posted by Kate on .
A fisherman's scale stretches 2.8 cm when a 3.7-kg fish hangs from it. (a) What is the spring constant? (b) What will be the amplitude and frequency of vibration fi the fish is pulled down 2.5 cm more and released so that it vibrates up and down?
I have a question about finding the frequency. For some reason I keep on getting ((3.7 / 129.5)^.5) / (2 * pi) = 0.0269020955 that for my answer but the back of the book says 3.0 hz
please put the subject in the subject line.
i did for some reason it changes it
3.7kg * 9.8 = 36.26 N
stretch distance = .028 m
k = 36.26/.028 = 1295 N/m
F = m a
-kx = m a
if x = A cos 2pi f t
a = -A (2 pi f)^2 cos 2 pi f t
-k A cos 2 pi f t = -mA (2 pi f )^2 cos 2 pi f t
(2 pi f)^2 =k/m
(2 pi f)^2 = 1295/3.7 = 350
2 pi f = 18.7
f = 2.98 Hz
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