A fisherman's scale stretches 4.2 cm when a 3.0 kg fish hangs from it.
What will be the amplitude and frequency of vibration if the fish is pulled down 2.3 cm more and released so that it vibrates up and down?
F = -k x
k = 3 *9.8 / .042 Newtons/meter
assume x = .023 sin 2 pi f t
then acceleration = - .023 (2 pi f)^2 sin 2 pi f t
so
-k x = -.023 (2 pi f)^2 sin 2 pi f t
- 3 * 9.8/.042 * .023 sin 2 pi f t =
- 3* .023 (2 pi f)^2 * sin 2 pi f t
so
(2 pi f)^2 = 9.8/.042
or you could use the formula
f = (1/2 pi)sqrt (k/m)
Note that m canceled out for me because it was in the spring constant k
To determine the amplitude and frequency of vibration of the fish, you need to apply Hooke's Law and the equation for simple harmonic motion.
1. Find the spring constant (k) of the fisherman's scale:
Use Hooke's Law: F = k * x, where F is the force applied, k is the spring constant, and x is the displacement.
Rearrange the equation to solve for k: k = F / x.
Given:
- Force (F) = mass (m) * acceleration due to gravity (g)
- Mass (m) = 3.0 kg
- Acceleration due to gravity (g) = 9.8 m/s^2
- Displacement (x) = 4.2 cm = 0.042 m
Calculate the spring constant:
k = (m * g) / x = (3.0 kg * 9.8 m/s^2) / 0.042 m = 703.90 N/m
2. Find the new displacement (y):
Given:
- Additional downward displacement (y) = 2.3 cm = 0.023 m
The total displacement (D) is the sum of the original displacement (x) and the additional downward displacement (y):
D = x + y = 0.042 m + 0.023 m = 0.065 m
3. Find the amplitude (A):
The amplitude (A) is half of the total displacement (D):
A = D / 2 = 0.065 m / 2 = 0.0325 m
Therefore, the amplitude of vibration is 0.0325 m.
4. Find the frequency (f):
The frequency (f) of vibration is given by the equation: f = (1 / 2π) * √(k / m), where π is pi, k is the spring constant, and m is the mass.
Given:
- Mass (m) = 3.0 kg
- Spring constant (k) = 703.90 N/m
Calculate the frequency:
f = (1 / 2π) * √(k / m) = (1 / (2 * 3.14)) * √(703.90 N/m / 3.0 kg) = 3.183 Hz
Therefore, the frequency of vibration is approximately 3.183 Hz.