Oscillation in spring mass system questions

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SUMMARY

The discussion centers on solving a problem related to the oscillation of a spring mass system, specifically focusing on the equation T = 2π√(m/k). Participants clarify that squaring the period (T) leads to T² = (4π²m)/k, establishing a linear relationship between T² and 1/k. This method allows for easier calculation of the spring constant (k) or mass (m) by graphing T² against 1/k. The conversation emphasizes the importance of understanding these relationships for accurate data interpretation in experiments.

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  • Understanding of harmonic motion principles
  • Familiarity with the spring mass system equations
  • Basic graphing skills for linear relationships
  • Knowledge of variables m (mass) and k (spring constant)
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  • Study the implications of squaring equations in physical contexts
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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?
 

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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.
 
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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!
 

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