Will the Bug Ever Reach the End of the Stretching Rubber Band?

[ehild]

$x=(L+vt)\frac { u\log { (L+vt) } }{ v } +C$

don't you miss parentheses?

ehild

 

[ehild]

I would have never thought to simplify the problem in such a way! That is quite clever!
(Is this what orodruin was hinting at in post #3?)

Looking at that post again, Orodruin's hint was very much more clever: Write equation for the ratio of the lengths: y= x/(L+vt) where L is the initial length of the rope. You get a very simple differential equation.

But it is not bad that you have understood how to solve a linear first-order differential equation.

ehild

 

[Nathanael]

Substitute f into the remaining part of the equation:
$fg'+Q(t)=0$: $g'=-Q(t)/f=0$
Integrate, but add the integration constant now. Substitute back into y=fg.

So:

$g=\frac{a}{b}ln(bt+1)+C$

And:

$y=(bt+1)(\frac{a}{b}ln(bt+1)+C)$

From the initial condition we see that C=0

Replacing a, b, and y, with their respective definitions (and multiplying by L) I get:

$x=(vt+L)(\frac{u}{v}ln(\frac{vt}{L}+1))=L+vt$

Which yields:

$t=e^{100000}-1$

Which is the correct answer[Edit: much easier when I'm not tired hehe ]

 

[Nathanael]

Looking at that post again, Orodruin's hint was very much more clever: Write equation for the ratio of the lengths: y= x/(L+vt) where L is the initial length of the rope. You get a very simple differential equation.

It's amazing how you guys have such an intuitive grasp of differential equations. It is very impressive to me, I don't understand it at all.

But it is not bad that you have understood how to solve a linear first-order differential equation.

I would hardly say I've understood it! I would have never done that in a million years! I only "understood" because you were nice enough to walk me through it like a baby



I have one more question(s) about this method:

I usually solve first order linear differential equations y'+P(t)y+Q(t) =0 by a method I learned when I was a student.
I assume that y is a product of two functions y=fg. y'=f'g+fg' . Substituting back,
f'g+fg'+P(t)fg+Q(t)=0. I have the liberty to choose f and g so f'g+P(t)fg=0, that is f'+P(t)f=0.

Does this always work? (For first order linear DE's)
If not then how often does it work?

Are you always supposed to do f'g+P(t)fg=0?
Is that the 'trick' that makes it useful?
(Or do you have to be clever about it?)

 

[ehild]

I have one more question(s) about this method:


Does this always work? (For first order linear DE's)
If not then how often does it work?

This method always works for linear first order differential equations, just like the integrating factor method.

Are you always supposed to do f'g+P(t)fg=0?
Is that the 'trick' that makes it useful?

That belongs to the method. As you wrote, '"it is the trick" of the method.

Now what about Orodruin's hint? What differential equation do you get for y=x/(L+vt)?

ehild

 

[Nathanael]

Now what about Orodruin's hint? What differential equation do you get for y=x/(L+vt)?

ehild

$\frac{dy}{dt}=\frac{\dot x}{L+vt}-\frac{xv}{(L+vt)^2}$

(With initial value y=0 when t=0)
Which could be written as:

$\frac{dy}{dt}=\frac{\dot x-yv}{L+vt}$

(I got stuck here for a few minutes but then thought of this )
Maybe I should plug in $\dot x=u+vy$ and then I get:

$\frac{dy}{dt}=\frac{u}{L+vt}$

$\int dy=\int \frac{u}{L+vt}dt=\frac{u}{v}ln(L+vt)+C=y$

From our initial condition we see that C=0

Now we just set $y=1$ and solve for $t$!

$ln(1+t)=100000$

$t=e^{100000}-1$

 

[ehild]

Is not Maths fun?

I would have started it with x=y(L+vt)
x'=y'(L+vt)+yv. Substituting back into the original equation:
y'(L+vt)+yv=u+yv--->y'(L+vt)=u---> y'=1/(L+vt)

 

[Nathanael]

Is not Maths fun?

Very fun! But not as fun as physics! I have no idea why I suddenly understood how to do it that way. All you did was say "now what about orodruin's hint"

It feels much better when someone doesn't have to walk you through every single step
(even though I would've never done it without orodruin's hint!)

 

[Orodruin]

Looking at that post again, Orodruin's hint was very much more clever: Write equation for the ratio of the lengths: y= x/(L+vt) where L is the initial length of the rope. You get a very simple differential equation.

It's amazing how you guys have such an intuitive grasp of differential equations. It is very impressive to me, I don't understand it at all.

I would call it occupational injury. This problem is very reminiscent of an expanding Universe (for the bug it probably is an expanding universe). In cosmology, $y = \frac{x}{L+vt} \equiv \frac{x}{a(t)}$, where $a(t)$ is the scale factor, is called the comoving distance. Don't worry, understanding will come with learning experience. I have been known to give students similar problems on relativity exams ...

 

[ehild]

Very fun! But not as fun as physics!


I have no idea why I suddenly understood how to do it that way. All you did was say "now what about orodruin's hint"

It feels much better when someone doesn't have to walk you through every single step
(even though I would've never done it without orodruin's hint!)

