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a vertical spring scale can measure weights up to 175N the scale extends by an amount of 13.0cm from its equilibrium position at 0N to the 175 N mark. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.35 Hz. Ignore the mass of the spring, what is the mass Mof the fish

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Answer:

Approximately 0.024 kg

Step-by-step explanation:

We can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. Mathematically, this can be written as:

F = -kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. The negative sign indicates that the force is in the opposite direction of the displacement.

We can use the given information to find the spring constant:

k = F/x = 175 N / 0.13 m = 1346.15 N/m

The fish is oscillating vertically, which means that the force of gravity is acting on it. The weight of the fish can be calculated as:

W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity (9.81 m/s^2).

The oscillation frequency of the fish can be related to its mass and the spring constant using the formula:

f = 1/2π * sqrt(k/m)

where f is the frequency of oscillation, π is a constant (approximately 3.14), and sqrt is the square root function.

We can rearrange this equation to solve for the mass of the fish:

m = k/(4π^2 * f^2)

Substituting the given values, we get:

m = 1346.15 N/m / (4 * 3.14^2 * (2.35 Hz)^2) ≈ 0.024 kg

Therefore, the mass of the fish is approximately 0.024 kg.

User Mike El Jackson
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