Oscillation in spring mass system questions

In summary, the equation would give you a linear relationship between T and \sqrt{\frac{1}{k}} if you square both sides, but it would be easier to solve for k if you graphed T^{2} vs. \frac{1}{k} instead.
  • #1
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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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  • #2
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.
 
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  • #3
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!
 

1. What is a spring mass system?

A spring mass system is a physical system consisting of a mass attached to a spring that is able to oscillate back and forth when displaced from its equilibrium position.

2. What is the equation of motion for a spring mass system?

The equation of motion for a spring mass system is given by Newton's Second Law: F = ma, where F is the net force on the mass, m is the mass, and a is the acceleration. For a spring mass system, this can be written as F = -kx, where k is the spring constant and x is the displacement from equilibrium.

3. What factors affect the oscillation of a spring mass system?

The factors that affect the oscillation of a spring mass system include the mass of the object attached to the spring, the spring constant, and the amplitude and frequency of the oscillation.

4. How do you calculate the period of oscillation for a spring mass system?

The period of oscillation for a spring mass system can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.

5. How does the amplitude affect the oscillation of a spring mass system?

The amplitude of oscillation affects the maximum displacement of the mass from equilibrium and the energy of the system. A larger amplitude will result in a larger maximum displacement and more energy being transferred between kinetic and potential energy during oscillation.

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