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.
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
To determine the weight of the fish using the information provided, we can make use of Hooke's Law and the formula for the period of oscillation of a mass-spring system.
First, let's calculate the spring constant (k) of the spring. According to Hooke's Law, the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, which can be represented by the equation F = kx.
We are given that a 5.50 N weight attached to the spring causes it to oscillate 20 times in 64 seconds. This means that the force applied to the spring by the weight is equal to the spring force at maximum displacement (x). So, we can write the equation as follows:
k * x = 5.50 N ... (1)
Next, we need to find the period of oscillation (T) of the fish attached to the spring. We are given that it takes 213 seconds for the spring to oscillate 20 times. The period (T) can be calculated using the formula:
T = t / n,
where T is the period, t is the time taken (213 s), and n is the number of oscillations (20). Substituting the values, we get:
T = 213 s / 20 = 10.65 s ... (2)
Now, using the formula for the period of a mass-spring system:
T = 2π√(m / k),
where T is the period, m is the mass attached to the spring, and k is the spring constant. Substituting the known values, we get:
10.65 s = 2π√(m / k) ... (3)
From equation (1), we can rearrange it to solve for x:
x = 5.50 N / k.
Substituting this value in equation (3), we get:
10.65 s = 2π√(m / (5.50 N / x)).
Now, we can simplify it further:
10.65 s = 2π√(m * x / 5.50 N).
To isolate the unknowns, we can square both sides:
(10.65 s)^2 = (2π)^2 * (m * x / 5.50 N).
Simplifying it, we have:
m * x = (10.65 s)^2 * (5.50 N) / (2π)^2.
Substituting the values, we can calculate the product of m * x.
Once we know the value of m * x, we still need to find the individual values of m and x. However, with the given information, we cannot directly calculate the weight or displacement separately.
Without additional information, it is not possible to determine the weight of the fish.