Oscillation in spring mass system questions

Click For Summary
The discussion revolves around solving a problem related to the oscillation of a spring-mass system, specifically addressing the relationship between the period (T), mass (m), and spring constant (k). It highlights that squaring the equation T = 2π√(m/k) can clarify the relationships between these variables, as both T and T² can be graphed against 1/k to reveal linear relationships. The conversation emphasizes that squaring the equation may simplify solving for k or m, depending on the available data. The question's wording is noted as confusing, but the method of squaring is acknowledged as a useful approach for future reference. Overall, understanding these relationships is crucial for analyzing spring-mass systems effectively.
xtrubambinoxpr
Messages
86
Reaction score
0

Homework Statement



I am having an issue with answering number 4 in the attached image.

Homework Equations



Relevant equations are given in question 3.

The Attempt at a Solution



Squaring the equation would only make it easier to solve for any of the other variables, or the period squared. I cannot see the correlation with the square of the period. Would it just make it more accurate?
 

Attachments

  • Screen Shot 2014-02-04 at 12.22.31 AM.png
    Screen Shot 2014-02-04 at 12.22.31 AM.png
    33.8 KB · Views: 586
  • Screen Shot 2014-02-04 at 12.22.42 AM.png
    Screen Shot 2014-02-04 at 12.22.42 AM.png
    17 KB · Views: 493
  • Screen Shot 2014-02-04 at 12.30.20 AM.png
    Screen Shot 2014-02-04 at 12.30.20 AM.png
    4.6 KB · Views: 457
Physics news on Phys.org
It's an oddly worded question -- I think the only thing it's getting at is that, like you said, it would allow you to solve for m or k, and it would also allow you to see the relationships of the variables more clearly.

With the original equation you have:
T = 2\pi\sqrt{\frac{m}{k}}

So that means T \propto \sqrt{\frac{1}{k}}, so graphing T vs. \sqrt{\frac{1}{k}} would give you a linear relationship, and you could solve for k.

If you square both sides, you get:
T^{2} = \frac{4\pi^{2}m}{k}, so T^{2} \propto \frac{1}{k}, so graphing T^{2} vs. \frac{1}{k} would also give you a linear relationship, and you could solve for k that way too. So maybe that would be little easier? It really just depends on what data you have.
 
  • Like
Likes 1 person
jackarms said:
It's an oddly worded question -- I think the only thing it's getting at is that, like you said, it would allow you to solve for m or k, and it would also allow you to see the relationships of the variables more clearly.

With the original equation you have:
T = 2\pi\sqrt{\frac{m}{k}}

So that means T \propto \sqrt{\frac{1}{k}}, so graphing T vs. \sqrt{\frac{1}{k}} would give you a linear relationship, and you could solve for k.

If you square both sides, you get:
T^{2} = \frac{4\pi^{2}m}{k}, so T^{2} \propto \frac{1}{k}, so graphing T^{2} vs. \frac{1}{k} would also give you a linear relationship, and you could solve for k that way too. So maybe that would be little easier? It really just depends on what data you have.


I see where you are coming from! just makes no sense to me to square it given I had all the variables in the experiment that we did in class. Maybe for future reference know this is something great to remember. Thanks Jack!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
17
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
1
Views
1K
Replies
9
Views
4K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
4
Views
861
Replies
17
Views
1K
  • · Replies 58 ·
2
Replies
58
Views
3K
Replies
24
Views
3K