Simple Harmonic Motion Problems

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Homework Help Overview

The discussion revolves around solving for the spring constant \( k \) in the context of simple harmonic motion, specifically using the equation \( T = 2\pi \sqrt{\frac{m}{k}} \). Participants are exploring algebraic manipulations related to this equation.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss squaring both sides of the equation to isolate \( k \). There are questions about the proper application of algebraic rules, particularly regarding squaring terms on both sides of the equation.

Discussion Status

Several participants have provided algebraic transformations and corrections to earlier attempts. There is an ongoing exploration of how to correctly manipulate the equation while maintaining balance, with some participants suggesting alternative forms for \( k \).

Contextual Notes

Participants are navigating the constraints of algebraic manipulation rules and ensuring that operations applied to one side of the equation are consistently applied to the other side. There is an emphasis on maintaining the integrity of the equation throughout the discussion.

03MGCobra
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I am having a hard time solving for k in the equation
T =
2 x pi sq. rt. (m/k)

I know this is simple algebra but its keeping me from solving this.
 
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03MGCobra said:
I am having a hard time solving for k in the equation
T =
2 x pi sq. rt. (m/k)

I know this is simple algebra but its keeping me from solving this.

What happens if you square both sides of the equation?
 
the sq rt goes away on the side with 2 x pi (m/k) and then T^2. so T^2 = 2pi x (m/k) now can be simplified into k = (2pi x m) / T^2?
 
03MGCobra said:
the sq rt goes away on the side with 2 x pi (m/k) and then T^2. so T^2 = 2pi x (m/k) now can be simplified into k = (2pi x m) / T^2?

No, you can't just square part of a side. The [itex]2 \pi[/itex] has to be squared, too.
 
T^2 = 2pi^2 x (m/k) then k = 2pi ^2 x m/ T^2
 
03MGCobra said:
T^2 = 2pi^2 x (m/k) then k = 2pi ^2 x m/ T^2

What about the 2? You must do the same operation to all parts of a given side if the equation is to remain balanced.
 
k =((2pi)^2 x m)/T^2
 
03MGCobra said:
k =((2pi)^2 x m)/T^2

That looks better!

If you want to clean it up a bit, you could make it:
[tex]k = \left( \frac{2 \pi}{T} \right)^2 m[/tex]
 

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