 

 

When I said distance as mass, I really should have said radians since this would be the case of distance being = 0..

But never quite figured out its multi-purpose uses...

 

Much wave phenomena uses radians to describe a wave disturbance, I figured the mass on the spring " and spring in general" system had something to do with waves phenomena since the experiment is basically =0

 

 

Like I said I am finding " now " the very basics becoming quite difficult to understand.

[unwarranted pedantry]

 

it's Hooke with an E -> http://en.wikipedia.org/wiki/Robert_Hooke

 

[/unwarranted pedantry]

 

 

 

pedantry "excessive concern with minor details and rules"

 

https://www.google.com/search?q=pedantry&oq=pedantry&aqs=chrome..69i57j0l5.727j0j4&sourceid=chrome&es_sm=93&ie=UTF-8

 

 

 

 

727 runs out of gas in mid air because of unit error

Fuel loading was miscalculated due to a misunderstanding of the recently adopted metric system which replaced the imperial system.

 

http://en.wikipedia.org/wiki/Gimli_Glider

 

 

 

 

""Feynman avoids exposing the reader to the mathematics of complex numbers by using a simple but accurate representation of them as arrows on a piece of paper or screen""

http://en.wikipedia.org/wiki/Quantum_electrodynamics

 

 

 

 

 

 

 

 

NASA's metric confusion caused Mars orbiter loss - CNN.com

 

http://www.cnn.com/TECH/space/9909/30/mars.metric/

 

 

I think you get the point

Edited April 24, 2014 by Iwonderaboutthings

[swansont]

No time, no frequency, no electricity at all in this experiment.

 

 

Then what was the point of this experiment and Hook's Law??

To confirm the validity of Hooke's law, maybe, and learn some experimental and analysis techniques, perhaps? The basic reason you do a school lab?

 

Everyone makes such a big deal about how famous he became after this law..

 

It's not a priori obvious that the restoring force should behave in such a linear fashion.

 

 

I assume the restoring force would be g 9.8 m/s on the earth's surface then?

g is not a force, it's an acceleration, and as such has units of acceleration, not speed.

 

What does g on the earth's havto do do with this "No time, no frequency, no electricity at all in this experiment"

In the setup, there are two forces on the mass: that of the spring, and that of gravity. When the mass is at rest there is no net force on it, so one knows that the force from the spring and that from gravity must cancel, i.e. they have an equal magnitude.

 

mg = kx

 

Since m, g, and x are known, that allows you to determine the spring constant k, and by doing multiple measurements, confirm that it is indeed a constant.

 

 

[Iwonderaboutthings]

To confirm the validity of Hooke's law, maybe, and learn some experimental and analysis techniques, perhaps? The basic reason you do a school lab?

 

 

It's not a priori obvious that the restoring force should behave in such a linear fashion.

 

 

g is not a force, it's an acceleration, and as such has units of acceleration, not speed.

 

In the setup, there are two forces on the mass: that of the spring, and that of gravity. When the mass is at rest there is no net force on it, so one knows that the force from the spring and that from gravity must cancel, i.e. they have an equal magnitude.

 

mg = kx

 

Since m, g, and x are known, that allows you to determine the spring constant k, and by doing multiple measurements, confirm that it is indeed a constant.

 

 

YOUR 100% CORRECT AND THANKS!

I was wondering if this was the case, but rather be told by professionals, now things are much clearer..

 

 

states it here too:

 

 

http://en.wikipedia.org/wiki/G-force

Since such a force is perceived as a weight, any g-force can be described as a "weight per unit mass"

 

 

Many online resources refer to g as 9.8 m/s on the earth's surface so it was a tid but confusion, I should have remember this from some time back...

 

SO THEN??????????

 

What is the earth's surface then in relation to this g force of acceleration??

[swansont]

 

What is the earth's surface then in relation to this g force of acceleration??

 

The value of gravitational acceleration varies with the mass of the planet and the distance from the center (a= GM/r^2). So the location of the surface determines the value of g

[Iwonderaboutthings]

 

The value of gravitational acceleration varies with the mass of the planet and the distance from the center (a= GM/r^2). So the location of the surface determines the value of g

So then earth would need to be calculated relative to another planet?

[swansont]

So then earth would need to be calculated relative to another planet?

 

No, it's an absolute number for our unit system, but you could compare it to other planets. G is a constant, and M and r are determined by the planet. A different celestial body would have its own surface gravity determined by its mass and radius.

[Iwonderaboutthings]

 

No, it's an absolute number for our unit system, but you could compare it to other planets. G is a constant, and M and r are determined by the planet. A different celestial body would have its own surface gravity determined by its mass and radius.

Is it true that in the space shuttle, 9.8 m/s is still the affect of acceleration ??

 

I don't remember the link, but they stated as per special relativity, that if you are inside a closed system " per say" that g = 9.8 m/s still holds true in empty space as it still would on the earth's surface if and only if you are inside this " system" IE space ship, rocket space suit even I assume...

 

In the case of the space suite, would a naked human body in outer space " Their Skin" be applicable as a system on itself??

