Sign of potential energy of SHO

[devang2]

Suppose the particle is at distance x from mean position and moving away . The standard formula for calculating potential U is U=-w here wis the work w=kxdx when the particle is moving away .On integration U=1/2Kx^2 . When the particle is moving towards mean position w=-kxdx on integration U=-1/2kx^2 thus the sign changes with direction Is it correct

 

[devang2]

sorry i omitted in my previous post that work is calculated by dot product formula

 

[Tanya Sharma]

Suppose the particle is at distance x from mean position and moving away . The standard formula for calculating potential U is U=-w here wis the work w=kxdx when the particle is moving away .On integration U=1/2Kx^2 . When the particle is moving towards mean position w=-kxdx on integration U=-1/2kx^2 thus the sign changes with direction Is it correct

Suppose there is a spring with its left end attached to a wall and right end having a block of mass m .Now we pull it slowly towards right ,then the work done by external agent is stored as potential energy.

W = ∫ F.dx
= ∫kxdxcos0 (since force is parallel to the displacement )
= [(1/2)kx²]₀^x
= (1/2)kx²

So , U = (1/2)kx²

 

[devang2]

Thank you for the reply . So you mean that whether the spring is stretched or compressed the potential energy is always positive.While calculating the potential of particle executing simple harmonic oscillations the energy is always positive . I have used basic relation dU=-w to calculate the potential energy which gives opposite signs of potential energy when the particle is moving away or towards the mean position . Please take some trouble to point out where i have gone wrong while using the above formula

 

[Tanya Sharma]

See...understand one thing dU = -dW,if dW is work done by the conservative force,spring force in our spring block example . But dU = dW if dW is work done by external agent .

Do you agree with this ?

 

[Philip Wood]

This is correct. The reason is that a system's potential energy is the work it can do because of configuration (i.e. the relative positions of its parts). So if a system DOES work dW, it loses potential energy dU.

Suppose we consider force and displacement components in the x direction, that is the direction which tends to stretch the spring. If the spring is already extended, F_x will be negative, so for positive dx, dW = F_x dx will be negative and dU will be positive.
If the spring is compressed F_x will be positive, and if we compress it further, dx will be negative, so dW = F_x dx will be negative and dU will (again) be positive.

 

[devang2]

Thank you for the reply .AS you know for oscillator force is always -kx and always directed towards mean position . i calculate the work dw by applying dot product of force and displacement. While dot product formula you know scalar values of force and displacement are used . When the particle is moving away from mean position angle is 180 because force and displacement are in opposite direction hence dw is negative but when particle is moving towards mean position force and displacement are in the same direction hence angle is 0 therefore according to product dw is positive. I shall be very thankful if you point out where i have gone wrong

 

[Nugatory]

Thank you for the reply .AS you know for oscillator force is always -kx and always directed towards mean position . i calculate the work dw by applying dot product of force and displacement. While dot product formula you know scalar values of force and displacement are used . When the particle is moving away from mean position angle is 180 because force and displacement are in opposite direction hence dw is negative but when particle is moving towards mean position force and displacement are in the same direction hence angle is 0 therefore according to product dw is positive. I shall be very thankful if you point out where i have gone wrong

The potential energy is at a maximum when the displacement is at a maximum, and it's zero at the mean position. So it's not at all surprising that the sign of $dW$ is different when the displacement is increasing and when the displacement is decreasing; in one case we're putting energy into the spring and in the other case we're taking energy out of it.

 

[Philip Wood]

My method also uses dot product. If two vectors, A and B, have no y or z components, then it is easy to show that A.B = A_xB_x. The components, A_x and B_x, are scalar, but can be positive or negative.

Suggest you read my post carefully again, with this in mind. Or, if you prefer, use magnitudes (scalar and non-negative) of vectors, and put in the cosine value, +1 or -1, separately as you follow my argument. You'll find it gives the same result as my method each time. The argument shows that the more you deform the spring - either by extending it or compressing it - the more energy it stores. If you repeat the argument, changing the sign of dx, you'll find that if you reduce the deformation - either the extension or the compression - the spring stores less energy.

What I found confusing in your last post - and it may have some bearing on your own difficulty - is what you mean by 'displacement'. Are you using it to mean displacement of the mass from its equilibrium position (that's the usual meaning) or are you using it to mean the incremental displacement, dx?

