F = ma
-kx = m a
if x = A sin wt
a = -Aw^2 sin wt = -w^2 x
-kx = m (-w^2 x)
or
w^2 = k/m
w = sqrt(k/m)
but w = 2pi/T where T is the period
2pi/T = sqrt (k/m)
T = 2pi/sqrt(k/m)
(all of that is probably in your text)
first problem is to find k of this spring
5.5 = m g = m (9.8)
m = .561 kg
T =64/20 = 3.2 s
so
3.2 = 2 pi /sqrt(k/.561)
sqrt(k/.561) = 2 pi/3.2
k/.561 = 3.855
k = 2.16 N/m
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Now the second half, new T = 213/20
= 10.65 s
T = 2pi/sqrt(k/m)
10.65 = 2 pi/sqrt(2.16/m)
solve that for m, the fish mass
then multiply by 9.8 to get weight in Newtons
Martin caught a fish and wanted to know how much it weighed, but he didn't have a scale. He did, however, have a stopwatch, a spring, and a 5.50 N weight. He attached the weight to the spring and found that the spring would oscillate 20 times in 64 s. Next he hung the fish on the spring and found that it took 213 s for the spring to oscillate 20 times.
1 answer
1 answer