 

 

Since I am here, if a placed a rock in my hand and held it still, " here on earth" is g still constant with no acceleration??

 

Meaning that so long I don't move my hand while holding the rock " but feeling its weight" does g still register as 9.8 m/s ?

 

 

Does 9.8 m/s also apply at the atomic scale as well.

Meaning that do residual forces, strong and weak " within the atom" also register 9.8 m/s

 

 

From what I am gathering gravity seems to be very complex and I am rethinking many things over from what I had understood years ago...

Edited April 27, 2014 by Iwonderaboutthings

[swansont]

Is it true that in the space shuttle, 9.8 m/s is still the affect of acceleration ??

 

I don't remember the link, but they stated as per special relativity, that if you are inside a closed system " per say" that g = 9.8 m/s still holds true in empty space as it still would on the earth's surface if and only if you are inside this " system" IE space ship, rocket space suit even I assume...

 

In the case of the space suite, would a naked human body in outer space " Their Skin" be applicable as a system on itself??

 

 

Since I am here, if a placed a rock in my hand and held it still, " here on earth" is g still constant with no acceleration??

 

Meaning that so long I don't move my hand while holding the rock " but feeling its weight" does g still register as 9.8 m/s ?

 

 

Does 9.8 m/s also apply at the atomic scale as well.

Meaning that do residual forces, strong and weak " within the atom" also register 9.8 m/s

 

 

From what I am gathering gravity seems to be very complex and I am rethinking many things over from what I had understood years ago...

 

The space shuttles (when they were flying) were not on the earth's surface. You would have to use the value for r for wherever they were; that would reduce their acceleration due to gravity slightly, since r increased slightly.

 

g does not apply in empty space as an acceleration; whatever you read or heard was either wrong or you are mis-remembering it. The only connection to empty space is the value you'd have to recreate as some other acceleration so that you could simulate earth's gravity, because we can't tell the difference between the sources of accelerations.

[Endy0816]

[latex]g=\frac{GM_{Earth}}{r^{2}} =(6.6742\textup{x}10^{-11})\frac{5.9736\textup{x}10^{24}}{(6.37101\textup{x}10^{6})^{2}}=9.822\textup{m}/\textup{s}^{2} [/latex]

 

The equation for determining force due to gravity is:

 

[latex]F=\frac{Gm_{1}m_{2}}{r^{2}} [/latex]

 

So as long as you are on Earth's surface you can use:

 

[latex]F=gm [/latex]

 

If you change your radius or the amount of mass involved, g is no longer relevant.

[Iwonderaboutthings]

[latex]g=\frac{GM_{Earth}}{r^{2}} =(6.6742\textup{x}10^{-11})\frac{5.9736\textup{x}10^{24}}{(6.37101\textup{x}10^{6})^{2}}=9.822\textup{m}/\textup{s}^{2} [/latex]

 

The equation for determining force due to gravity is:

 

[latex]F=\frac{Gm_{1}m_{2}}{r^{2}} [/latex]

 

So as long as you are on Earth's surface you can use:

 

[latex]F=gm [/latex]

 

If you change your radius or the amount of mass involved, g is no longer relevant.

Earth is very very huge and yet still we experiment weight phenomena on the surface of earth, not earth as a whole??

 

 

why?????

 

This formula describes a huge planet, and yet that same value can be found on the surface of earth..

 

why?????

 

 

Accelerometers prove this:

 

 

http://en.wikipedia.org/wiki/Accelerometer

 

Yes I know it deals with acceleration, but it registers 1 g and deduces motion from there...

 

 

Why??

[swansont]

Earth is very very huge and yet still we experiment weight phenomena on the surface of earth, not earth as a whole??

 

 

why?????

 

This formula describes a huge planet, and yet that same value can be found on the surface of earth..

 

why?????

 

 

Because the mass distribution is roughly constant and r doesn't change much. (In reality, small changes in g will occur, because the numbers aren't constant.) That's what the equation is telling you: for a given mass and distance, you will get a certain value for g. Investigating further, we see that m and r are large compared to differences we might normally experience, and small changes in them will have only small changes in the answer.

[Iwonderaboutthings]

 

Because the mass distribution is roughly constant and r doesn't change much. (In reality, small changes in g will occur, because the numbers aren't constant.) That's what the equation is telling you: for a given mass and distance, you will get a certain value for g. Investigating further, we see that m and r are large compared to differences we might normally experience, and small changes in them will have only small changes in the answer.

If you change your radius or the amount of mass involved, g is no longer relevant.

 

Is this what causes, weight phenomena??

 

I look at this as angular momentum, but could be wrong..

 

However, if this is the case, then g seems to have a relation to time...

[swansont]

If you change your radius or the amount of mass involved, g is no longer relevant.

 

Is this what causes, weight phenomena??

 

I look at this as angular momentum, but could be wrong..

 

However, if this is the case, then g seems to have a relation to time...

 

As long as you're on (or very near) the earth's surface, r is basically constant, and so is the mass.

 

This is not angular momentum.