 

[devang2]

Thank you for the trouble you have taken for the clarification . By displacement i mean the distance of the mass of the particle from the mean position executing harmonic oscillation.
suppose spring is stretched by certain distance , the work involved is w(!) and when the spring is released to restore the equilibrium original position work involved is W(2) because spring is in original state no net work is done therefore w(!)+w(2)=0 which leads to the conclusion the sign of work during stretching and decompressing are opposite .Similarly when the particle executing harmonic oscillation is moving away from mean position it is like stretching and when it is moving towards mean position it is like decompressing hence sign of work when it is moving away from mean position is negative and when it moves towards mean position it is positive which finally leads to the conclusion that sign of potential energy when particle is moving away is positive and it is negative when motion of particle is towards mean position I shall be very thankful if you let me know if i am right or wrog

 

[Philip Wood]

suppose spring is stretched by certain distance , the work involved is w(!) and when the spring is released to restore the equilibrium original position work involved is W(2) because spring is in original state no net work is done therefore w(!)+w(2)=0 which leads to the conclusion the sign of work during stretching and decompressing are opposite .

Yes, that's right. The work done BY the spring is negative when it is being compressed or extended, and positive when being decompressed or de-extended. But, since I can see you like precision, I'll point out that your last sentence isn't quite correct...

sign of potential energy when particle is moving away is positive and it is negative when motion of particle is towards mean position

What you should say is that the sign of the CHANGE in potential energy when particle is moving away is positive and it is negative when the motion of the particle is towards the mean position. [The potential energy itself is always positive (or zero) if we adopt the usual convention of assigning zero to the potential energy when the spring is undeformed.]

 

[vanhees71]

I don't understand the problem. It's pretty simple. The potential of a force (if it exists!) is defined by
$$\vec{F}(\vec{r})=-\vec{\nabla} V(\vec{r}).$$
Given that, for the isotropic harmonic oscillator,
$$\vec{F}(\vec{r})=- m \omega^2 \vec{r},$$
you find by simple integration or taking the appropriate line integral (e.g., a straight line from the origin to $\vec{r}$) you get
$$V(\vec{r})=\frac{m \omega^2}{2} \vec{r}^2.$$
You easily check that indeed this is the potential, i.e., that $\vec{F}$ is indeed a conservative field.

Of course it also satisfies the local condition
$$\vec{\nabla} \times \vec{F}=0,$$
and since it is defined and analytic everywhere in $\mathbb{R}^3$ there must exist a potential.

 

[devang2]

thank you for the analytical reply. My problem is very basic and simple ,it is about the sign of potential which changes sign with change in direction of motion of the particle executing harmonic oscillation though magnitude remains same .There is change in sign if potential is calculated by applying the basic formula ,dU=-W u is the potential and w is the work which i calcule by using dot product. Please calculate the potential by using this method , you will will see that sign of potential is positive when particle is moving away from mean position while it is negative when particle is moving towards mean position.Please pay attention to the sign. I shall be very thankful if you let me know your views about the change in sign .All the textbooks assign positive sign yo the potential disregarding the change in direction of motion .You know when mass is shifted to infinity from Earth against gravity potential is positive but it is negative when it is brought from infinity moving in the direction of gravity thus it is seen that there is change in sign with the direction so same thing is applicable to motion of particle executing harmonic scillation because it is continusuosly changing direction of motion

 

[Philip Wood]

sign of potential is positive when particle is moving away from mean position while it is negative when particle is moving towards mean position.

You're still making the mistake that I pointed out in my last post! It's not the potential energy, but the CHANGE in potential energy which is positive when the particle is moving away from the mean position and negative when the particle is moving towards the mean position. The potential energy itself is always positive (if we assign zero to it at the mean position, when the spring is unstretched). It is more positive the more the spring is stretched.

Say if you're still not clear.

[PS: You've started to write 'potential' instead of 'potential energy'. Stick with potential energy!]

 

[devang2]

Thank you very much indeed for the clarification and pointing out ther mistake from which it can be concluded that work is negative when the particle is moving away from mean posiytion whereas it is positive whan the particle is moving towards mean position .

