A fisherman's scale stretches 2.5cm when a 2.1kg fish hangs from it. What is the spring constant andwhat will be the frequency of vibration if the fish is pulled down and released so that it vibrates up and down?
To determine the spring constant, you can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law equation is given as:
F = -kx
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement of the spring.
In this case, we are given that the fisherman's scale stretches 2.5cm (0.025m) when a 2.1kg fish hangs from it. The force exerted by the spring is equal to the weight of the fish.
Step 1: Calculate the force exerted by the spring:
Weight = mass × acceleration due to gravity
Weight = 2.1kg × 9.8m/s² (acceleration due to gravity)
Weight = 20.58N
Since the stretch of the spring is in the opposite direction of the gravitational force, we need to use the negative sign in the equation.
Force exerted by the spring, F = -20.58N
Step 2: Calculate the spring constant:
Using Hooke's Law equation:
F = -kx
Substituting the values:
-20.58N = -k × 0.025m
Simplifying the equation:
k = 20.58N / 0.025m
k = 823.2 N/m
Therefore, the spring constant is 823.2 N/m.
To calculate the frequency of vibration when the fish is pulled down and released, you can use the formula for the frequency of a mass-spring system.
Frequency (f) = 1 / (2π) * √(k/m)
Where:
k is the spring constant,
m is the mass, and
π is a mathematical constant approximately equal to 3.14.
Step 3: Calculate the frequency of vibration:
Given: m = 2.1kg and k = 823.2 N/m
Frequency, f = 1 / (2π) * √(823.2 N/m / 2.1kg)
f ≈ 1 / (2π) * √(391.52 N/kg)
f ≈ 1 / (2π) * 19.79 s⁻¹
f ≈ 3.15 Hz
Therefore, the frequency of vibration of the fish will be approximately 3.15 Hz.