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September 30, 2014

September 30, 2014

Posted by **Justin** on Thursday, February 18, 2010 at 7:00pm.

What will be the amplitude and frequency of vibration if the fish is pulled down 2.3 cm more and released so that it vibrates up and down?

- Physics -
**Damon**, Thursday, February 18, 2010 at 7:16pmF = -k x

k = 3 *9.8 / .042 Newtons/meter

assume x = .023 sin 2 pi f t

then acceleration = - .023 (2 pi f)^2 sin 2 pi f t

so

-k x = -.023 (2 pi f)^2 sin 2 pi f t

- 3 * 9.8/.042 * .023 sin 2 pi f t =

- 3* .023 (2 pi f)^2 * sin 2 pi f t

so

(2 pi f)^2 = 9.8/.042

or you could use the formula

f = (1/2 pi)sqrt (k/m)

Note that m canceled out for me because it was in the spring constant k

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