sidharath Posted July 25, 2014 Share Posted July 25, 2014 the basic relation between potential energy and work energy is given below dU=-w Uis potential energy and w is work . Suppose the particle is moving away from mean position. work w can be calculated by applying dot product of force and distance. if x is the displacement w=modulus of force x modulus of distance x cos (angle between direction of force and distance) when the particle is moving away from mean position angle is 180 .therefore dU=kx .dx kx is the value of force and xis displacement because cos180=-1. On integrating the above relation U=1/2 k x^2 which shows that value of potential energy is positive. if the particle is moving away from mean position the angle between force and displacement is o the value of cos 0=1 which leads to dU=--kx dx on integrating U=-1/2kx^2 which shows that sign of potential energy changes to negative sign. The change in sign with change in direction is not mentioned in textbooks Please express your opinion Link to comment Share on other sites More sharing options...
imatfaal Posted July 25, 2014 Share Posted July 25, 2014 Depends on the direction of the force as well remember - and yes you can either be lifting the weight against gravity ( or compressing the spring with a pellet etc) or the weight can be dropping with gravity (or the spring firing the pellet). One direction will increase potential energy and the other will see pe diminish and be converted in these cases to kinetic. You add to the pe when you do work over a distance against a force Link to comment Share on other sites More sharing options...
sidharath Posted August 5, 2014 Author Share Posted August 5, 2014 Let us take the case of sign of potential energy of SHO.There is two way motion of the particle in SHO therefore the sign of potential energy is expected to change with direction of motion of particle.When the particle is moving away work is being done on the particle while when it i moving towards mean position particle is doing work hence sign of potential energy should be reversed. The follo wing discussion conclusively proves it the basic formula to calculate potential energy is dU=-w ,U is the potential energy and w is work. Work is calculated by applying dot product formula .w=scalar value of force x scalar value of displacement x cos(angle between direction of force and displacement) when the particle is moving away from mean position w==-kxdx because angle is 18o dU=kxdx on integration U=1/2kx^2 because potential energy at mean position is 0. when the particle is moving towards mean position w=kxdx because angle is 0 so dU=-kxdx on integration U= -1/2KX^2 The above discussion proves that sign of potential is negative when particle moves towards mean position. This basic fact is not mentione in text books Please let me know where i have gone wrong . Link to comment Share on other sites More sharing options...
imatfaal Posted August 5, 2014 Share Posted August 5, 2014 Signs of Potential Energy under gravity are just convention - you really just need to make sure you have your KE and PE sorted out. We decided that that at very large distance PE tends to zero and as gravity does work on a falling particle then PE must decline (and KE etc increase) thus PE starts at zero (infinitely far away) and gets more and more negative as the particle moves towards the gravity source. And I haven't double checked but I think you are getting confused on the direction of the force doing work in your pendulum example Link to comment Share on other sites More sharing options...
studiot Posted August 5, 2014 Share Posted August 5, 2014 No you need to think again. The sign of potential energy (PE) itself is always positive, unless it is zero in which case it is neither positive nor negative. But you are talking about change of PE, which can indeed be positive or negative, depending upon whether the objects are gaining or loosing PE. Link to comment Share on other sites More sharing options...
imatfaal Posted August 5, 2014 Share Posted August 5, 2014 No you need to think again. The sign of potential energy (PE) itself is always positive, unless it is zero in which case it is neither positive nor negative. But you are talking about change of PE, which can indeed be positive or negative, depending upon whether the objects are gaining or loosing PE. I don't know if you are responding to me or to OP - but gravitational PE is never positive. As I said above it is zero at infinity and grows negative at it approaches the massive object. http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html#gpt The general expression for gravitational potential energy arises from the law of gravity and is equal to the work done against gravity to bring a mass to a given point in space. Because of the inverse square nature of the gravity force, the force approaches zero for large distances, and it makes sense to choose the zero of gravitational potential energy at an infinite distance away. The gravitational potential energy near a planet is then negative, since gravity does positive work as the mass approaches. This negative potential is indicative of a "bound state"; once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape. The general form of the gravitational potential energy of mass m is: Link to comment Share on other sites More sharing options...
swansont Posted August 5, 2014 Share Posted August 5, 2014 ! Moderator Note Similar subjects merged. Please, only one thread per topic, and do not respond to this modnote Link to comment Share on other sites More sharing options...
