- #1

- 43

- 0

... I tried using KEo + PEo + SPEo = KEf + PEf + SPEf ... but, I can't get anywhere ...

... I found out that the frequency is 18.05hz and it's relative time is .056 seconds - and that's where I'm stuck!

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- Thread starter sailordragonball
- Start date

- #1

- 43

- 0

... I tried using KEo + PEo + SPEo = KEf + PEf + SPEf ... but, I can't get anywhere ...

... I found out that the frequency is 18.05hz and it's relative time is .056 seconds - and that's where I'm stuck!

- #2

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 7

Use Hooke's law and conservation of energy.

- #3

- 43

- 0

... I don't understand the .64 meter compression part ... does that equal "X" in Hooke's Law and in ????

- #4

- 926

- 3

BTW -- you are right in that the compression WILL be X in hooke's Law.

- #5

- 43

- 0

A = 0.064m (given in the problem)

omega = 11.5 (rad/s) - given in the problem

x = 0.037m (given in the problem)

V(max) = A(omega)

V(max) = 0.064(11.5)

V(max) = .736(m/s)

V = V(max) * [the square root of the quantity of (1- {[x^2]/[A^2]})]

V = .736 * [the square root of the quantity of (1- {[0.037^2]/[0.064^2]})]

V = 0.601 (m/s)

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