Mass attached on two springs between the walls

[BruceW]

Bruce, you explain things for yourself quite nicely. I meant that the equation first-order in ε is

\ddot \epsilon+\frac{2k}{m} \left( 1- \left(1+\frac{D^2}{L^2} \right)^{-1.5} \right )\epsilon=0
with D and L and everyting else constant. It is an equation for SHM...

yeah... hehe I understand what you mean now. We must take the limit $\varepsilon \rightarrow 0$ first, before we think about taking a limit of D and L. (since we want the first-order motion, it is most important that $\varepsilon$ be small).

But you can also isolate $\left(1+\frac{D^2}{L^2} \right)^{-0.5}$ from f(D)=0 and substitute it into the equation for ω² and expand after.

ehild

ah, this is genius. you don't even need to expand after, it is already as simple as it could be, even without expanding it.

edit: ah no wait, I jumped the gun again. It is not as nice as it could be. The power is not the same. There is a power to 1.5 in the equation for frequency and a power to 0.5 in the f(D)=0 equation.

edit again: but, it does get rid of L, even without using the approximation. So that's pretty good. Although, to calculate D itself would require using L. But luckily, the question only asks for the frequency in terms of D.

 

[ehild]

yeah... hehe I understand what you mean now. We must take the limit $\varepsilon \rightarrow 0$ first, before we think about taking a limit of D and L. (since we want the first-order motion, it is most important that $\varepsilon$ be small).


ah, this is genius. you don't even need to expand after, it is already as simple as it could be, even without expanding it.

edit: ah no wait, I jumped the gun again. It is not as nice as it could be. The power is not the same. There is a power to 1.5 in the equation for frequency and a power to 0.5 in the f(D)=0 equation.

You can simplify... and then expand.


ehild

 

[BruceW]

yeah. we have
\left(1+\frac{D^2}{L^2} \right)^{-0.5} = 1 - \frac{gm}{2kD}
and the equation for frequency contains the cube of this. So then we just plug in the cube of the right hand side, into our equation for frequency, and take first order of the variable $gm/2kD$, since it will be small if $D/l$ is small.

Having said all this, I don't think the OP'er is meant to use the approximation $D/l<<1$. Even though it is an interesting approximation to make.

 

[BvU]

Wow, goes to show what an ambiguously formulated exercise can bring about. Why on Earth would someone ask something like that in such a way that in can be turned into such a hideous monstrum. You give l but don't give a clue as to whether it is anywhere near L.

I (sulk sulk) liked the $\ \ f = {1\over 2\pi} \sqrt{3g\over D}\$ a lot better. I now even have a hard time showing the b&e result goes to $\ \ f = {1\over 2\pi} \sqrt{g\over D}\$ for L going to zero...

 

[ehild]

In case (D/L)²<<1,

ω^2≈\frac{2k}{m} \left( 1- (1-1.5\frac{D^2}{L^2}) \right )=\frac{3k}{m} \frac{D^2}{L^2}

Expanding f(D) f(D)=g+\frac{2kD}{m}\left( \left(1+\frac{D^2}{L^2} \right)^{-0.5}-1 \right)= g+\frac{2kD}{m} ( 1-0.5 \frac{D^2}{L^2} -1)\rightarrow g=\frac {kD^3}{mL^2}

Divide the equations ω^2=\frac{3k}{m} \frac{D^2}{L^2}
and g=\frac {kD^3}{mL^2}

You get ω^2=\frac{3g}{D}.
...............
Using the original equation for ω²and f(D)=0, you can eliminate m before the approximation, and get a more accurate expression for ω². With the notation
η=\left(1+\frac{D^2}{L^2}\right)^{-0.5}:ω^2=\frac{2k}{m} \left( 1- \left(1+\frac{D^2}{L^2} \right)^{-1.5} \right ) \rightarrow
ω^2=\frac{2k}{m} ( 1- η^{3} )
and f(D)=g+\frac{2kD}{m}\left( \left(1+\frac{D^2}{L^2} \right)^{-0.5}-1 \right)=0\rightarrow g=\frac{2kD}{m}(1-η)
Dividing the equations, k and m cancel.

ω^2=\frac{g}{D}\frac{1-η^3}{1-η}=\frac{g}{D}(1+η+η^2)

With the approximation $η=1-0.5\frac{D^2}{L^2}$ when (D/L)²<<1,

ω^2=\frac{g}{D}(1+η+η^2)=\frac{3g}{D}(1-0.5\frac{D^2}{L^2})

With the given numerical data (L=0.5 m, k=500 N/m, m=0.05 kg), D≈0.063 m, (D/L)²≈ 0.016. Ignoring it means about half percent error in the frequency.

ehild

 

[BvU]

The given numerical data mention the length of the spring l. They say nothing at all about the distance between the walls, 2L. It could be zero for all we know... Or 20 meter. That's why I don't like the formulation of the exercise.

ehild: yes for D << L I can follow $\ ω^2=\frac{3g}{D}\$: in your picture L < l (lower case l, that is).

