Asked by Anonymous
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.
if you answer please show your work.
Answers
Answered by
Damon
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.
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.
Answered by
bobpursley
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