Mass attached on two springs between the walls

In summary: I think you'll find you have an extra factor of m in the final expression. If you want to match the standard form of the simple harmonic oscillator, the coefficient should be 3k/m, not just k. Note that the term in the brackets is the period, and it appears in the exponent of the exponential in the standard form of the solution.Chet
  • #36
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).
 
Physics news on Phys.org
  • #37
BvU said:
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 :

skrat said:
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.
 
  • #38
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.
 
  • #39
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?
 
  • #40
BruceW said:
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.
 
  • #41
How do you know and why didn't that appear in the exercise ?
 
  • #42
BvU said:
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.
 
  • #43
You did well. Thanks for bringing up such a nice exercise (and kudos for sitting it out...) !
 
  • Like
Likes 1 person
  • #44
ehild said:
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.
 
  • #45
BruceW said:
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.
 
  • #46
You are right, approximations should be handled with care. Now you know it:smile:


ehild
 
  • #47
skrat said:
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.
 
  • #48
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 :smile:)
 

Similar threads

Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
927
  • Introductory Physics Homework Help
Replies
15
Views
266
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
201
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
234
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top