Simple Harmonic Motion Problems

  • Thread starter 03MGCobra
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  • #1
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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.
 

Answers and Replies

  • #2
gneill
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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.

What happens if you square both sides of the equation?
 
  • #3
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0
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?
 
  • #4
gneill
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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.
 
  • #5
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T^2 = 2pi^2 x (m/k) then k = 2pi ^2 x m/ T^2
 
  • #6
gneill
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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.
 
  • #7
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k =((2pi)^2 x m)/T^2
 
  • #8
gneill
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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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