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
so
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
so
-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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To find the frequency of vibration for the fisherman's scale in this scenario, we can follow these steps:
Step 1: Calculate the spring constant (k).
The spring constant (k) can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement (stretch or compression) of the spring from its equilibrium position.
Hooke's Law equation: F = -kx
Given:
Initial displacement (x) = 2.8 cm = 0.028 m
Weight of the fish (F) = 3.7 kg * 9.8 m/s^2 (gravitational acceleration) = 36.26 N
Substituting the values into Hooke's Law equation:
36.26 N = -k * 0.028 m
To isolate the spring constant (k), rearrange the equation:
k = -(36.26 N) / 0.028 m
Calculate k:
k = -1295 N/m
Step 2: Determine the new displacement (amplitude).
The fish is pulled down an additional 2.5 cm (0.025 m). Therefore, the new displacement or amplitude (A) is the sum of the initial displacement (0.028 m) and the additional displacement (0.025 m):
A = 0.028 m + 0.025 m
A = 0.053 m
Step 3: Calculate the frequency (f).
The frequency of vibration (f) can be found using the formula: f = (1 / 2π) * sqrt(k / m)
Given:
Mass of the fish (m) = 3.7 kg
Substituting the values into the formula:
f = (1 / 2π) * sqrt(-k / m)
f = (1 / (2 * 3.14)) * sqrt(-(-1295 N/m) / 3.7 kg)
f = (1 / 6.28) * sqrt(1295 N/m / 3.7 kg)
f = 0.159 / sqrt(3.7 / 1295)
f ≈ 0.159 / 0.108301361
Calculating:
f ≈ 1.469 Hz
Therefore, the approximate frequency of vibration for the fisherman's scale would be 1.469 Hz.
It seems there may have been an error in your calculation. Please recheck the steps and ensure that all values are entered correctly.