F = k x
F = m g
x = 0.021 meter
m g = 2.3 * 9.81
so
k (0.021) = 2.3 * 9.81
then
omega = 2 pi f = sqrt (k/m)
the 3.8 cm has nothing to do with it.
A fisherman's scale stretches 2.1 cm when a 2.3-kg fish hangs from it. What will be the frequency of vibration if the fish is pulled down 3.8 cm more and released so that it vibrates up and down?
if you answer please show your work.
2 answers
force=kx
2.3*9.8N=k*.021m
solve for k
period=2PI sqrt(mass/k) and period= 1/freq so
freq= sqrt(k/mass) / 2PI
2.3*9.8N=k*.021m
solve for k
period=2PI sqrt(mass/k) and period= 1/freq so
freq= sqrt(k/mass) / 2PI
1 answer