Look at the previous post, I edited it.

Sometimes you need to be guided, but next time you will figure it out by yourself. If you practice enough, your brain will find connections among things you have learnt, and suddenly something new occurs to you that you never heard or read before : You discover something.

Yes, Physics is connected to the world we live in and it is fun when the things work in the way as it is predicted by Physics. But Physics is nothing without Maths.

You set up a model, do the Maths and predict the outcome of an experiment. It was always a great surprise and also a great pleasure for me when my measurement results agreed with theory.

ehild

 

[ehild]

I would call it occupational injury. This problem is very reminiscent of an expanding Universe (for the bug it probably is an expanding universe). In cosmology, $y = \frac{x}{L+vt} \equiv \frac{x}{a(t)}$, where $a(t)$ is the scale factor, is called the comoving distance. Don't worry, understanding will come with learning experience. I have been known to give students similar problems on relativity exams ...

Bug in an expanding Universe ... Amazing! Never would have occurred to me, but I never taught Relativity Theory.

ehild

 

[Nathanael]

I would call it occupational injury. This problem is very reminiscent of an expanding Universe (for the bug it probably is an expanding universe). In cosmology, $y = \frac{x}{L+vt} \equiv \frac{x}{a(t)}$, where $a(t)$ is the scale factor, is called the comoving distance. Don't worry, understanding will come with learning experience. I have been known to give students similar problems on relativity exams ...

Thanks orodruin! Thank you ehild too!

I wish to give you both more thanks, but, apparently, I've given too many in the past 24 hours (I've only given like 5 )

Look at the previous post, I edited it.

That works too. Both ways are good

Yes, Physics is connected to the world we live in and it is fun when the things work in the way as it is predicted by Physics. But Physics is nothing without Maths.

I've always liked math, long before I knew what physics was, but the main thing I enjoyed was word problems! I think that is why I enjoy physics; it's essentially mathematical word problems
To me, the fun part of math is figuring out how to figure it out, and that's why I find physics much more fun. Physics, to me, is all the fun parts of math without the boring parts! (Maybe one day I'll develop more of a liking for pure math.)

It was always a great surprise and also a great pleasure for me when my measurement results agreed with theory.

I bet!


Thanks for helping me out ehild, you're awesome




Edit:

I have been known to give students similar problems on relativity exams ...

but I never taught Relativity Theory.

So you both are or have been teachers? (Or professors, sorry if it's rude to say teacher )
I think I might want to be a teacher when I'm older (if I have what it takes) that would be a fun job!

 

[Orodruin]

Yes, Physics is connected to the world we live in and it is fun when the things work in the way as it is predicted by Physics. But Physics is nothing without Maths.

You know, Feynman put it the other way around in slightly other words ...

Also, while on the subject:

 

[ehild]

You know, Feynman put it the other way around in slightly other words ...

May I ask, what he said? Somehow I am not fond of Feynman, so I am not familiar with his sayings.

ehild

 

[ehild]

I think I might want to be a teacher when I'm older (if I have what it takes) that would be a fun job!

I am sure you will become a very good teacher.

ehild

 

[Orodruin]

May I ask, what he said? Somehow I am not fond of Feynman, so I am not familiar with his sayings.

ehild

Something to the effect of "Physics is to mathematics what sex is to masturbation" if I don't misremember.

But it is good that everyone has different preferences or physics jobs would be even scarcer than they are now ...

 

[ehild]

I think I start to like Feynman

ehild

 

[Nathanael]

Something to the effect of "Physics is to mathematics what sex is to masturbation"

I think I start to like Feynman

 

[ehild]

Well, we are a bit offtopic. Better to stop it before we get infractions.

ehild

 

[Satvik Pandey]

Thank you ehild,Orodruin,Nathanael I got the answers too.

Thank you very much Nathanael for posting this question.

 

[Satvik Pandey]

Suppose if the bug would have started walking from the end of the rubber band which is in our hand to end which is attached to wall? Will it reach the other end of the rubber band?
Assuming that other conditions of question remains same.

 

[Orodruin]

Suppose if the bug would have started walking from the end of the rubber band which is in our hand to end which is attached to wall? Will it reach the other end of the rubber band?
Assuming that other conditions of question remains same.

What are your thoughts?

 

[Satvik Pandey]

What are your thoughts?

I think that bug will reach the other end of the band.
I think relative velocity of bug w.r.t Earth at any point $x$ (measured from the wall)
is $u-\frac { xv }{ L+vt }$.

 

[D H]

$\frac{dy}{dt}=\frac{\dot x}{L+vt}-\frac{xv}{(L+vt)^2}$

(With initial value y=0 when t=0)
Which could be written as:

$\frac{dy}{dt}=\frac{\dot x-yv}{L+vt}$

(I got stuck here for a few minutes but then thought of this )
Maybe I should plug in $\dot x=u+vy$ and then I get:

$\frac{dy}{dt}=\frac{u}{L+vt}$

Excellent! This is what I wanted you to do way back in post #7. It's such a simple result that it's hard to give a hint (other than try it, it's easy!) that doesn't tell the answer.