Is it also obvious that $\ ω^2 \rightarrow \frac{g}{D}\$ for L $\downarrow$ 0 ? (l stays 0.5).

 

[skrat]

The given numerical data mention the length of the spring l. They say nothing at all about the distance between the walls, 2L. It could be zero for all we know... Or 20 meter. That's why I don't like the formulation of the exercise.

Post #3 :

My inability to speak English properly is the answer to the last question. The problem also states that the springs are NOT deformed before the mass is applied.

 

[BvU]

My inability to properly read proper english is reflected in the fact that I didn't let that one reach my little grey cells. If I read it now, to me it still means that all we know about L is that L $\le$ l. It can be 1 cm or 50 cm. At least the 20 meter is excluded.

 

[BruceW]

yeah it could have been written more clearly. But it is implied. Because (before the mass is attached), we know that the springs un-stretched length is 0.5m each. And the springs are initially attached together in the middle. And the springs are initially not deformed. So this would imply that the walls are 1m apart... Unless the springs are initially not horizontal... In which case, the walls could be closer than 1m. But I'm guessing the springs are meant to be horizontal initially. Is that right skrat?

 

[skrat]

yeah it could have been written more clearly. But it is implied. Because (before the mass is attached), we know that the springs un-stretched length is 0.5m each. And the springs are initially attached together in the middle. And the springs are initially not deformed. So this would imply that the walls are 1m apart... Unless the springs are initially not horizontal... In which case, the walls could be closer than 1m. But I'm guessing the springs are meant to be horizontal initially. Is that right skrat?

Exactly.

 

[BvU]

How do you know and why didn't that appear in the exercise ?

 

[skrat]

How do you know and why didn't that appear in the exercise ?

I don't. But I do agree that the problem is a bit loose. It only says that the vertical walls are not more than $2l$ apart (because the springs are not deformed), if $l$ is the distance of each spring, however the walls could easily be anything from $0$ to $2l$ apart.

Basically I just focused myself on the case when the distance between the walls is $2l$. Don't ask me why. It just seemed to be the right thing to do at that moment. However, doing so, of course doesn't bring me to a general solution of the original problem but to a specific one where the walls are $2l$ apart.

 

[BvU]

You did well. Thanks for bringing up such a nice exercise (and kudos for sitting it out...) !

 

[BruceW]

With the given numerical data (L=0.5 m, k=500 N/m, m=0.05 kg), D≈0.063 m, (D/L)²≈ 0.016. Ignoring it means about half percent error in the frequency.

ehild

this is good. So for the numerical calculation in part b, skrat is fine to use the approximation. But I still think that for part a, skrat should give the exact equations. Maybe they will also accept the approximate answer, but by the wording of the question, I don't think so. Especially since they say:

Hint: There is no need to explicitly write the initial offset as function of mass. Just use the offset as independent parameter using which mass can be calculated.

Surely they say this because it is too difficult to re-write the exact D as a function of the other variables. And if they were expecting skrat to give the approximate answer for part a, then there would be no need to say the hint above, because using the approximation, D is a simple function of the other parameters.

 

[skrat]

Surely they say this because it is too difficult to re-write the exact D as a function of the other variables. And if they were expecting skrat to give the approximate answer for part a, then there would be no need to say the hint above, because using the approximation, D is a simple function of the other parameters.

Nobody expects anything from me. This is just me, sitting behind the desk, learning new things and preparing myself for the exam I don't want to barely pass, but I want to pass it with a result that I can be happy about (or proud of, as you wish).

So this problem was once on the exams (year 2011) and I tried to solve it. Since I ran into problems, I decided not to just skip it like most do but to fight it, and with your help I not only solved the problem I also learned a lot. I learned a lot about where I got it all wrong and what could be done a lot better (like ignoring that $D<<L$ in part a). By doing this, you can be sure that the next time I will ask myself if it is really completely legit to make any approximations before the problem says so. I should have known that by now, but hey I am only a student.

 

[ehild]

You are right, approximations should be handled with care. Now you know it


ehild

 

[BruceW]

So this problem was once on the exams (year 2011) and I tried to solve it. Since I ran into problems, I decided not to just skip it like most do but to fight it, and with your help I not only solved the problem I also learned a lot. I learned a lot about where I got it all wrong and what could be done a lot better (like ignoring that $D<<L$ in part a). By doing this, you can be sure that the next time I will ask myself if it is really completely legit to make any approximations before the problem says so. I should have known that by now, but hey I am only a student.

we're all students, in the lecture course that is life. (cheesy line of the year award goes to me). Also, thanks for sharing this problem with us. It's an interesting one.

 

[BvU]

Good luck with your exams. Like the attitude, but make sure it doesn't grow too serious. After all, the poor blokes who think up the exam questions generally don't ask for things you can't know (I used to 'help' my own kids with tidbits of 'wisdom' like that if they got nervous before tests -- long time ago )