Maximizing Spring Elongation in Charged Object System

[Tanya Sharma]

Homework Statement

Two equal and opposite charged objects of magnitude 2C and mass 2kg are attached to a spring of spring constant 4 N/m in the natural length of the spring. The system is suddenly placed in a uniform electric field of field strength 5 N/C in the direction of electric field such that the -ve charge is towards the origin of electric field. assuming the natural length of the spring to be large, find the maximum compression and elongation in the spring length.

Homework Equations

The Attempt at a Solution

First I tried to find the maximum extension.

Let charge -2 C be at a distance x₁ and charge +2 be at a distance x₂ from the origin .

Let the length of the spring be l .

The length of the spring at any instant is x₂-x₁ .

The extension in the spring would be x₂-x₁-l .

The force of attraction between the charges would be kq²/(x₂-x₁)² i.e 16/(x₂-x₁)²

Force due to the external electric field will be of magnitude 10N .

EOM for mass 1 (i.e -2C ) will be k(x₂-x₁-l)-10+16/(x₂-x₁)² = 2d²x₁/dt²

EOM for mass 2 (i.e +2C ) will be 10-k(x₂-x₁-l)-16/(x₂-x₁)² = 2d²x₂/dt²

From the above we get ,

2(d²x₂/dt²-d²x₁/dt²)=20-2k(x₂-x₁-l)-32/(x₂-x₁)²

or , d²x₂/dt²-d²x₁/dt²=10-k(x₂-x₁-l)-16/(x₂-x₁)²

d²(x₂-x₁)/dt²=10-k(x₂-x₁-l)-16/(x₂-x₁)²

Now putting x₂-x₁ = z ,

We have d²z/dt²=10-k(z-l)-16/z² .Now I need to maximize z .

Is it the correct way to approach the problem ?

I would be grateful if someone could help me with the problem .

 

[Saitama]

Hello Tanya!

I am not sure but I think you are required to use the assumption that the natural length of spring is large to cancel one of the factors.

Also, how do you get kq^2=20?

 

[voko]

For this kind of problems (two interacting bodies) it is usually simplest to convert to the CoM frame and use conservation of total momentum in the CoM frame. This separates coordinates into the coordinates of the CoM and the distance between the bodies.

 

[Tanya Sharma]

I am not sure but I think you are required to use the assumption that the natural length of spring is large to cancel one of the factors.

Does that mean the force between the charges doesn't come in picture ? Do I have to neglect this force ?

Also, how do you get kq^2=20?

I have fixed the error .kq²=16

Have I approached it correctly ?

 

[Saitama]

Does that mean the force between the charges doesn't come in picture ? Do I have to neglect this force ?

This is what I think but as I said, I am not sure. :)

I have fixed the error .kq²=16

Still incorrect. It looks to me as if you are confusing the Coulomb's constant with the spring constant.

 

[Tanya Sharma]

Still incorrect. It looks to me as if you are confusing the Coulomb's constant with the spring constant.

. Forgive me for that .What next ?

 

[Saitama]

. Forgive me for that .What next ?

You can use the substitution $x_2-x_1-l=z$ but your substitution is fine too.

What do you get if you solve the differential equation? I suggest to denote everything by symbols and substitute the values later.

Btw, voko's method is a nice one, you can use energy conservation in the CM frame, no need of differential equations. :)

 

[consciousness]

Work Energy theorem is the best approach according to me.
Since the magnitude of charge and the masses are same, these objects will move symmetrically in opposite directions. Therefore, when maximum compression and elongation occur, both the objects will be at rest. So you just have to consider
(i)Spring potential energy
(ii)Interaction potential energy
(iii)Work done by external force
while making the equations.

 

[Tanya Sharma]

I understand work energy approach is better .But since I started off with DE's ,want to finish it first.

The DE is

d²(x₂-x₁)/dt²=10-k(x₂-x₁-l)

Now putting x₂-x₁ = z ,

We have d²z/dt²=10-k(z-l)

or , d²z/dt²+kz-(kl+10) = 0The solution is z = C₁cos(√kt) + C₂sin(√kt) + (l + 10/k)

Does that make sense ?

