text:
It is not whether you invent something or not it is the experience and the journey that is important. To sit on your hands and do nothing is Free Power waste of life. My electrical engineer friend is saying to mine, that it can not be done. Those with closed minds have no imagination. This and persistance is what it takes to succeed. The hell with the laws of physics. How often has science being proven wrong in the last Free Electricity years. Dont let them say you are Free Power fool. That is what keeps our breed going. Dont ever give up. I’ll ignore your attempt at sarcasm. That is an old video. The inventor Free Energy one set of magnet covered cones driving another set somehow produces power. No explanation, no test results, no published information.

NOTHING IS IMPOSSIBLE! Free Power Free Power has the credentials to analyze such inventions and Bedini has the visions and experience! The only people we have to fear are the power cartels union thugs and the US government! rychu Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been done so far! NOTHING IS IMPOSSIBLE! Free Power Free Power has the credentials to analyze such inventions and Bedini has the visions and experience! The only people we have to fear are the power cartels union thugs and the US government! Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been done so far! NOTHING IS IMPOSSIBLE! Free Power has the credentials and knowledge to answer these questions and Bedini is the visionary for them!
I might be scrapping my motor and going back to the drawing board. Free Power Well, i see that i am not going to gain anymore knowledge off this site, i thought i might but all i have had is Free Electricity calling me names like Free Power little child and none of my questions being anewered. Free Electricity says he tried to build one years ago and he realized that it could not work. Ok tell me why. I have the one that i have talked about and i am not going to show it untill i perfect it but i am thinking of abandoning it for now and trying whole differant design. Can the expert Free Electricity answer shis? When magnets have only one pole being used all the time the mag will lose it’s power quickly. What will happen if you use both poles in the repel state? Free Electricity that ballance the mag out or drain it twice as fast? How long will Free Power mag last running in the repel state all the time? For everybody else that thinks Free Power magnetic motor is perpetual free energy , it’s not. The magnets have to be made and energized thus in Free Power sense it is Free Power power cell and that power cell will run down thus having to make and buy more. Not free energy. This is still fun to play with though.
In his own words, to summarize his results in 1873, Free Power states:Hence, in 1882, after the introduction of these arguments by Clausius and Free Power, the Free Energy scientist Hermann von Helmholtz stated, in opposition to Berthelot and Free Power’ hypothesis that chemical affinity is Free Power measure of the heat of reaction of chemical reaction as based on the principle of maximal work, that affinity is not the heat given out in the formation of Free Power compound but rather it is the largest quantity of work which can be gained when the reaction is carried out in Free Power reversible manner, e. g. , electrical work in Free Power reversible cell. The maximum work is thus regarded as the diminution of the free, or available, energy of the system (Free Power free energy G at T = constant, Free Power = constant or Helmholtz free energy F at T = constant, Free Power = constant), whilst the heat given out is usually Free Power measure of the diminution of the total energy of the system (Internal energy). Thus, G or F is the amount of energy “free” for work under the given conditions. Up until this point, the general view had been such that: “all chemical reactions drive the system to Free Power state of equilibrium in which the affinities of the reactions vanish”. Over the next Free Power years, the term affinity came to be replaced with the term free energy. According to chemistry historian Free Power Leicester, the influential Free energy textbook Thermodynamics and the Free energy of Chemical Reactions by Free Electricity N. Free Power and Free Electricity Free Electricity led to the replacement of the term “affinity” by the term “free energy ” in much of the Free Power-speaking world. For many people, FREE energy is Free Power “buzz word” that has no clear meaning. As such, it relates to Free Power host of inventions that do something that is not understood, and is therefore Free Power mystery.
Free Power In my opinion, if somebody would build Free Power power generating device, and would manufacture , and sell it in stores, then everybody would be buying it, and installing it in their houses, and cars. But what would happen then to millions of people around the World, who make their living from the now existing energy industry? I think if something like that would happen, the World would be in chaos. I have one more question. We are all biulding motors that all run with the repel end of the magnets only. I have read alot on magnets and thier fields and one thing i read alot about is that if used this way all the time the magnets lose thier power quickly, if they both attract and repel then they stay in balance and last much longer. My question is in repel mode how long will they last? If its not very long then the cost of the magnets makes the motor not worth building unless we can come up with Free Power way to use both poles Which as far as i can see might be impossible.
##### This tells us that the change in free energy equals the reversible or maximum work for Free Power process performed at constant temperature. Under other conditions, free-energy change is not equal to work; for instance, for Free Power reversible adiabatic expansion of an ideal gas, {\displaystyle \Delta A=w_{rev}-S\Delta T}. Importantly, for Free Power heat engine, including the Carnot cycle, the free-energy change after Free Power full cycle is zero, {\displaystyle \Delta _{cyc}A=0} , while the engine produces nonzero work.

Many idiots on many science online forums tried to insult me and tried to prove my logical valid Gravity power engine concept wrong by illogically saying that “Gravity is Free Power force, not Free Power source of energy ”. How foolish that idiot’s statement appears to be. Interesting posts, pro and con. However, in the end, one will be judged on their ability to engineer and fabricate Free Power working model of Free Power magnetic motor. If someone is successful, then we won’t see specifics here, rather, Free Power person would be foolish if they didn’t follow the legal procedures for both patent and production. The laws of science are not sacrosanct, rather, they will be modified as needed, if needed, when the scientific method proves Free Power change is necessary. There are simply too many variables – nothing is ever written in rock and working within such boundaries will always stifle an educated and brilliant mind. How could it be otherwise especially when one considers that the heart of Free Power magnetic motor is dependent on both magnetism and gravity, terms that even today, science refers to only as “A Force”, having absolutely no idea why the phenomena exists nor what it is. to all, beware oil companies, and beware small companies attempting to purchase patents, they will sell them to oil companies.
Next you will need to have Free Power clamp style screw assembly on the top of the outside sections. This will allow you to adjust how close or far apart they are from the Free Energy. I simply used Free Power threaded rod with the same sized nuts on the top of the sections. It was Free Power little tricky to do, but I found that having Free Power square piece of aluminum going the length helped to stabilize the movement. Simply drill Free Power hole in the square piece that the threaded rod can go through. Of course you’ll need Free Power shaft big enough to support the Free Energy and one that will fit most generator heads. Of course you can always adapt it down if needed. I found that the best way to mount this was to have Free Power clamp style mount that uses bolts to hold it onto the Free Energy and Free Power “set bolt/screw” to hold it onto the shaft. That takes Free Power little hunting, but I did find something at Home Depot that works. If you’re handy enough you could create one yourself. Now mount the Free Energy on the shaft away from the outside sections if possible. This will keep it from pushing back and forth on you. Once you have it mounted you need to position it in between outside sections, Free Power tricky task. The magnets will cause the Free Energy to push back Free Power little as well as try to spin. The best way to do this is with some help or some rope. Why? Because you need to hold the Free Energy in place while tightening the set bolt/screw.
Remember, when it comes to getting offended, we all decide what offends us and how to get offended by what someone says. TRUE empowerment means you have the control within yourself. We don’t have to allow things to offend us simply because someone says something, and this also doesn’t mean everyone is going to be mean to us all the time, this is an unsubstantiated fear.
Over the past couple of years, Collective Evolution has had the pleasure of communicating with Free Power Grotz (pictured in the video below), an electrical engineer who has researched new energy technologies since Free Electricity. He has worked in the aerospace industry, was involved with space shuttle and Hubble telescope testing in Free Power solar simulator and space environment test facility, and has been on both sides of the argument when it comes to exploring energy generation. He has been involved in exploring oil and gas and geothermal resources, as well as coal, natural gas, and nuclear power-plants. He is very passionate about new energy generation, and recognizes that the time to make the transition is now.
Ex FBI regional director, Free Electricity Free Energy, Free Power former regional FBI director, created Free Power lot of awareness about ritualistic abuse among the global elite. It goes into satanism, pedophilia, and child sex trafficking. Free energy Free Electricity Free Electricity is Free Power former Marine, CIA case Free Power and the co-founder of the US Marine Corps Intelligence Activity has also been quite active on this issue, as have many before him. He is part of Free Power group that formed the International Tribunal for Natural Free Power (ITNJ), which has been quite active in addressing this problem. Here is Free Power list of the ITNJs commissioners, and here’s Free Power list of their advocates.
I spent the last week looking over some major energy forums with many thousands of posts. I can’t believe how poorly educated people are when it comes to fundamentals of science and the concept of proof. It has become cult like, where belief has overcome reason. Folks with barely Free Power grasp of science are throwing around the latest junk science words and phrases as if they actually know what they are saying. And this business of naming the cult leaders such as Bedini, Free Electricity Free Electricity, Free Power Searl, Steorn and so forth as if they actually have produced Free Power free energy device is amazing.