 

[Philip Wood]

That's right: work done by the spring (or other body responsible for the restoring force) is negative when the particle moves away from the mean position, so the spring gains potential energy (dU = -dW). If we stretch the spring by X, then potential energy stored is

$$U = \int_{x = 0}^{X}{dU} = -\int_{x =0}^{X}{dW} = -\int_{0}^{X}{-kx dx} =\frac{1}{2}k X^2$$

You won't find it done quite like this in the textbooks. I'm using dW to mean an increment of work done BY the spring, following - I hope - what you were doing. If you consider the work done ON the spring as it is stretched, you can dispense with the minus signs in the above derivation.

 

[devang2]

Your way of expressing makes it easy for me to understand the problem.If the particle is moving towards the mean position,applying the method as adopted when the particle is moving away from mean position potential energy of the particle in SHO is negative that is -1/2kx^2. I shall be very thankful if you can clarify it . My contention is that the potential energy changes sign with change in direction of motion

 

[vanhees71]

thank you for the analytical reply. My problem is very basic and simple ,it is about the sign of potential which changes sign with change in direction of motion of the particle executing harmonic oscillation though magnitude remains same.

No, it doesn't. It doesn't even depend on the velocity. The potential is $V(\vec{r})=k \vec{r}^2/2$ and thus independent on the direction of the position vector and doesn't depend on velocity at all. It's the force, which is
$$\vec{F}=-k \vec{r}$$
that changes it's direction with the position vector, not the potential!

 

[Philip Wood]

deVang2. Same mistake AGAIN ! When the particle is moving TOWARDS the mean position, the CHANGE in potential energy is negative, so the spring is LOSING the potential energy that it had when it had a greater displacement. By the time it has got back to x = 0 it has lost ALL the potential energy. So at x = 0, it has zero potential energy, not a negative potential energy. Its potential energy never goes negative.

 

[Philip Wood]

Same mistake AGAIN ! As the particle moves towards the mean position the CHANGE in potential energy is negative, so the spring is losing the potential energy which it had at displacement X. When it has got back to the mean position it has lost ALL the potential energy. Its potential energy is now zero. The potential energy is never negative.

 

[Tanya Sharma]

.If the particle is moving towards the mean position,applying the method as adopted when the particle is moving away from mean position potential energy of the particle in SHO is negative that is -1/2kx^2. I shall be very thankful if you can clarify it . My contention is that the potential energy changes sign with change in direction of motion

Please understand that potential energy is a function of position such that U = (1/2)kx² ,where 'x' is the displacement of the object .So it doesn't matter whether the object at a particular displacement 'x' is moving towards the mean position or away from it.Its potential energy is (1/2)kx²

Also note that whether the object is at +x or -x the potential energy is (1/2)kx²,always a positive quantity .

In the derivation of the expression of potential energy ,important thing is the displacement of the object from the mean position .Whether the object is moving towards or away from the mean position is irrelevant.

 

[devang2]

The confusion about the sign is still there . I .quote example. If certain mass is displaced away from the gravitating mass potential energy is positive but it is negative when the sane mass is displaced towards gravitating mass . You can check by actual calculation . The above problem is similar to that of particle executing harmonic oscillations.Why in case of SHO potential energy is always taken to be positive independent of direction of displacement. there is changein sign if dU=-dw or w is used to calculate potential energy .Please let me know where i have gone wring .

 

[Philip Wood]

You've gone wrong in the same way that you've gone wrong in all your posts. You are still confusing CHANGE in potential energy (which DOES change sign according to whether you're going towards or away from the zero of extension, and potential energy itself, which doesn't change sign.

Suggest you read my last post (20) again VERY CAREFULLY.

 

[Tanya Sharma]

The confusion about the sign is still there . I .quote example. If certain mass is displaced away from the gravitating mass potential energy is positive but it is negative when the sane mass is displaced towards gravitating mass . You can check by actual calculation . .

The GPE(Gravitational potential energy) is always negative .Whether you move the object towards or away from the mass .It is the change in GPE which is negative or positive .If you move towards the mass ,the change is negative and GPE becomes more negative , whereas if you move away ,the change in GPE is positive and GPE becomes less negative .

Similarly in case of spring mass system , when the mass moves towards the mean position ,the change in EPE(Elastic Potential energy) is negative ,but EPE remains positive .On the other hand ,if the mass moves away from the mean position,the change in EPE is positive.Again the EPE is positive.

 

[Philip Wood]

devang2: Don't want to be patronising, but may I suggest that you take a 'cooling off period' of a few days, during which you read very carefully what posters have said in response to your question. Understanding may dawn!

