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
I did this below:
http://www.jiskha.com/display.cgi?id=1263835390
To find the frequency of the fish's vibration, we first need to calculate the spring constant (k) of the fisherman's scale.
The equation that relates the spring constant (k) to the amount the scale stretches (x) and the weight attached (m) is:
k = (m * g) / x
where:
k = spring constant
m = mass of the fish = 3.7 kg
g = acceleration due to gravity = 9.8 m/s²
x = amount the scale stretches = 2.8 cm = 0.028 m
Substituting the given values into the equation, we get:
k = (3.7 * 9.8) / 0.028 ≈ 1295 N/m
Therefore, the spring constant is approximately 1295 N/m.
Now, let's move on to finding the amplitude and frequency of the fish's vibration.
The amplitude (A) is the maximum displacement from the equilibrium position, which in this case, is the additional 2.5 cm that the fish is pulled down before being released. So, the amplitude (A) is 2.5 cm = 0.025 m.
The frequency (f) of vibration can be determined using the formula:
f = (1 / (2 * pi)) * sqrt(k / m)
where:
f = frequency
k = spring constant = 1295 N/m
m = mass of the fish = 3.7 kg
Substituting the values into the equation, we get:
f = (1 / (2 * pi)) * sqrt(1295 / 3.7) ≈ 3.0 Hz
Hence, the frequency of vibration is approximately 3.0 Hz, as stated in the back of the book.