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Motion of a mass on a spring suspended vertically

  • Thread starter brayrbob
  • Start date
  • #1
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Okay I have part of this problem right but am not sure how to proceed on the last part.

A 50 g spring is made from a new steel/titanium alloy. Engineers determine that if a 500 g mass is hung from the spring it oscillates with a period of exactly 1.00 seconds. What is the spring's constant?
Equation used is = system mass = hanging mass + (1/3) spring mass
system mass = 500 + (1/3)50 = 516.6666667
Now I have to use the equation Period^2 = 4pi^2 system mass/spring constant.
I have to solve for mass and don't know how to turn this last equation around.
 

Answers and Replies

  • #2
Doc Al
Mentor
44,912
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You start with this:
[tex]T^2 = \frac{4 \pi^2 m}{k}[/tex]

To solve for the spring constant, try this: First multiply both sides by [itex]k[/itex], then divide both sides by [itex]T^2[/itex]. That will isolate k.
 
  • #3
24
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So then that equation should be k = 4pi^2/T^2?

4pi^2(516.6666667)/1.00^2 = 20397.8243 is the spring constant?
 

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