A fisherman's scale stretches 3.6 cm when a 2.7-kg fish hangs from it.
(A) What is the spring stiffness constant?
(B) What will be the amplitude and frequency of vibration if the fish is pulled down 2.5 cm more and released so that it vibrates up and down?
What formulas do I need to use? Thanks!!
To answer part (A) of the question, you need to use Hooke's law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's law is:
F = -kx
Where:
F is the force applied to the spring
k is the spring stiffness constant
x is the displacement of the spring from its equilibrium position
In this case, the displacement of the spring is 3.6 cm (or 0.036 m), and the force exerted on the spring is the weight of the fish, which can be calculated using the formula:
F = mg
Where:
m is the mass of the fish
g is the acceleration due to gravity (which is approximately 9.8 m/s^2)
Using this information, you can solve for the spring stiffness constant:
k = -F / x = -(m*g) / x
Substituting the given values, we have:
k = -(2.7 kg * 9.8 m/s^2) / 0.036 m = -72.55 N/m
So the spring stiffness constant is approximately 72.55 N/m.
Now, moving on to part (B) of the question, to calculate the amplitude and frequency of vibration, we need to consider the motion of a simple harmonic oscillator. The formula for the period of a simple harmonic oscillator is:
T = 2π√(m/k)
Where:
T is the period of the oscillation (time taken for one complete cycle)
m is the mass of the object
k is the spring stiffness constant
Since the fish is pulled down an additional 2.5 cm (or 0.025 m), the new amplitude of vibration is the sum of the initial displacement and the additional displacement:
A = 0.036 m + 0.025 m = 0.061 m
The frequency of vibration (f) can be calculated as the reciprocal of the period (T):
f = 1 / T
Substituting the given values, we have:
T = 2π√(m/k) = 2π√(2.7 kg / 72.55 N/m) ≈ 0.724 seconds
f = 1 / T = 1 / 0.724 s ≈ 1.378 Hz
So, the amplitude of vibration is approximately 0.061 m and the frequency of vibration is approximately 1.378 Hz.
To solve this problem, you will need to use Hooke's law and the formulas for amplitude and frequency of vibration.
(A) Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be written as:
F = -kx
Where:
F is the force applied to the spring
k is the spring stiffness constant
x is the displacement of the spring
In this case, we have the displacement (3.6 cm) and the mass of the fish (2.7 kg). We can use these values to solve for the spring stiffness constant.
(B) The amplitude (A) of the vibration is the maximum displacement from the equilibrium position, and the frequency (f) is the number of complete oscillations per unit of time.
The formulas for amplitude and frequency of vibration are:
A = x + Δx
f = 1 / T
Where:
A is the amplitude
x is the initial displacement of the spring
Δx is the additional displacement
f is the frequency
T is the period (time for one complete oscillation)
Now let's solve the problem step by step:
(A) To find the spring stiffness constant (k), we can rearrange Hooke's law:
k = F / x
We know the force (weight of the fish), which is F = mg, where m is the mass of the fish, and g is the acceleration due to gravity. Plugging in the values, we get:
k = (2.7 kg * 9.8 m/s^2) / 0.036 m
Simplifying, we find:
k ≈ 73.5 N/m
(B) To find the amplitude and frequency of vibration, we can use the given additional displacement (Δx = 2.5 cm) and the formulas:
Amplitude (A) = x + Δx = 3.6 cm + 2.5 cm = 6.1 cm
Frequency (f) = 1 / T
To find the period (T), we can use the formula for the period of an oscillating mass-spring system:
T = 2π√(m / k)
Plugging in the known values, we get:
T = 2π√(2.7 kg / 73.5 N/m)
Simplifying, we find:
T ≈ 0.95 s
Finally, we can calculate the frequency:
f = 1 / T ≈ 1 / 0.95 s ≈ 1.05 Hz
So, the amplitude of vibration is 6.1 cm and the frequency is approximately 1.05 Hz.
W = m g = 9.81*2.7
x = .036 meter
F = kx
9.81*2.7 = k * .036
solve for k
x = A sin w t
A = .25 meter
w = sqrt (k/m) = sqrt (k/2.7)
w = 2 pi f
so f = w/2pi