 

[voko]

"assuming the natural length of the spring to be large" means that you are supposed to approach the problem assuming that $s = z - l << l$. It does not mean that the term for the electrical interaction between the charges is completely neglected. It means that it must be approximated with a term linear in $s$.

 

[Tanya Sharma]

Hi voko...

Suppose we neglect the force between the charges , does post #9 makes sense ?

 

[voko]

In the sense whether the solution of the equation in #9 is correct, yes.

 

[Tanya Sharma]

Okay...

So ,z = C₁cos(√kt) + C₂sin(√kt) + (l + 10/k)

Now ,using z(0) = l we get C₁ = -10/k .

How do I find C₂ ?

 

[voko]

What about $\dot z(0)$?

 

[Tanya Sharma]

That gives C₂ = 0 .

So,z =(-10/k)cos(√kt) + (l+10/k) . Is it correct ?

 

[voko]

The equation from #9, which you have solved, is not a correct equation for the original problem. Yet you used the initial conditions from the original problem to determine the coefficients in its solution. That does not look correct to me.

 

[Saitama]

"assuming the natural length of the spring to be large" means that you are supposed to approach the problem assuming that $s = z - l << l$. It does not mean that the term for the electrical interaction between the charges is completely neglected. It means that it must be approximated with a term linear in $s$.

Hi voko!

This is the DE we have:
$$-2k(x_2-x_1-l)-\frac{2Kq^2}{(x_2-x_1)^2}+2qE=m\left(\frac{d^2x_2}{dt^2}-\frac{d^2x_1}{dt^2}\right)$$

If we go with Tanya's substitution $x_2-x_1=z$, I don't see how we get z-l in the denominator of electric force.

 

[voko]

I suggest that you first rewrite that equation in terms of $s$. Then think how you can simplify the equation if $s \ll l$.

 

[Saitama]

I suggest that you first rewrite that equation in terms of $s$. Then think how you can simplify the equation if $s \ll l$.

OK, so I switch to the substitution $x_2-x_1-l=s$, then I have $\ddot{x_2}-\ddot{x_1}=\ddot{s}$, hence
$$-2ks-\frac{2Kq^2}{(s+l)^2}+2qE=m\frac{d^2s}{dt^2}$$
$$\Rightarrow -2ks-\frac{2Kq^2}{l^2}\left(1+\frac{s}{l}\right)^{-2}+2qE=m\frac{d^2s}{dt^2}$$
$$\Rightarrow m\frac{d^2s}{dt^2}=-2ks+\frac{4Kq^2s}{l^3}+2qE-\frac{2Kq^2}{l^2}$$
Do I have to solve the above DE?

 

[voko]

The problem does not want you to solve differential equations. It wants you to find the minimum and maximum elongations.

Note that the equation you obtained can be re-interpreted as an equation for a massive particle in a force field, where the force depends linearly on the position. That force is potential, so you could find the corresponding potential energy. Then you can write down the equation for conservation of energy, and determine the min/max elongation from it.

 

[Saitama]

The problem does not want you to solve differential equations. It wants you to find the minimum and maximum elongations.

Note that the equation you obtained can be re-interpreted as an equation for a massive particle in a force field, where the force depends linearly on the position. That force is potential, so you could find the corresponding potential energy. Then you can write down the equation for conservation of energy, and determine the min/max elongation from it.

But still, I would like to know if the DE is correct.

Okay, I try the conservation of energy. The acceleration of COM is zero. I switch to COM frame and use conservation of energy. Initially, let both the masses move away from each other, then
$$\frac{-Kq^2}{l}=qE(x_1+x_2)+\frac{1}{2}k(x_1+x_2)^2+\frac{-Kq^2}{(l+x_1+x_2)}$$
where $x_1$ and $x_2$ are the distances moved by A and B. Also, at max elongation, kinetic energy of blocks is zero. Let $x_1+x_2=s$, then
$$\Rightarrow Kq^2\left(\frac{1}{l+s}-\frac{1}{l}\right)=qEs+\frac{1}{2}ks^2$$
$$\Rightarrow -\frac{Kq^2}{l^2}\left(1-\frac{s}{l}\right)=qE+\frac{1}{2}ks$$

Is the above correct?