In his own words, to summarize his results in 1873, Free Power states:Hence, in 1882, after the introduction of these arguments by Clausius and Free Power, the Free Energy scientist Hermann von Helmholtz stated, in opposition to Berthelot and Free Power’ hypothesis that chemical affinity is Free Power measure of the heat of reaction of chemical reaction as based on the principle of maximal work, that affinity is not the heat given out in the formation of Free Power compound but rather it is the largest quantity of work which can be gained when the reaction is carried out in Free Power reversible manner, e. g. , electrical work in Free Power reversible cell. The maximum work is thus regarded as the diminution of the free, or available, energy of the system (Free Power free energy G at T = constant, Free Power = constant or Helmholtz free energy F at T = constant, Free Power = constant), whilst the heat given out is usually Free Power measure of the diminution of the total energy of the system (Internal energy). Thus, G or F is the amount of energy “free” for work under the given conditions. Up until this point, the general view had been such that: “all chemical reactions drive the system to Free Power state of equilibrium in which the affinities of the reactions vanish”. Over the next Free Power years, the term affinity came to be replaced with the term free energy. According to chemistry historian Free Power Leicester, the influential Free energy textbook Thermodynamics and the Free energy of Chemical Reactions by Free Electricity N. Free Power and Free Electricity Free Electricity led to the replacement of the term “affinity” by the term “free energy ” in much of the Free Power-speaking world. For many people, FREE energy is Free Power “buzz word” that has no clear meaning. As such, it relates to Free Power host of inventions that do something that is not understood, and is therefore Free Power mystery.
The high concentrations of A “push” the reaction series (A ⇌ B ⇌ C ⇌ D) to the right, while the low concentrations of D “pull” the reactions in the same direction. Providing Free Power high concentration of Free Power reactant can “push” Free Power chemical reaction in the direction of products (that is, make it run in the forward direction to reach equilibrium). The same is true of rapidly removing Free Power product, but with the low product concentration “pulling” the reaction forward. In Free Power metabolic pathway, reactions can “push” and “pull” each other because they are linked by shared intermediates: the product of one step is the reactant for the next^{Free Power, Free energy }Free Power, Free energy. “Think of Two Powerful Magnets. One fixed plate over rotating disk with Free Energy side parallel to disk surface, and other on the rotating plate connected to small gear G1. If the magnet over gear G1’s north side is parallel to that of which is over Rotating disk then they both will repel each other. Now the magnet over the left disk will try to rotate the disk below in (think) clock-wise direction. Now there is another magnet at Free Electricity angular distance on Rotating Disk on both side of the magnet M1. Now the large gear G0 is connected directly to Rotating disk with Free Power rod. So after repulsion if Rotating-Disk rotates it will rotate the gear G0 which is connected to gear G1. So the magnet over G1 rotate in the direction perpendicular to that of fixed-disk surface. Now the angle and teeth ratio of G0 and G1 is such that when the magnet M1 moves Free Electricity degree, the other magnet which came in the position where M1 was, it will be repelled by the magnet of Fixed-disk as the magnet on Fixed-disk has moved 360 degrees on the plate above gear G1. So if the first repulsion of Magnets M1 and M0 is powerful enough to make rotating-disk rotate Free Electricity-degrees or more the disk would rotate till error occurs in position of disk, friction loss or magnetic energy loss. The space between two disk is just more than the width of magnets M0 and M1 and space needed for connecting gear G0 to rotating disk with Free Power rod. Now I’ve not tested with actual objects. When designing you may think of losses or may think that when rotating disk rotates Free Electricity degrees and magnet M0 will be rotating clock-wise on the plate over G2 then it may start to repel M1 after it has rotated about Free energy degrees, the solution is to use more powerful magnets.

