A fisherman's scale stretches 3.9 cm when a 2.8 kg fish hangs from it.
(a) What is the spring stiffness constant?
N/m
(b) What will be the amplitude and frequency of vibration if the fish is pulled down 2.7 cm more and released so that it vibrates up and down?
amplitude cm
frequency Hz
To find the spring stiffness constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
(a) The spring stiffness constant, also known as the spring constant or the force constant, can be calculated using the formula:
k = F / x
where k is the spring stiffness constant, F is the force exerted by the spring, and x is the displacement.
Given that the scale stretches 3.9 cm (which is the displacement) when a 2.8 kg fish hangs from it, we can calculate the force exerted by the spring using the gravitational force formula:
F = m * g
where F is the force, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values into the formulas:
F = 2.8 kg * 9.8 m/s²
F ≈ 27.44 N
Now we can calculate the spring stiffness constant:
k = 27.44 N / 3.9 cm
To convert centimeters to meters, divide by 100:
k ≈ 27.44 N / (3.9 cm / 100 cm/m)
k ≈ 27.44 N / (0.039 m)
k ≈ 703.59 N/m
Therefore, the spring stiffness constant is approximately 703.59 N/m.
(b) To determine the amplitude and frequency of vibration when the fish is pulled down 2.7 cm more and released, we need to consider the properties of simple harmonic motion.
The amplitude of the vibration is the maximum displacement from the equilibrium position. In this case, the fish is pulled down an additional 2.7 cm, so the total amplitude is 3.9 cm + 2.7 cm = 6.6 cm.
The frequency of vibration is determined by the spring stiffness constant and the mass of the fish. The formula for the frequency of vibration is:
f = (1 / 2π) * √(k / m)
where f is the frequency, k is the spring stiffness constant, and m is the mass.
Given the spring stiffness constant (k) and the mass of the fish (2.8 kg), we can calculate the frequency:
f = (1 / 2π) * √(703.59 N/m / 2.8 kg)
Simplifying the equation:
f = (1 / 2π) * √(703.59 N/m / 2.8 kg)
f = (1 / 2π) * √(703.59 N / (2.8 kg * 1 m²/s²))
f ≈ (1 / 2π) * √(703.59 / 2.8)
f ≈ (1 / 2π) * √(251.28)
f ≈ (0.159) * 15.86
f ≈ 2.52 Hz
Therefore, the amplitude of vibration is 6.6 cm and the frequency of vibration is approximately 2.52 Hz.
a) 3.9cm = .039m
k = F/x = mg/x
= ((2.7)(9.8))/.039
= 680 N/m