Beads on a thread - What stops the acceleration?

[haruspex]

Okay, so the expression is $v_{N+1} = \frac{N}{N+1} * \sqrt{w_N + 2k/N}$
I could get rid of the root-sign by squaring both sides, but that would yield several variables with the power of two which probably is not desired.
The other way is to rewrite the expression under the root-sign as a square, but I can't see how.

Squaring both sides is exactly right. Turn all the v terms into corresponding w terms.

 

[Alettix]

Squaring both sides is exactly right. Turn all the v terms into corresponding w terms.

Oh, yes! So now we have:
$w_{N+1} = (\frac{N}{N+1})^2 * (w_N + \frac{2k}{N})$
And when N → ∞ we got $\frac{N}{N+1} → 1$ and $2k/N → 0$
thus: $w_{N+1} → w_N$
is this right? :)

 

[haruspex]

Oh, yes! So now we have:
$w_{N+1} = (\frac{N}{N+1})^2 * (w_N + \frac{2k}{N})$
And when N → ∞ we got $\frac{N}{N+1} → 1$ and $2k/N → 0$
thus: $w_{N+1} → w_N$
is this right? :)

Yes, but that doesn't show w_N converges. Consider a_n+1=(1+1/n)a_n. That does not converge. (Indeed, solution is a_n=n.)
Multiply through the equation by (N+1)². This should suggest another change of variable.

 

[Alettix]

Yes, but that doesn't show w_N converges. Consider a_n+1=(1+1/n)a_n. That does not converge. (Indeed, solution is a_n=n.)

But if $w_{N+1} = w_N$, it means that the velocity does not change when more beads are added, thus it is constant, isn't it? I also graphed your function, and for big n-s, $a_{n+1}$ always approached the constant $a_n$, how comes this does not count as a convergance?

Multiply through the equation by (N+1)². This should suggest another change of variable.

If we say: $m_N=w_N*N^2$, we now have:
$m_{N+1} = m_N + 2kN$
but in this case, $m_{N+1}$ does not approach $m_N$ when $N→∞$
Have I done the wrong change of varibale?

 

[haruspex]

But if $w_{N+1} = w_N$, it means that the velocity does not change when more beads are added, thus it is constant, isn't it?

You have not proved that equality. You only showed the ratio tends to 1. That is not the same thing.

I also graphed your function, and for big n-s, $a_{n+1}$ always approached the constant $a_n$, how comes this does not count as a convergance?

a_n is not a constant. As I posted, the solution to my a_n recurrence relation is a_n=n. The sequence {n} does not converge to a constant.

If we say: $m_N=w_N*N^2$, we now have:
$m_{N+1} = m_N + 2kN$

That's the right step. Can you see how to rewrite that as m_N= sum of a series? You can easily sum that series and find the exact algebraic form of v_N.

 

[Alettix]

That's the right step. Can you see how to rewrite that as m_N= sum of a series? You can easily sum that series and find the exact algebraic form of v_N.

Should it be rewritten: $m_N = m_1 + m_2 + m_3$ ?
No, that can't be right, can it? I am sorry, I can't see which sum $m_N$ equlas.
Should I go back to some of the "basic" equations and use v instead of w to find a sum?

 

[haruspex]

Should it be rewritten: $m_N = m_1 + m_2 + m_3$ ?
No, that can't be right, can it? I am sorry, I can't see which sum $m_N$ equlas.
Should I go back to some of the "basic" equations and use v instead of w to find a sum?

No, persevere with $m_{N+1}=m_N+2kN$.
This tells you what to add to go from m_N to m_N+1. So what do you need to add to go from m₁ to m_N? If it's still not clear, find m₂ in terms of m₁, then m₃, etc.

 

[Alettix]

$m_N = m_1 + (N-1)N *k$
is what I find, is that right?
Now, I tried finding the difference between $m_{N+1}$ and $m_N$
--> $m_{N+1} - m_N = (N+1)N - (N-1)N = 2kN$
Now, using $m_N = w_N * N^2$ , we can obtain:
$w_{N+1} = \frac{2kN + w_N * N^2}{(N+1)^2}$
if now N → ∞ , we can apporoximate: $w_{N+1} → \frac{2kN+w_N * N^2}{N^2} = 2k/N + w_N → 0 + w_N = w_N$
because $w_N = v_N^2$, this proves that $v_{N+1} → v_N$ when N → ∞. Or am I wrong?

 

[haruspex]

$m_N = m_1 + (N-1)N *k$
is what I find, is that right?

Good, but the next step is to clean up that m₁. The whole system starts from rest, so m₁=0.
m_N=N(N-1)k.
Use that to find v_N² as a function of N and k.

 

[Alettix]

Well in that case, I get:
$v_N = (k - k/N)^{1/2}$
If then N → ∞ $v_N = k^{1/2} = (\frac{Fd}{m})^{1/2}$ which is the correct answer.

Thank you very much for your kind help and patience Sir!

 

[haruspex]

Well in that case, I get:
$v_N = (k - k/N)^{1/2}$
If then N → ∞ $v_N = k^{1/2} = (\frac{Fd}{m})^{1/2}$ which is the correct answer.

Thank you very much for your kind help and patience Sir!

You got it.