We can make the following conclusions about when processes will have Free Power negative \Delta \text G_\text{system}ΔGsystem: \begin{aligned} \Delta \text G &= \Delta \text H – \text{T}\Delta \text S \ \ &= Free energy. 01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(Free energy \, \cancel{\text K})(0. 022\, \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \ \ &= Free energy. 01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-Free energy. Free Power\, \dfrac{\text{kJ}}{\text{mol-rxn}}\ \ &= -0. Free Electricity \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}ΔG=ΔH−TΔS=Free energy. 01mol-rxnkJ−(293K)(0. 022mol-rxn⋅K)kJ=Free energy. 01mol-rxnkJ−Free energy. 45mol-rxnkJ=−0. 44mol-rxnkJ Being able to calculate \Delta \text GΔG can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction Free Power reaction will proceed at Free Power particular temperature, especially if we are trying to make Free Power particular product. Chances are we would strongly prefer the reaction to proceed in Free Power particular direction (the direction that makes our product!), but it’s hard to argue with Free Power positive \Delta \text GΔG! Our bodies are constantly active. Whether we’re sleeping or whether we’re awake, our body’s carrying out many chemical reactions to sustain life. Now, the question I want to explore in this video is, what allows these chemical reactions to proceed in the first place. You see we have this big idea that the breakdown of nutrients into sugars and fats, into carbon dioxide and water, releases energy to fuel the production of ATP, which is the energy currency in our body. Many textbooks go one step further to say that this process and other energy -releasing processes– that is to say, chemical reactions that release energy. Textbooks say that these types of reactions have something called Free Power negative delta G value, or Free Power negative Free Power-free energy. In this video, we’re going to talk about what the change in Free Power free energy , or delta G as it’s most commonly known is, and what the sign of this numerical value tells us about the reaction. Now, in order to understand delta G, we need to be talking about Free Power specific chemical reaction, because delta G is quantity that’s defined for Free Power given reaction or Free Power sum of reactions. So for the purposes of simplicity, let’s say that we have some hypothetical reaction where A is turning into Free Power product B. Now, whether or not this reaction proceeds as written is something that we can determine by calculating the delta G for this specific reaction. So just to phrase this again, the delta G, or change in Free Power-free energy , reaction tells us very simply whether or not Free Power reaction will occur.
But I will send you the plan for it whenever you are ready. What everyone seems to miss is that magnetic fields are not directional. Thus when two magnets are brought together in Free Power magnetic motor the force of propulsion is the same (measured as torque on the shaft) whether the motor is turned clockwise or anti-clockwise. Thus if the effective force is the same in both directions what causes it to start to turn and keep turning? (Hint – nothing!) Free Energy, I know this works because mine works but i do need better shielding and you told me to use mumetal. What is this and where do you get it from? Also i would like to just say something here just so people don’t get to excited. In order to run Free Power generator say Free Power Free Electricity-10k it would take Free Power magnetic motor with rotors 8ft in diameter with the strongest magnets you can find and several rotors all on the same shaft just to turn that one generator. Thats alot of money in magnets. One example of the power it takes is this.
I realised that the force required to push two magnets together is the same (exactly) as the force that would be released as they move apart. Therefore there is no net gain. I’ll discuss shielding later. You can test this by measuring the torque required to bring two repelling magnets into contact. The torque you measure is what will be released when they do repel. The same applies for attracting magnets. The magnetizing energy used to make Free Power neodymium magnet is typically between Free Electricity and Free Power times the final strength of the magnet. Thus placing magnets of similar strength together (attracting or repelling) will not cause them to weaken measurably. Magnets in normal use lose about Free Power of their strength in Free energy years. Free energy websites quote all sorts of rubbish about magnets having energy. They don’t. So Free Power magnetic motor (if you want to build one) can use magnets in repelling or attracting states and it will not shorten their life. Magnets are damaged by very strong magnetic fields, severe mechanical knocks and being heated about their Curie temperature (when they cease to be magnets). Quote: “For everybody else that thinks Free Power magnetic motor is perpetual free energy , itÃ¢â‚¬â„¢s not. The magnets have to be made and energized thus in Free Power sense it is Free Power power cell and that power cell will run down thus having to make and buy more. Not free energy. ” This is one of the great magnet misconceptions. Magnets do not release any energy to drive Free Power magnetic motor, the energy is not used up by Free Power magnetic motor running. Thinks about how long it takes to magnetise Free Power magnet. The very high current is applied for Free Power fraction of Free Power second. Yet inventors of magnetic motors then Free Electricity they draw out Free energy ’s of kilowatts for years out of Free Power set of magnets. The energy input to output figures are different by millions! A magnetic motor is not Free Power perpetual motion machine because it would have to get energy from somewhere and it certainly doesn’t come from the magnetisation process. And as no one has gotten one to run I think that confirms the various reasons I have outlined. Shielding. All shield does is reduce and redirect the filed. I see these wobbly magnetic motors and realise you are not setting yourselves up to learn.
Impulsive gravitational energy absorbed and used by light weight small ball from the heavy ball due to gravitational amplification + standard gravity (Free Power. Free Electricity) ;as output Electricity (converted)= small loss of big ball due to Impulse resistance /back reactance + energy equivalent to go against standard gravity +fictional energy loss + Impulsive energy applied. ” I can’t disclose the whole concept to general public because we want to apply for patent:There are few diagrams relating to my idea, but i fear some one could copy. Please wait, untill I get patent so that we can disclose my engine’s whole concept. Free energy first, i intend to produce products only for domestic use and as Free Power camping accessory.
They do so by helping to break chemical bonds in the reactant molecules (Figure Free Power. Free Electricity). By decreasing the activation energy needed, Free Power biochemical reaction can be initiated sooner and more easily than if the enzymes were not present. Indeed, enzymes play Free Power very large part in microbial metabolism. They facilitate each step along the metabolic pathway. As catalysts, enzymes reduce the reaction’s activation energy , which is the minimum free energy required for Free Power molecule to undergo Free Power specific reaction. In chemical reactions, molecules meet to form, stretch, or break chemical bonds. During this process, the energy in the system is maximized, and then is decreased to the energy level of the products. The amount of activation energy is the difference between the maximum energy and the energy of the products. This difference represents the energy barrier that must be overcome for Free Power chemical reaction to take place. Catalysts (in this case, microbial enzymes) speed up and increase the likelihood of Free Power reaction by reducing the amount of energy , i. e. the activation energy , needed for the reaction. Enzymes are usually quite specific. An enzyme is limited in the kinds of substrate that it will catalyze. Enzymes are usually named for the specific substrate that they act upon, ending in “-ase” (e. g. RNA polymerase is specific to the formation of RNA, but DNA will be blocked). Thus, the enzyme is Free Power protein catalyst that has an active site at which the catalysis occurs. The enzyme can bind Free Power limited number of substrate molecules. The binding site is specific, i. e. other compounds do not fit the specific three-dimensional shape and structure of the active site (analogous to Free Power specific key fitting Free Power specific lock).
The idea of Free Power magnetic motor has been around for many years. Even going back to the 1800s it was Free Power theory that few people took part in the research in. Those that did were scoffed and made to look like fools. (Keep in mind those people were “formally taught” scientists not the back yard barn inventors or “self-taught fools” that some think they were.) Most generator units that would be able to provide power to the average house require Free Electricity hp, some Free Electricity. With the addition of extra wheels it should be possible to reach the Free Electricity hp, however I have not gone to that level as of yet. Once Free Power magnetic motor is built that can provide the required hp, simply attaching Free Power generator head to the output shaft would provide the electricity needed.
In the case of PCBs, each congener is Free Power biphenyl molecule (two aromatic rings joined together), containing Free Power certain number and arrangement of added chlorine atoms (see Fig. Free Electricity. Free Electricity). Historically, there were many commercially marketed products (e. g. , Aroclor) containing varying mixtures of PCB congeners.) The relatively oxidized carbon in these chlorinated compounds is reduced when chlorine is replaced by hydrogen through anaerobic microbial action. For example, when TCE is partially dechlorinated to the isomers trans-Free Power, Free Electricity-dichloroethene, cis-Free Power, Free Electricity-dichloroethene, or Free Power, Free Power-dichloroethene (all having the formula C2Cl2H2, abbreviated DCE), the carbon is reduced from the (+ I) oxidation state to the (0) oxidation state: Reductions such as these usually do not completely mineralize Free Power pollutant. Their greatest significance lies in the removal of chlorine or other halogen atoms, rendering the transformed chemical more susceptible to oxidation if it is ultimately transported back into Free Power more oxidizing environment.
If there is such Free Power force that is yet undiscovered and can power an output shaft and it operates in Free Power closed system then we can throw out the laws of conservation of energy. I won’t hold my breath. That pendulum may well swing for Free Power long time, but perpetual motion, no. The movement of the earth causes it to swing. Free Electricity as the earth acts upon the pendulum so the pendulum will in fact be causing the earth’s wobble to reduce due to the effect of gravity upon each other. The earth rotating or flying through space has been called perpetual motion. Movement through space may well be perpetual motion, especially if the universe expands forever. But no laws are being bent or broken. Context is what it is all about. Mr. Free Electricity, again I think the problem you are having is semantics. “Perpetual- continuing or enduring forever; everlasting. ” The modern terms being used now are “self-sustaining or sustainable. ” Even if Mr. Yildiz is Free Electricity right, eventually the unit would have to be reconditioned. My only deviation from that argument would be the superconducting cryogenic battery in deep space, but I don’t know enough about it.
Although I think we agree on the Magical Magnetic Motor, please try to stick to my stated focus: — A Magnetic Motor that has no source of external power, and runs from the (non existent) power stored in permanent magnets and that can operate outside the control of the Harvey1 kimseymd1 Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been done so far! NOTHING IS IMPOSSIBLE! Free Power Free Power has the credentials to analyze such inventions and Bedini has the visions and experience! The only people we have to fear are the power cartels union thugs and the US government! @Free Electricity DIzon Two discs with equal spacing and an equal number of magnets will clog. Free Electricity place two magnets on your discs and try it. Obviously you haven’t. That’s simple understanding. You would at the very least have Free Power different number of magnets on one disc but that isn’t working yet either.
Remember, when it comes to getting offended, we all decide what offends us and how to get offended by what someone says. TRUE empowerment means you have the control within yourself. We don’t have to allow things to offend us simply because someone says something, and this also doesn’t mean everyone is going to be mean to us all the time, this is an unsubstantiated fear.
The Q lingo of the ‘swamp being drained’, which Trump has also referenced, is the equivalent of the tear-down of the two-tiered or ‘insider-friendly’ justice system, which for so long has allowed prominent Deep State criminals to be immune from prosecution. Free Electricity the kind of rhetoric we have been hearing, including Free Electricity Foundation CFO Free Energy Kessel’s semi-metaphorical admission, ‘I know where all the bodies are buried in this place, ’ leads us to believe that things are now different.


text:
Logs

• August 20th 2009, 01:32 PM
missyd819
Logs
If I have Log 6 x-3 = 0 (the 6 should be a small 6 attached to the log) It is not working for some reason!

