A fisherman's scale stretches 2.3 cm. when a 2.5 kg. fish hangs from it. What will be the frequency of vibration if the fish is pulled down and released so that it vibrates up and down?

To find the frequency of vibration of the fish, we can use Hooke's Law for springs. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. In this case, the displacement is the stretching of the scale.

The formula for the force exerted by a spring is: F = k * x

Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement.

Given that the scale stretches 2.3 cm (or 0.023 m) when the fish hangs from it, and the mass of the fish is 2.5 kg, we can find the spring constant.

The weight of the fish (force) can be calculated using the formula: F = m * g

Where:
F is the force (weight),
m is the mass of the fish, and
g is the acceleration due to gravity (approximately 9.8 m/s²).

Calculating the weight of the fish:
F = 2.5 kg * 9.8 m/s² = 24.5 N

We know that the force exerted by the spring (24.5 N) is equal to the spring constant (k) multiplied by the displacement (0.023 m).

24.5 N = k * 0.023 m

Rearranging for k:
k = 24.5 N / 0.023 m ≈ 1065.22 N/m

Now we have the spring constant (k), and we can use it to find the angular frequency (ω) using the formula: ω = √(k / m)

Where:
ω is the angular frequency, and
m is the mass of the fish.

Calculating the angular frequency:
ω = √(1065.22 N/m / 2.5 kg) ≈ 20.57 rad/s

The frequency (f) of the fish's vibration is related to the angular frequency (ω) by the equation: f = ω / (2π)

Calculating the frequency:
f = 20.57 rad/s / (2π) ≈ 3.27 Hz

Therefore, the frequency of vibration of the fish is approximately 3.27 Hz.