 

[voko]

But still, I would like to know if the DE is correct.

Yes, it was correct. I did not mean that you should drop that equation. Rather, I meant that you could use that equation to restore the potential energy for the fictitious massive particle. The equation was $m \ddot s = F(s)$, so find $U(s)$ such that $F(s) = - U'(s)$, then use conservation of energy.

Okay, I try the conservation of energy. The acceleration of COM is zero. I switch to COM frame and use conservation of energy. Initially, let both the masses move away from each other, then
$$\frac{-Kq^2}{l}=qE(x_1+x_2)+\frac{1}{2}k(x_1+x_2)^2+\frac{-Kq^2}{(l+x_1+x_2)}$$
where $x_1$ and $x_2$ are the distances moved by A and B.

I do not follow you here. The LHS is apparently the initial potential energy of electric interaction of the charges. What the about the initial potential energy of the interaction with the field?

I am having difficulty understanding the RHS. It seems that $x_1$ and $x_2$ are measured from the CoM, but they use different sign conventions. Is that correct? Secondly, the RHS must have terms for kinetic energy in general.

But, again, I did not mean that you should drop the DE and redo everything from scratch.

 

[Saitama]

Yes, it was correct. I did not mean that you should drop that equation. Rather, I meant that you could use that equation to restore the potential energy for the fictitious massive particle. The equation was $m \ddot s = F(s)$, so find $U(s)$ such that $F(s) = - U'(s)$, then use conservation of energy.



I do not follow you here. The LHS is apparently the initial potential energy of electric interaction of the charges. What the about the initial potential energy of the interaction with the field?

I am having difficulty understanding the RHS. It seems that $x_1$ and $x_2$ are measured from the CoM, but they use different sign conventions. Is that correct? Secondly, the RHS must have terms for kinetic energy in general.

But, again, I did not mean that you should drop the DE and redo everything from scratch.

Sorry for so many errors voko.

I asked Tanya for the answer to the given problem and the given answer is $2qE/k$. If we neglect the attractive Coulomb force, this answer is easily reached. Is there an error in the question?

I have to leave now, I will make a correction post when I return. Thank you for all the help voko!

 

[voko]

Well, if that is the answer, then the problem was already solved.

I would not say there is an error in the question, but it is definitely ambiguous. It should have said "ignore the interaction between the charges".

 

[Saitama]

Well, if that is the answer, then the problem was already solved.

I would not say there is an error in the question, but it is definitely ambiguous. It should have said "ignore the interaction between the charges".

Thanks voko! :)

So we have the following DE:
$$-2ks+2qE=m\frac{d^2s}{dt^2}$$
The solution of the above differential equation is:
$$s(t)=\frac{qE}{k}+A\sin(\omega t)+B\cos(\omega t)$$
where $\omega=\sqrt{2k/m}$.
Using the initial conditions,
$$s(t)=\frac{qE}{k}(1-\cos(\omega t) )$$
The question asks the maximum compression and maximum elongation. Maximum elongation is maximum of $s(t)$ i.e $2qE/k$, how to find the maximum compression?

 

[voko]

Minimum elongation does not mean there will be any compression. From the final formula for $s$, it is obvious that the minimum elongation is zero.

That could also have been understood from conservation of energy. Initially, kinetic energy was zero, so potential energy must have been at a maximum, thus the initial state must correspond to either min or max elongation.

 

[Saitama]

Minimum elongation does not mean there will be any compression. From the final formula for $s$, it is obvious that the minimum elongation is zero.

That could also have been understood from conservation of energy. Initially, kinetic energy was zero, so potential energy must have been at a maximum, thus the initial state must correspond to either min or max elongation.

I too thought that the minimum elongation should be zero as seen from the equation but I just wanted to be sure. Thanks a lot for clearing it up voko!