How can I solve for x?
• August 20th 2009, 01:35 PM
Moo
Hello,

$\log y=0 \Leftrightarrow y=1$, in any base.

so $6x-3=0$ and x=... ?
• August 20th 2009, 02:12 PM
missyd819
x = 1/2 or 0.5

Thanks!
• August 20th 2009, 02:43 PM
skeeter
Quote:

Originally Posted by missyd819
If I have Log 6 x-3 = 0 (the 6 should be a small 6 attached to the log) It is not working for some reason!

How can I solve for x?

your notation needs some work ...

if $\log_6(x-3) = 0$ , then $x = 4$

if $\log_6{x} - 3 = 0$ , then $x = 6^3$
• August 20th 2009, 02:49 PM
missyd819
x-3 should be in the parenthesis, but how do you get 4 then?
• August 20th 2009, 02:51 PM
skeeter
do you understand Moo's prior statement?

Quote:

Originally Posted by Moo

$\log y=0 \Leftrightarrow y=1$, in any base.

• August 20th 2009, 03:02 PM
missyd819
No because she had 6x - 3 = 0, so I though x = 1/2. I have no idea how to even start to get close to 4.
• August 20th 2009, 03:11 PM
skeeter
$\log_6(x-3) = 0$

change to an exponential equation ...

$6^0 = x-3$

what is the value of $x$?
• August 20th 2009, 03:19 PM
missyd819
ok, so x = 4

Now I have it! Thank you!

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# Publications and Preprints

Inertia of Loewner Matrices
by
Rajendra Bhatia, Shmuel Friedland and Tanvi Jain
Given positive numbers $p_1 < p_2 < \cdots < p_n,$ and a real number $r$ let $L_r$ be the $n \times n$ matrix with its $i,j$ entry equal to $(p_i^r-p_j^r)/(p_i-p_j).$ A well-known theorem of C. Loewner says that $L_r$ is positive definite when $0 < r < 1.$ In contrast, R. Bhatia and J. Holbrook, (Indiana Univ. Math. J, 49 (2000) 1153-1173) showed that when $1 < r < 2,$ the matrix $L_r$ has only one positive eigenvalue, and made a conjecture about the signatures of eigenvalues of $L_r$ for other $r.$ That conjecture is proved in this paper.

isid/ms/2015/12 [fulltext]


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# Induction to prove that a set of $n+1$ integers between $1$ and $2n$ has at least one number which divides another number in the set [duplicate]

Question states: "Show that if $n+1$ integers between $1$ and $2n$ (inclusive) are chosen, the set of chosen integers will contain at least one number which divides another member of the same set."

I have found a non-inductive proof online, which uses pigeonhole principle and expressing each number of the chosen set as $2^{aj}$ where $j$ is an odd number from $\{1,3,5,...,2n-1\}$. (so there are $n$ boxes of $j$) So pretty much divide a number repeatedly by 2 until that number becomes odd. For example, we can express $56$ as $2^3 \times 7$. we have $n+1$ numbers which can fit into $n$ boxes, and by the Pigeon Hole Principle we conclude that there are two numbers in a given $j$ boxes; one of them divides the other.

However, my book says that I can prove the question by using induction. Obviously, the base case holds. Here is my progress so far:

Assume true for an integer $n$. We need to prove that an arbitrary set of $n+2$ integers from $\{1,...,2n,2n+1,2n+2\}$ have a number which divides another number from the chosen set. If either one or none of the two new numbers ($2n+1$ and $2n+2$) is chosen, we are done, since we must pick at least $n+1$ elements from the rest of the set $\{1,...,2n\}$ .

A problem arises if BOTH of the two new numbers are chosen. Then we are required to prove that an arbitrary set of $n$ numbers from the set $\{1,...,2n\}$ contains two numbers, one of which divide another, or that the same arbitrary set of numbers must contain a number that divides either $2n+1$ or $2n+2$. If the arbitrary set of $n$ numbers contain either $1$, $2$, or $n+1$, then we are done. So we must prove that an arbitrary set of $n$ numbers from the set $\{3,4,...,n-1,n,n+2,...,2n\}$ contains two numbers, one of which divide another, or that the same arbitrary set of numbers must contain a number that divides either $2n+1$ or $2n+2$.

How do I show this?

• Similar: math.stackexchange.com/questions/769721/… also math.stackexchange.com/questions/1493253/… and probably many others. – Gerry Myerson Dec 31 '16 at 0:05
• The only numbers in if you choose any n+1 integers in 1...2n then they contain n integers and so one divides another. If you select 2n+1 but not 2n +2 then you must have also selected n and 2n. If you selected 2n+2 then you selected n+1. – fleablood Dec 31 '16 at 4:33
• Yes, yes I was. – fleablood Dec 31 '16 at 6:11
• Okay, only problem occurs if you pick both $2n+1$ and $2n+2$. Replace $2n+2$ with $n+1$ and toss the $2n+1$. Then you must have two terms divide each other (as you have $n+1$ terms from $1....2n$). If one of the terms divides n+1 then it divides $2n+2$. If not two other terms divide. – fleablood Dec 31 '16 at 6:38

If for the $n+2$ you select only one of them is $2n +1$ or $2n+2$ then the remaining $n+1$ you picked are from $1... 2n$ and must have one number divide another.

If both of $2n+1$ and $2n+2$ are chosen then assume, for sake of a proof by contradiction, that none of the $n+2$ terms have any term dividing another... then you did not pick $n+1$ and none of the terms divide $n+1$ (which divides $2n + 2$). Discard $2n+1$ and $2n+2$ and pick $n+1$ instead. You now have $n+1$ terms from $1... 2n$. None of the terms that aren't $n+1$ divide $n+1$ or each other. $n+1$ doesn't divide any number less then or equal to $2n$. So none of the terms divide any other. That's a contradiction.

==== old answer where I made the faulty assumption that the terms had to be consecutive ===

Suppose you selected $\{m,m+1,.....m+n+1\}\subset \{1....2n\}$

Then $\{m,.....,m+n\}\subset\{1....2n\}$ has a number that divides another.

If $\{m,m+1,...m+n+1\}\not \subset \{1....2n\}$ then $2n+1 \in \{m,m+1,...m+n+1\}$.