studiot Posted August 5, 2014 Share Posted August 5, 2014 (edited) With respect. Imatfaal, sidharath specified simple harmonic oscillation, not other forms of PE. I have posted the quadratic diagram here before and I see from my settings page that it is still there. If I knew how to do it I would refer to it here, can you help? Edited August 5, 2014 by studiot Link to comment Share on other sites More sharing options...
swansont Posted August 5, 2014 Share Posted August 5, 2014 Let us take the case of sign of potential energy of SHO.There is two way motion of the particle in SHO therefore the sign of potential energy is expected to change with direction of motion of particle. PE is defined by position, not direction of motion. The sign is NOT expected to change by anyone using the standard definitions. Link to comment Share on other sites More sharing options...
imatfaal Posted August 5, 2014 Share Posted August 5, 2014 I read a pendulum that wasn't there ;-) Which diagram are you looking for Studiot? Will see if I can find and insert Link to comment Share on other sites More sharing options...
studiot Posted August 5, 2014 Share Posted August 5, 2014 I would still like to be able to post the diagram if someone could help please. Link to comment Share on other sites More sharing options...
imatfaal Posted August 5, 2014 Share Posted August 5, 2014 I would still like to be able to post the diagram if someone could help please. Per above message - which one? If you know which topic you posted in (or even rough subject area) I can probably do so. Link to comment Share on other sites More sharing options...
studiot Posted August 5, 2014 Share Posted August 5, 2014 (edited) This whole discussion appears to be a rerun of this thread http://www.scienceforums.net/topic/77993-selective-nature-of-schrodinger-equation/#entry761150 The diagram I mentioned is called harmonic1 and appears in my post#13 there. Edited August 5, 2014 by imatfaal Link to comment Share on other sites More sharing options...
sidharath Posted August 6, 2014 Author Share Posted August 6, 2014 I request all of you intelligent persons that you specifically point out the error in my calculation . The calculations are based on standard relations and conventions.If the final result got by using the acknowledged relations is wrong where i have gone wrong in using the relations.. .. Are the standard relations used by me wrong ? Dear Studiot s' result is based on graphs .while my result is based on analytically calculations. Again i beg your help to pin point where lies the error in my calculations. Let this matter be conclusively settled . I shall be thankful to you Link to comment Share on other sites More sharing options...
studiot Posted August 6, 2014 Share Posted August 6, 2014 (edited) Dear Studiot s' result is based on graphs .while my result is based on analytically calculations Actually my graphs are merely a plot of the correct formulae, which are also stated on my diagram. A body executing simple harmonic motion is subject to two forces. A disturbing force and a restoring force. The equations that lead to SHM and the energies involved are as a result of the restoring force only. It is a common mistake to try to calculate with the disturbing force. The restoring force always acts towards a fixed point. That is the restoring force changes direction halfway through the cycle. We normally place the origin of coordinates at this point. In your analysis the restoring force is kx for half the cycle and -kx for the other half. You do not need to worry about the angle between the restoring force and the displacement since they are along the same line (the x axis). Because the restoring force direction reverses you need to separate the work integral into two integrals, each with its correct sign. See if you can work through this now and come up with the correct formulae. This is a bit like asking the question What is the integral of sinxdx from x=0 to x=2pi? The answer is zero, but the area between the curve and the x axis is obviously not zero. Again you have to split the integral to obtain the area. So far this is all mathematical, but thinking about the Physics may also help. What exactly is x? ie what does it represent? Well not only is SHM motion under the action of a restoring force always directed towards a fixed point, but further the value of that restoring force is proportional to the distance from thatpoint ie the displacement. x is the displacement from the mean point. Now if you wish to consider work and energy we can say that When the body is moving towards the fixed point the restoring force and displacement are in the same direction. So the restoring force is doing work on the body, accelerating it, ie increasing its KE. When the body is moving away from the fixed point the body is doing work against the restoring force, storing the work energy in the source of the restoring force as PE. This energy comes from diminishing the KE of the body. Edited August 6, 2014 by studiot Link to comment Share on other sites More sharing options...
