ok i'm putting in physics it'll most likely change it watch
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
well know it's posting it twice :O
Three times you posted it
see:
http://www.jiskha.com/display.cgi?id=1263835390
To find the frequency of vibration, we can use Hooke's law and the formula for the period of a mass-spring system.
(a) Finding the spring constant:
According to Hooke's law, the force exerted by the spring is directly proportional to its displacement from the equilibrium position. Mathematically, we can express this as F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, the spring stretches 2.8 cm (0.028 m) when a 3.7 kg fish hangs from it. Using the given values, we can solve for the spring constant:
F = -kx
mg = kx
k = mg/x = (3.7 kg)(9.8 m/s^2)/(0.028 m)
Calculating this value, we find the spring constant is approximately 129.5 N/m.
(b) Finding the frequency of vibration:
The frequency of vibration for a mass-spring system is given by the equation f = 1/T, where f is the frequency and T is the period.
To find the period T, we need to calculate the time it takes for the fish to complete one full oscillation. The period can be calculated as the time it takes for a particle to move from the maximum displacement to the minimum displacement and back.
Given that the fish is pulled down an additional 2.5 cm (0.025 m) from its equilibrium position and then released, we can find the amplitude of vibration. The amplitude is the maximum displacement from the equilibrium position, which is the sum of the initial displacement (2.8 cm) and the additional displacement (2.5 cm). Therefore, the amplitude is 5.3 cm (0.053 m).
Using the formula T = 2π√(m/k), where m is the mass and k is the spring constant, we can calculate the period T:
T = 2π√(m/k) = 2π√(3.7 kg / 129.5 N/m)
Calculating this value, we find the period T is approximately 2.026 seconds.
Finally, we can calculate the frequency f using the formula f = 1/T:
f = 1/T = 1/2.026 s
Calculating this value, we find the frequency is approximately 0.494 Hz.
It seems that you made an error in your calculation. The correct frequency, rounded to one decimal place, is 0.5 Hz, not 0.03 Hz. Therefore, the answer provided in the back of the book, which states that the frequency is 3.0 Hz, seems to be incorrect.