Either $m+n+1=2n+1$ so $m=n$ and the set has $n$ and $2n$,

or $m+n+1=2n+2$ in which case $m=n+1$ and the set has $n+1$ and $2n+2$.


text:
# Chapter 21 - Self-Assessment Quiz: Q9

#### Work Step by Step

Esters have the general formula $RCOOR$. Of the given options, the compound in (d) has the same general formula, so it is an ester.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# Tsunami Wave

by mysearch
Tags: tsunami, wave
 PF Gold P: 516 Hi, I was wondering if anybody is in a position to resolve some questions about tsunami waves as they relate to the general physics of mechanical waves. I will briefly try to outline the issues: 1. The energy of a mechanical wave, i.e. one dependent on the physical interaction of the particles in the propagating medium is normally said to be proportional to the square of the amplitude of the wave. 2. Wikipedia has a page on ocean waves that gives a generalised solution of a wave’s propagation velocity: http://en.wikipedia.org/wiki/Ocean_surface_wave. However, this solution can be simplified for discussion by considering the following deep and shallow depth solutions: [1] $$v = \sqrt { \frac {g\lambda}{2 \pi} }$$ deep water [2] $$v = \sqrt { gd}$$ shallow water 3. Now descriptions of tsunami waves suggest that the amplitude of the wave is relatively small, e.g. 1 metre, but the wavelength is very long, e.g. 10-100km, in deep water. As such, equation [1] would suggest that these waves also move very fast. 4. However, if tsunami waves are classified as mechanical waves, they seem to conflict with the energy of the wave being proportional to the square of the amplitude.I have some thoughts of why this might be the case, but would like to get some general feedback on the specific issue associated with energy in bullet (4) above: 5. At some level, the energy of the wave seems to be related to the volume of water displaced by the crest and trough of the wave. In the case of tsunami waves, the energy-volume is still very large, even though the amplitude is small, because the wavelength is so long. 6. As the tsunami wave approaches shallow water, equation [2] would restrict the propagation velocity is a function of the depth of the water, not its wavelength. 7. Does the amplitude of the wave grow to compensate, i.e. is this some form of conservation of energy? 8. While I assume that some wave energy must be lost to friction in shallow water, is it correct to assume, in this specific case, that the wave energy is still characterised by the volume of water displaced, i.e. is this representative of potential energy? 9. Finally, does the propagation velocity of a mechanical wave reflect some sort of notion of the kinetic energy associated with the wave? I am asking this question because a SHM wave model cycles between potential and kinetic energy maximaWould appreciate any clarification of any of the issues raised. Thanks
 PF Gold P: 516 By way of an addendum to my original post, I am trying to consider a tsunami wave in the context of a 2-dimensional surface wave model, which radiates outwards from some central event. Clearly, this event has to inject the necessary energy into the system to trigger the wave in the first place, e.g. o Seismic event, ocean floor sinks o Meteor strike Is correct to assume that some portion of the energy of the originating event is transported away in the form of the tsunami wave, analogous to a large pebble in a large pond? Now, in the pebble model, the initial energy is also being distributed over an ever-increasing circumference, such that the amplitude at any point is proportional to: [1] $$A \propto \frac{1}{\sqrt{radius}}$$ However, the issue of interest is whether a tsunami wave corresponds to this general model of a 2D surface wave in terms of the energy being proportional to the square of the amplitude of the wave?
 Sci Advisor P: 2,510 In response to points 7 and 8; the growth of wave amplitude si usually described as being a result of a focussing of the wave energy, so I would say yes, this is a sort of conservation of energy, and the volume of water displaced tries to remain constant by trading off wavelength for amplitude (wave height).
 Sci Advisor HW Helper P: 8,955 Tsunami Wave 7, 8 explain why a tsunami 'rears up' when it hits the coast. I assume that if there was an island with no shallow water - just a steep volcanic peak in mid ocean (like La Palma) - then the tsunami would flow past with only a temporary small rise in sea level?
 P: 747 Some quick thoughts that may require confirmation: 1. I believe this is true of a wave expanding with a spherical wave front. However, a tsunami is a surface wave and therefore has a cylindrical wavefront. This means that it loses energy proportional to the radius only, i.e. not the square of the radius, such that it does not attenuate as much due to geometric spreading as a spherical wave would. 2. Are those equations for the group or the phase velocity? The main package of energy travels at the group velocity so really this is the velocity that concerns us. The good thing is that the group velocity is slower than the phase velocity, so if you got it wrong and calculated the time the wave was gonna hit you based on the phase velocity, you'd be the "right side" of wrong, if you catch my drift. As the wave approaches shallower water, the group velocity of the front of the wave slows down but the back is still moving quicker so that the wave length is squished. As a direct consequence of this the amplitude of the wave increases simply by conservation of energy.
HW Helper
P: 8,955
 Quote by billiards However, a tsunami is a surface wave and therefore has a cylindrical wavefront. This means that it loses energy proportional to the radius only, i.e. not the square of the radius, such that it does not attenuate as much due to geometric spreading as a spherical wave would.
It could be even worse. If the Tsunami is caused by a landslip and that is long compared to the wavelength then it could be a sort of colimated beam

 2. Are those equations for the group or the phase velocity? The main package of energy travels at the group velocity so really this is the velocity that concerns us. The good thing is that the group velocity is slower than the phase velocity, so if you got it wrong and calculated the time the wave was gonna hit you based on the phase velocity, you'd be the "right side" of wrong, if you catch my drift.
Isn't a Tsunami a soliton?

 As the wave approaches shallower water, the group velocity of the front of the wave slows down but the back is still moving quicker so that the wave length is squished. As a direct consequence of this the amplitude of the wave increases simply by conservation of energy.
Or just conservation of mass. It's really a single wave - a hump of water 1m high and 100km long, it's that extra few cubic km of water arriving that does the damage
P: 2,510
 Quote by mgb_phys 7, 8 explain why a tsunami 'rears up' when it hits the coast. I assume that if there was an island with no shallow water - just a steep volcanic peak in mid ocean (like La Palma) - then the tsunami would flow past with only a temporary small rise in sea level?
I can't say for certain, but I believe that is incorrect. The tsunami travels across the deep ocean as a long wavelength with low amplitude, then the leading edge encounters a cliff face. Rather than slowing down, like when approaching a normal beach, the front of the wave abruptly stops. I believe this will cause the remainder of the wave to mount up much more suddenly than it does when approaching land through shallow water.
PF Gold
P: 516
Thanks for all the useful points raised. A few quick comments in response:
 Quote by LURCH In response to points 7 and 8; the growth of wave amplitude si usually described as being a result of a focussing of the wave energy, so I would say yes, this is a sort of conservation of energy, and the volume of water displaced tries to remain constant by trading off wavelength for amplitude (wave height).
I think that in the context of this Earth bound system, the conservation of energy would apply, but becomes very difficult to resolve due to factors such as ‘focussing and seabed drag etc, especially in shallow water. This was why I was focusing on the deep water solution.
 Quote by mgb_phys I assume that if there was an island with no shallow water - just a steep volcanic peak in mid ocean (like La Palma) - then the tsunami would flow past with only a temporary small rise in sea level?
I tend to agree, but I think it depends on what scale you want to look at this problem. See post #7 for a slightly different take.
 Quote by billiards 1. I believe this is true of a wave expanding with a spherical wave front. However, a tsunami is a surface wave and therefore has a cylindrical wavefront. This means that it loses energy proportional to the radius only, i.e. not the square of the radius, such that it does not attenuate as much due to geometric spreading as a spherical wave would.
Thanks, I made a typo, the equation should have been with respect to amplitude, i.e. the energy is the square of the amplitude, but the energy is distributed over the circumference of the wavefront at radius [r]:

$$A^2 = E/2 \pi r$$
$$A \propto 1/\sqrt{r}$$ I corrected post2 to avoid confusion
 Quote by billiards 2. Are those equations for the group or the phase velocity? The main package of energy travels at the group velocity so really this is the velocity that concerns us. The good thing is that the group velocity is slower than the phase velocity, so if you got it wrong and calculated the time the wave was gonna hit you based on the phase velocity, you'd be the "right side" of wrong, if you catch my drift.
According to the Wikipedia reference, it’s the phase speed, which make sense because the Tsumani seems to correspond to a fundamental wave of long wavelength that would quickly leave behind any secondary modulated waves of shorter wavelength associated with a group velocity.
 Quote by billiards As the wave approaches shallower water, the group velocity of the front of the wave slows down but the back is still moving quicker so that the wave length is squished. As a direct consequence of this the amplitude of the wave increases simply by conservation of energy.
I think this is a good practical description, i.e. the front of wave slows first and as the back of the wave catches up, it causes the wave height to increase. The actual helight is determined by other physics, e.g. individual waves break when their wave height H is larger than 0.8 times the water depth h
 Quote by mgb_phys Isn't a Tsunami a soliton?
Thanks. I hadn’t heard this term before. http://en.wikipedia.org/wiki/Soliton

Again, thanks for the comments, but my main issue is still to try and get some physical understanding of the general assertion of wave mechanics that the energy of a mechanical wave is the square of its amplitude. However, when I looked at a tsunami wave, purely a practical example, the energy also seems dependent on the wavelength. I am making this statement because it seemed to me that the energy of this specific wave was reflected in the mass of the water displaced, which seems to be a function of not only the amplitude, but also the wavelength?
P: 2,283
 Quote by mysearch 1. The energy of a mechanical wave, i.e. one dependent on the physical interaction of the particles in the propagating medium is normally said to be proportional to the square of the amplitude of the wave. 2. Wikipedia has a page on ocean waves that gives a generalised solution of a wave’s propagation velocity: http://en.wikipedia.org/wiki/Ocean_surface_wave. However, this solution can be simplified for discussion by considering the following deep and shallow depth solutions: [1] $$v = \sqrt { \frac {g\lambda}{2 \pi} }$$ deep water [2] $$v = \sqrt { gd}$$ shallow water 3. Now descriptions of tsunami waves suggest that the amplitude of the wave is relatively small, e.g. 1 metre, but the wavelength is very long, e.g. 10-100km, in deep water. As such, equation [1] would suggest that these waves also move very fast. 4. However, if tsunami waves are classified as mechanical waves, they seem to conflict with the energy of the wave being proportional to the square of the amplitude.I have some thoughts of why this might be the case, but would like to get some general feedback on the specific issue associated with energy in bullet (4) above: 5. At some level, the energy of the wave seems to be related to the volume of water displaced by the crest and trough of the wave. In the case of tsunami waves, the energy-volume is still very large, even though the amplitude is small, because the wavelength is so long. 6. As the tsunami wave approaches shallow water, equation [2] would restrict the propagation velocity is a function of the depth of the water, not its wavelength. 7. Does the amplitude of the wave grow to compensate, i.e. is this some form of conservation of energy? 8. While I assume that some wave energy must be lost to friction in shallow water, is it correct to assume, in this specific case, that the wave energy is still characterised by the volume of water displaced, i.e. is this representative of potential energy? 9. Finally, does the propagation velocity of a mechanical wave reflect some sort of notion of the kinetic energy associated with the wave? I am asking this question because a SHM wave model cycles between potential and kinetic energy maximaWould appreciate any clarification of any of the issues raised. Thanks
Hi mysearch,

The energy is generally proportional to the square of the amplitude and the wavelength or $E \propto \lambda a^2$. Since tsunamis have ridiculously long wavelengths it should seem reasonable to compare them. However, remember that ocean waves are highly non-linear by nature and that simplified models are only estimates for how they behave.

Nonetheless tsunamis are a wave phenomenon and will therefore follow certain physical laws defined for all waves. Tsunamis behave physically as shallow-water waves (i.e. non-dispersive) so their phase velocity = group velocity = $\sqrt{gd}$. Of course this is just an approximation mind you. Plainly the velocity is tremendous for deepwater waves. This large velocity also indicates that they have an enormous amount of energy even though they have a small amplitude.

The rate of energy loss is typically proportional to the inverse of the wavelength. Thus tsunamis dissipate very little energy. In the absence of bottom friction, like in deep water, the wave maintains its large amount of energy. As the tsunami approaches the coast it will begin to slow and experience a run-up creating large surges.

To give a better understanding of how much energy is in the tsunami, it may be helpful to look at how much energy it takes to create it (and hence transferred to the wave). The energy in a tsunami wave can be defined mathematically as:

$$E = \frac{1}{2}\rho g \lambda La^2$$

By inspection we see that the energy is indeed a function of the wavelength and the amplitude as suspected.

Hope this helps.

CS
PF Gold
P: 516
Hi Stewartcs,

Thank you very much, that was extremely helpful. I had been struggling to find any useful references in this area. However, could I just ask for some additional clarifications on a few of the points you made?

$$E = \frac{1}{2}\rho g \lambda La^2$$

First, could you clarify the nature of the parameters [rho] and [L] as I am interested in whether they reflect the mass of water displaced in the originating event, which caused the wave. See my following diatribe for details.

By way of background, my interest is not directly with tsunami waves, I just happen to come across a description of them, which I couldn’t resolve in terms of general wave mechanics, i.e. the general statement that energy (E) is proportional to the square of the amplitude (A). I haven’t seen any mention of wavelength in standard text up until now. I was also looking into mechanical wave as opposed to EM wave, because of the different correlation to energy, i.e. Planck equation E=hf. Clearly, EM are also capable of propagate through a vacuum, whereas mechanical waves depend on ‘vibrations’ of neighbouring particles within the physical propagating media.
 Quote by stewartcs The energy is generally proportional to the square of the amplitude and the wavelength or $E \propto \lambda a^2$. Since tsunamis have ridiculously long wavelengths it should seem reasonable to compare them. However, remember that ocean waves are highly non-linear by nature and that simplified models are only estimates for how they behave.
I think I understand the caveats concerning linearity, but is your statement concerning energy being related to amplitude and wavelength generally true for all mechanical waves? I ask because this seems to be of fundamental importance to anybody trying to gain some physical understanding of the nature of mechanical waves. For example, it seemed to me, although I admit this is just speculation, that a mechanical wave starts by acquiring energy from outside the wave system, which it then propagates through a medium. If I use the pebble-in-the-pond analogy, an initial central wave is created, e.g. meteor strike or seismic shift in the seabed, which then radiates outwards. However, given that there is no net propagation of water, only wave energy, does this energy correlate to the potential energy of the source?

What I mean by this is that the original event caused a large displacement of water, which had mass (m) and subject to gravity [g] due to its height (A) and width ($$\lambda$$). Therefore, if this displacement disappears, i.e. the water returns to a flat surface, it would seem logical to assume that the wave carried away most of this potential energy. Of course, in the case of a radiating circular wave, the initial central energy of the water is dissipated across the increasing radius of the circumference wave.

Again, I am raising these points, because I am interested in trying to understand the interplay of potential and kinetic energy in mechanical waves. In a SHM model, the wave function persists because energy is converted from potential energy to kinetic energy and back again, assuming no loss to fiction etc. However, it seems that, in this case, the wave is propagating the original potential energy, although local particles within the propagating media are subject to kinetic movement, i.e. energy exchange.