swansont Posted August 6, 2014 Share Posted August 6, 2014 I request all of you intelligent persons that you specifically point out the error in my calculation . The calculations are based on standard relations and conventions.If the final result got by using the acknowledged relations is wrong where i have gone wrong in using the relations.. .. Are the standard relations used by me wrong ? Dear Studiot s' result is based on graphs .while my result is based on analytically calculations. Again i beg your help to pin point where lies the error in my calculations. Let this matter be conclusively settled . I shall be thankful to you The direction change means the work changes sign, but you also need to reverse the limits of your integral. i.e. when theta is 0º you move from x1 to x2, but when theta is 180º you are moving from x2 to x1 The change in PE changes sign in the two cases, because your starting and ending points are reversed. The PE at each point is the same. Link to comment Share on other sites More sharing options...
sidharath Posted August 7, 2014 Author Share Posted August 7, 2014 According to stud idiot when the particle is moving away or towards mean position KE decteases and increases respectively.Let us see what it leads to . In SHO T+U=constant so dU=-dT when the particle is moving away from mean position KE decreases hence dT is negative so dU is positive it is possible only if expression for PE is +1/2kx^2 . If the particle is moving towards mean position KE increases .dT is positive .dU is negative which is possible only if U=-1/2kx^2. it can be concluded that there is change in sign of PE Link to comment Share on other sites More sharing options...
studiot Posted August 7, 2014 Share Posted August 7, 2014 (edited) Let us start with a body at rest at the mean position and then apply a disturbing force. For instance let a pendulum hang freely at rest, and be drawn aside a small distance and held at this side position. When at rest the (total) energy of the pendulum is zero. That is its KE and its PE are both zero (not equal and opposite). As it is drawn aside the disturbing force raises the pendulum against gravity imparting potential energy and kinetic energy to the pendulum. The kinetic energy is undefined, but when the pendulum is again at rest at the side position it has zero KE and a defined PE. Up to this point the pendulum is not executing SHM. If the pendulum is now released it will fall back to its original level under gravity. As the pendulum swings its PE is converted to the KE of motion until it reaches its lowest point at which time it again has zero PE, but now has substantial KE. So it swings on through the lowest point and start to climb back up. As it climbs it looses KE and gains PE, until in a frictionless system it has gained the same height on the other side of rest as it was originally released from. At this point it again comes to rest. All the KE has now been converted to PE, which is again positive. The pendulum is now executing SHM and the cycle continues indefinitely. At all times in this cycle the PE (and the KE) are non negative ie zero or positive. The pendulum starts off SHM with a specific PE and zero KE. Initially, the change in PE is negative, the change in KE is positive. Once this negative change in PE has reduced the PE to zero the change becomes positive and the PE increases again. If we can agree these factual observations we can proceed to discuss how this is implemented as a force-displacement-work calculation. Edited August 7, 2014 by studiot Link to comment Share on other sites More sharing options...
CaptainPanic Posted August 7, 2014 Share Posted August 7, 2014 ! Moderator Note sidharath, You will discuss without the use of any insults. If you fail to do so, we will make sure you will not discuss at all on our forum. Do not respond to this moderator note in the thread. Link to comment Share on other sites More sharing options...
studiot Posted August 10, 2014 Share Posted August 10, 2014 @sidharath I do not feel insulted by anything you have said to me, so if you wish to continue working through SHM I am happy to do so, continuing from my post#18, until we find which one of the various wrinkles about SHM has tripped you up. Link to comment Share on other sites More sharing options...
sidharath Posted September 3, 2014 Author Share Posted September 3, 2014 Dear studiot you express in your post 18 that as the pendulum moves towards mean position change in potential energy is negative,kinetic energy acquired by pendulum is equal to change in potential energy ,therefore if change in PE is negative KE will be negative but KE is always positive. If the sign of potential energy is negative as pendulum moves towards meanm position the change in PE will be positive hence KE will be positive,no rule will be violated Link to comment Share on other sites More sharing options...
studiot Posted September 3, 2014 Share Posted September 3, 2014 If I take a £1 coin out of my right pocket the change in the number of coins is my right pocket is negative. That is I have one less coin in my right pocket. If I place that coin into my left pocket the change is positive. That is I have one more coin in my left pocket. The total coins I have remains the same. The situation with KE and PE is identical. This is because the sign is associated with the change, not the coin or energy itself. Conservation laws of coins (I wish) and energy require that one man's loss is another's gain. Link to comment Share on other sites More sharing options...
swansont Posted September 3, 2014 Share Posted September 3, 2014 IOW, x ≠ ∆x Link to comment Share on other sites More sharing options...
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