Oscillation in spring mass system questions

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
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: 614
  • 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: 512
  • 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: 477
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:
[itex]T = 2\pi\sqrt{\frac{m}{k}}[/itex]

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

If you square both sides, you get:
[itex]T^{2} = \frac{4\pi^{2}m}{k}[/itex], so [itex]T^{2} \propto \frac{1}{k}[/itex], so graphing [itex]T^{2}[/itex] vs. [itex]\frac{1}{k}[/itex] 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   Reactions: 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:
[itex]T = 2\pi\sqrt{\frac{m}{k}}[/itex]

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

If you square both sides, you get:
[itex]T^{2} = \frac{4\pi^{2}m}{k}[/itex], so [itex]T^{2} \propto \frac{1}{k}[/itex], so graphing [itex]T^{2}[/itex] vs. [itex]\frac{1}{k}[/itex] 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!