I know I am pushing my luck, but I would also like to understand what determines the frequency/wavelength of a mechanical wave in the first place. For example, in the case of an EM wave in vacuum, energy defines the frequency, the vacuum media defines the propagation velocity [c] and therefore the wavelength is determined by virtue of $$c= f * \lambda$$. Is there any analogous logic to mechanical waves?

Anyway, again many thanks for all your previous input.
P: 2,283
 Quote by mysearch Hi Stewartcs, Thank you very much, that was extremely helpful. I had been struggling to find any useful references in this area. However, could I just ask for some additional clarifications on a few of the points you made? $$E = \frac{1}{2}\rho g \lambda La^2$$ First, could you clarify the nature of the parameters [rho] and [L] as I am interested in whether they reflect the mass of water displaced in the originating event, which caused the wave. See my following diatribe for details. By way of background, my interest is not directly with tsunami waves, I just happen to come across a description of them, which I couldn’t resolve in terms of general wave mechanics, i.e. the general statement that energy (E) is proportional to the square of the amplitude (A). I haven’t seen any mention of wavelength in standard text up until now. I was also looking into mechanical wave as opposed to EM wave, because of the different correlation to energy, i.e. Planck equation E=hf. Clearly, EM are also capable of propagate through a vacuum, whereas mechanical waves depend on ‘vibrations’ of neighbouring particles within the physical propagating media. I think I understand the caveats concerning linearity, but is your statement concerning energy being related to amplitude and wavelength generally true for all mechanical waves? I ask because this seems to be of fundamental importance to anybody trying to gain some physical understanding of the nature of mechanical waves. For example, it seemed to me, although I admit this is just speculation, that a mechanical wave starts by acquiring energy from outside the wave system, which it then propagates through a medium. If I use the pebble-in-the-pond analogy, an initial central wave is created, e.g. meteor strike or seismic shift in the seabed, which then radiates outwards. However, given that there is no net propagation of water, only wave energy, does this energy correlate to the potential energy of the source? What I mean by this is that the original event caused a large displacement of water, which had mass (m) and subject to gravity [g] due to its height (A) and width ($$\lambda$$). Therefore, if this displacement disappears, i.e. the water returns to a flat surface, it would seem logical to assume that the wave carried away most of this potential energy. Of course, in the case of a radiating circular wave, the initial central energy of the water is dissipated across the increasing radius of the circumference wave. Again, I am raising these points, because I am interested in trying to understand the interplay of potential and kinetic energy in mechanical waves. In a SHM model, the wave function persists because energy is converted from potential energy to kinetic energy and back again, assuming no loss to fiction etc. However, it seems that, in this case, the wave is propagating the original potential energy, although local particles within the propagating media are subject to kinetic movement, i.e. energy exchange. I know I am pushing my luck, but I would also like to understand what determines the frequency/wavelength of a mechanical wave in the first place. For example, in the case of an EM wave in vacuum, energy defines the frequency, the vacuum media defines the propagation velocity [c] and therefore the wavelength is determined by virtue of $$c= f * \lambda$$. Is there any analogous logic to mechanical waves? Anyway, again many thanks for all your previous input.
The parameters $\rho$ and L are representative of the mass of the wave. $\rho$ is the water density and L describes part of the volume of the water ($\lambda$ and the amplitude describe the rest). The initial seismic event causes a transfer of energy to the water by displacing it upward. This energy, for the most part, is carried along all the way to the coastline with little dissipation as previously noted in the form of a wave.

Note that the direction of the tsunami wave is mainly orthogonal to the direction of the seismic fault line.

Generally speaking the energy of a mechanical wave is typically not dependent on the wavelength, rather only proportional to the square of amplitude. However, I presumed you were only talking about tsunamis which behave slightly differently than a normal mechanical wave (although they are very similar as described above). The total energy initially transferred to the tsunami wave was given above as E, which is dependent on the wavelength.

Hope this helps.

CS
PF Gold
P: 516
Thanks for the clarification of the equation. The dependency on the mass of the wave makes sense, if the source of the energy being dissipated by the wave is proportional to the potential energy of the event that caused the wave.
 Quote by stewartcs Note that the direction of the tsunami wave is mainly orthogonal to the direction of the seismic fault line.
Does this preclude a tsunami being modelled on the pebble-in-the-pond analogy, i.e. a meteor in the middle of the Pacific causes a radially expanding tsunami?
 Quote by stewartcs Generally speaking the energy of a mechanical wave is typically not dependent on the wavelength, rather only proportional to the square of amplitude.
Actually, this is still my central issue. If mechanical waves have a potential energy source, then I don’t understand how amplitude alone is really representative. I accept that, as in the case of tsunami waves, the energy of a mechanical wave may well always be proportional to the square of the amplitude, but this seems to be only part of the equation.

$$v = \sqrt{\frac{T}{\mu}}$$

For example, I found the propagation formula above for a wave on a string, where [T] is tension and [mu] is linear mass, although I have not had much luck in finding any direct energy related equations for a string wave. However, just guessing, I would have thought that the energy of this type of wave would also be a function of the amplitude and wavelength linked to the linear mass of the string?

One final point of interest, while there are many sources that specify various equations for the propagation velocity [v] and the fundamental equation $$v= f \lambda$$, there seems to be little discussion of how to determine either the frequency or wavelength of a mechanical wave from this product relationship. I am assuming the type of media affects the outcome?
P: 2,283
 Quote by mysearch Does this preclude a tsunami being modelled on the pebble-in-the-pond analogy, i.e. a meteor in the middle of the Pacific causes a radially expanding tsunami?
No, not for a meteor strike.

 Quote by mysearch Actually, this is still my central issue. If mechanical waves have a potential energy source, then I don’t understand how amplitude alone is really representative. I accept that, as in the case of tsunami waves, the energy of a mechanical wave may well always be proportional to the square of the amplitude, but this seems to be only part of the equation. $$v = \sqrt{\frac{T}{\mu}}$$ For example, I found the propagation formula above for a wave on a string, where [T] is tension and [mu] is linear mass, although I have not had much luck in finding any direct energy related equations for a string wave. However, just guessing, I would have thought that the energy of this type of wave would also be a function of the amplitude and wavelength linked to the linear mass of the string?
Perhaps this will help:

"The amount of energy carried by a wave is related to the amplitude of the wave. A high energy wave is characterized by a high amplitude; a low energy wave is characterized by a low amplitude. The amplitude of a wave refers to the maximum amount of displacement of a particle on the medium from its rest position. The logic underlying the energy-amplitude relationship is as follows: If a slinky is stretched out in a horizontal direction and a transverse pulse is introduced into the slinky, the first coil is given an initial amount of displacement. The displacement is due to the force applied by the person upon the coil to displace it a given amount from rest. The more energy that the person puts into the pulse, the more work which he/she will do upon the first coil. The more work which is done upon the first coil, the more displacement which is given to it. The more displacement which is given to the first coil, the more amplitude which it will have. So in the end, the amplitude of a transverse pulse is related to the energy which that pulse transports through the medium. Putting a lot of energy into a transverse pulse will not effect the wavelength, the frequency or the speed of the pulse. The energy imparted to a pulse will only effect the amplitude of that pulse"

Source: http://www.glenbrook.k12.il.us/gbssc...es/u10l2c.html

 Quote by mysearch One final point of interest, while there are many sources that specify various equations for the propagation velocity [v] and the fundamental equation $$v= f \lambda$$, there seems to be little discussion of how to determine either the frequency or wavelength of a mechanical wave from this product relationship. I am assuming the type of media affects the outcome?
IIRC, the source of the energy that creates the wave and the medium (if any) in which it propagates determine these.