 

[devang2]

Thanks a lot for the answer . No one has pointed so for where i have gone wrong if i use basic formula dU =-W to calculate potential energy and use dot product to calculate work . This method leads to positive and negative signs when the particle moves away and toward the mean position respectively

 

[Doc Al]

No one has pointed so for where i have gone wrong if i use basic formula dU =-W to calculate potential energy and use dot product to calculate work . This method leads to positive and negative signs when the particle moves away and toward the mean position respectively

I'd say that your error has been pointed out several times over.

The potential energy of a spring at any point is defined as the work done to move to that point starting from the equilibrium position (where x = 0). Thus the potential energy stored in the spring is always positive and equals 1/2kx².

When you move the spring towards equilibrium are you doing negative work? Sure. So what? As has been explained, you are just reducing the stored potential energy. At no point is the potential energy negative.

 

[sophiecentaur]

Perhaps defang has not realized that the reference point for describing PE for a spring is significantly different from the reference point which is used for describing GPE for a Planet. For a spring, the equilibrium point has Zero PE so it is convenient to use this point. The 'Equilibrium point' of an object in the gravitational field of a point mass is the point and the Potential is indeterminate - so we refer GPE to Infinity, again for convenience.

Wherever the reference point is taken, the closer you get to an attractor, the lower will be the Potential. The 'slope' is negative for an attractor (negative force = negative potential slope)
In fact he writes:

This method leads to positive and negative signs when the particle moves away and toward the mean position respectively

which looks OK to me - as long as he is referring to gradient of Potential. Perhaps a calm, cool look at what he has written and what he actually thinks would help.

 

[Philip Wood]

Perhaps a calm, cool look at what he has written and what he actually thinks would help.

Well, I for one, feel suitably chastened, having made no effort at all to understand what he wrote...

as long as he is referring to gradient of Potential.

... though I note that the gradient of potential energy (and distinguishing it from the potential energy itself) was the theme of several of my own - misguided - posts!

 

[sophiecentaur]

Well, I for one, feel suitably chastened, having made no effort at all to understand what he wrote...



... though I note that the gradient of potential energy (and distinguishing it from the potential energy itself) was the theme of several of my own - misguided - posts!

I am of the School of "do as I say and not as I do" No need to feel chastened, my friend.
This thread (like so many) has fallen foul of a failure to communicate the direction in which we (I mean you lot) are pointing. If we had had a diagram with the first Post, it would have all been done and dusted in three or four posts.

 

[Simon Bridge]

OP remains as convinced as ever that the HO potential changes sign, and is now asserting that there is some sort of global physicist conspiracy to ignore this inconvenient "fact". Please see:

https://www.physicsforums.com/showthread.php?t=730150

 

[Simon Bridge]

Thanks a lot for the answer . No one has pointed so for where i have gone wrong if i use basic formula dU =-W to calculate potential energy and use dot product to calculate work . This method leads to positive and negative signs when the particle moves away and toward the mean position respectively

That equation is incorrect - where did you get it from?
The correct relation is dW=-dU or dU=dW - depending on the context - as explained to you in post #5.
More precisely: http://en.wikipedia.org/wiki/Potential_energy#Work_and_potential_energy
... but it is not the definition.

One of the upshots of that equation is that there is no absolute zero for potential energy.
When we talk about a particular value of potential, we are using a shorthand for the difference in potential between two places.

There are conventions for some situations for where we put that zero by default.
For instance, for gravity, close to the Earth's surface, we put U=0 on the ground.
In that case you will see that U<0 below the ground, and U>0 above the ground.

If you start below the ground, and move upwards, then v>0, dU>0, and U goes from negative values to positive values.

But if we are talking about gravity at a long distance, it is more convenient to put U=0 at an infinite distance away. In that case, U<0 everywhere.

If you move towards a mass, dU<0, but U<0
If you stay still, or execute a circular orbit dU=0 but U<0
If you move away from the mass dU>0 but U<0

For a mass on a spring, it is convenient to put U=0 at the center of the motion simply because U increases to either side. This makes U>0 everywhere for the mass.
If the mass moves towards the center, then dU<0 and U>0
If the mass moves away from the center, then dU>0 and U>0

 

[jtbell]

Given the OP's history on this topic, there is no reason to continue either of these threads.