A fisherman's scale stretches 2.3 cm. when a 2.1 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, we need to use Hooke's Law, which states that the frequency of a vibrating object is determined by its mass and the stiffness of the spring. In this case, the fisherman's scale acts like a spring.
The equation for the frequency of vibration (f) is given by:
f = 1 / (2π) * √(k / m)
where k is the spring constant and m is the mass of the object.
First, let's calculate the spring constant (k) using the information given. The scale stretches 2.3 cm (or 0.023 m) when a 2.1 kg fish hangs from it. The spring constant is determined by the formula:
k = F / Δx
where F is the force applied to the spring, and Δx is the change in length of the spring.
In this case, F is the weight of the fish, which is the mass (m) multiplied by the acceleration due to gravity (g). Assuming the acceleration due to gravity is approximately 9.8 m/s², we can calculate F:
F = m * g
= 2.1 kg * 9.8 m/s²
≈ 20.58 N
Now we can calculate the spring constant:
k = F / Δx
= 20.58 N / 0.023 m
≈ 895.65 N/m
With the spring constant (k) and mass (m) known, we can calculate the frequency of vibration (f):
f = 1 / (2π) * √(k / m)
= 1 / (2π) * √(895.65 N/m / 2.1 kg)
≈ 2.67 Hz
So, the frequency of vibration of the fish when it is pulled down and released will be approximately 2.67 Hz.