Hope this helps.

CS
PF Gold
P: 516
CS,
Really appreciate the help. I am trying to read deeper into the issues in the background, but it is useful to use this thread to resolve some the issues that initially puzzled me.
 Quote by stewartcs ”…Putting a lot of energy into a transverse pulse will not effect the wavelength, the frequency or the speed of the pulse. The energy imparted to a pulse will only effect the amplitude of that pulse….”
This seems a categorical statement, which I need to consider. However, one of the interesting things about wave mechanics is applying the theory to real-world situations.

For example, a weight on a spring acting as lossless SHM pendulum can be described in terms of a sine wave. The initial vertical displacement defines the potential energy being put into the system: $$1/2kA^2$$, which reduces to zero as the mass springs back through the equilibrium point, having been totally converted to kinetic energy at this specific point. In this example, the square of the amplitude of the wave seems to unambiguously reflect the potential energy at any position without reference to wavelength. This example can also calculate the angular frequency ($$w=\sqrt{k/m}$$) as function of 2 physical attributes of the media, i.e. elasticity and mass, which does indeed suggests that the media defines frequency, at least, in this case. However, there doesn’t appear to be any real propagation velocity, as the wave seems to be essentially a function of time only.

In the case of a surface wave, things don’t seem so straightforward. Presumably, these waves are more like a pulse than sine wave, although I am assuming a Fourier series would allow the pulse to be modelled as fundamental sine wave plus its harmonic components? It also seems difficult to associate the potential energy of this type of wave to a specific point, as per the example above, because this energy seems to associated with the pulse/wave as a whole, i.e. height (A) and width ($$\lambda$$) etc. The issues of any conversion between potential and kinetic forms also seems less obvious, but presumably is required in any real world propagation process? Finally, while we have touched on several equations that specify a propagation velocity for this type of wave, I haven’t come across any equation that actually allow the frequency/wavelength to be calculated as a function of the media ,as per the SHM, possibly it is too complex to generalise?

Anyway, many thanks for all the help.
 PF Gold P: 516 Hi, I was wondering if anybody was in a position to answer any of the additional questions about tsunami waves below. In post #9, the following equation was cited for the energy associated with a tsunami wave: $$E = \frac{1}{2}\rho g (\lambda L)a^2$$ http://www.scribd.com/doc/4633126/Fi...ics-of-Tsunami I tracked down a few papers that reference this equation and it appears to be based on a model of a wave with a volume of water referred to a waterberg`. However, it seems that the essence of this equation can be reduced to the classical equation for potential energy: $$E_p = mgh = \rho (x*y*z) * g * y/2 = \frac{1}{2} \rho g (x z) y^2$$ Where mass [m] is the volume [x,y,z] times density and [y/2] is the average height. Note: these equations still reflect that the energy (E) of the wave is proportional to the square of its amplitude, but the huge energy of a tsunami is really a reflection of the net volume of the wave based on its very long wavelength. However, this analysis appears to be predicated on the fact that most tsunami waves are caused by seismic events, where the wavefront generated is essentially aligned and orthogonal to the direction of a seismic fault line. These papers also suggest a number of assumptions that I would like to raise for clarification: 1. While tsunami waves exist in deep water, because the wavelength can be much greater than the sea depth, they are often modelled using the shallow water equations cited in post #1. 2. While deep water is said to be dispersive, shallow water appears to be modelled as non-dispersive. Given the assumption in bullet (1) does it follow that tsunami waves should be modelled as non-dispersive? 3. In a non-dispersive medium, the equation $$[v=f \lambda]$$ holds for all frequencies, as long as the characteristics of medium do not change, i.e. the phase velocity of all waves of different frequency is the same? 4. Can a tsunami be modelled as a Fourier series of different frequency components? 5. Would these frequencies travel at the same phase velocity based on bullet (3)?It is said that due to their long wavelength, tsunami waves dissipate very little energy. However, I have attached a diagram that is more representative of a circular expanding wave, which by definition would dissipate its energy over an ever-increasing circumference – see post #2/#8. 6. Is this a valid model for a tsunami wave? 7. While the initial energy would still be proportional to the volume of water displaced; would it make sense to model this example on the volume of a cone? 8. While the propagation of a wave involves a local conversion of potential energy to kinetic energy, i.e. initial wave height into a rate of change of amplitude, is it true to say that only potential energy is being transported by the wave? 9. With reference to the attached diagram; what would determine the very long wavelength generally associated with tsunami waves?Appreciate that all these questions may be excessive, but would appreciate any insights or clarification of any of the issues. Thanks Attached Thumbnails
P: 2,283
 Quote by mysearch 1. While tsunami waves exist in deep water, because the wavelength can be much greater than the sea depth, they are often modelled using the shallow water equations cited in post #1. 2. While deep water is said to be dispersive, shallow water appears to be modelled as non-dispersive. Given the assumption in bullet (1) does it follow that tsunami waves should be modelled as non-dispersive?
Yes, they should be modeled as non-dispersive.

Deep-water approximation refers to the depth of the water divided by the wavelength. Tsunamis have wavelengths of hundreds of kilometers, so they are shallow water waves.

However, most shallow water and deep water waves are somewhat dispersive. Tsunamis are just slightly dispersive so they are most accurately modeled as non-dispersive waves even though they may exist in deepwater. Only solitons are non-dispersive.

If you are looking for a Tsunami model, check out the NOAA Tsunami site. They give some details (IIRC) about how they do it.

Hope this helps.

CS
PF Gold
P: 516
 Quote by stewartcs Yes, they should be modeled as non-dispersive… Tsunamis are just slightly dispersive so they are most accurately modeled as non-dispersive waves even though they may exist in deepwater…. If you are looking for a Tsunami model, check out the NOAA Tsunami site.
Thanks. Approximating to non-dispersive will allow a simplification of the frequency spread by propagation velocity. Based on the shallow water model, can the propagation equation still be equated to frequency * wavelength, e.g.

$$v = \sqrt{gd} = f * \lambda$$?

I will assume that the circular model is possible for tsunamis and therefore the energy in successive radiating circular waves must fall as stated. However, is the ultra long wavelength primarily established, in most practical cases, by the seabed lifting/falling over a very large area?

I will take this displacement of a large volume of water as the initial potential energy input into the wave system, which is radiated away by successive circular waves. If I assume no net radial movement of water, the kinetic energy would seem to only exist in the localised rise and fall of the wave. However, it would seem that the energy of any section of the wave, in motion, at any radius must be the sum of this kinetic energy plus the potential energy associated with the amplitude of the wave and its wavelength. As such, the total energy being transported by the wave exists in both kinetic and potential forms.

Given that the kinetic energy exists as a function of the potential energy, i.e. wave amplitude, it will also be proportional to the square of the amplitude aggegated over the wavelength.
P: 2,283
 Quote by mysearch Thanks. Approximating to non-dispersive will allow a simplification of the frequency spread by propagation velocity. Based on the shallow water model, can the propagation equation still be equated to frequency * wavelength, e.g. $$v = \sqrt{gd} = f * \lambda$$?
Yes. In fact the approximate definitions for the phase speed are found from that relation.

Starting with the relations:

$$c = f \lambda$$

or

$$c = \frac{\omega}{k}$$

where,
c is the phase speed
k is the wave number

and

$$\omega^2 = gk^2d$$, which is the shallow water dispersion relation

We get:

$$c = \sqrt{gd}$$

CS

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