Maximum speed of a spring mass is given by 2(pi)(frequency)(amptitude)?

  • Thread starter zeion
  • Start date
  • #1
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Homework Statement



Prove that the maximum speed (Vmax) of a mass on a spring is given by 2(pi)(f)(A)


Homework Equations



E(total) = kA^2

Vmax = sqrt[2E(total) / m]
(Because E(potential) = 0 when V is at max, so E(total) = mv^2 / 2 + 0)

f = sqrt(k/m) / 2(pi)
k = [(f)(2pi)]^2(m)

The Attempt at a Solution



Vmax = sqrt[2(kA^2) / m]
Vmax = sqrt [2[(f)(2pi)]^2 (m) A^2)] / m]
Square the [(f)(2pi)]^2, factor in m, factor in 2A^2 and cancel out all m's then sqrt I get:

2Af2(pi)

Did I make a mistake somewhere? Or did I go about this all wrong? :/
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
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Hi zeion,

Homework Statement



Prove that the maximum speed (Vmax) of a mass on a spring is given by 2(pi)(f)(A)


Homework Equations



E(total) = kA^2
I don't think this line is right; you appear to be missing something in this equation. Once you correct that I think you'll get the right answer.
 
  • #3
467
0
Oh is it kA^2 / 2?
Oh if it is then it all makes sense.
 

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