Question

A fisherman's scale stretches 4.2 cm when a 3.0 kg fish hangs from it.
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?

Answer 1

F = -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