Answer :
Answer:
2.23 Hz
Explanation:
From the attached diagram below; there exists a diagrammatic representation of the equilibrium position of the cylinder.
The equilibrium position of the spring is expressed as:
mg = K[tex]\delta _{st}[/tex]
where;
m = mass of the object
g = acceleration due to gravity
K = spring constant
[tex]\delta _{st}[/tex] = static deflection of the string
Given that:
m = 30 kg
g = 9.81 m/s²
[tex]\delta _{st}[/tex] = 50 mm = 50 × [tex]\frac{1 \ m}{1000 \ m}[/tex]
= 0.05 m
Then;
[tex]30 * 9.81= k * 0.05\\k = \frac{30*9.81}{0.05} \\k = 5886 N/m[/tex]
From here; let us find the angular velocity which will be needed to determine the natural frequency aftewards.
The angular velocity of the cylinder can be expressed by the formula:
[tex]\omega_{n} = \sqrt{\frac{k}{m}}[/tex]
[tex]\omega_{n} = \sqrt{\frac{5886}{30}}[/tex]
[tex]\omega_{n} = \sqrt{196.2}[/tex]
[tex]\omega_{n} = 14.007141 \ \ rad/s[/tex]
Finally; the natural frequency [tex]f_n[/tex] can be calculated by using the equation
[tex]f_n = \frac{\omega_n}{2 \ \pi }[/tex]
[tex]f_n = \frac{14.007141}{2 \ \pi }[/tex]
[tex]f_n[/tex]= 2.229305729
[tex]f_n[/tex] ≅ 2.23 Hz
Thus; the resulting natural frequency of the vertical vibration of the cylinder = 2.23 Hz
The resulting natural frequency is; ƒn = 2.228 Hz
In a spring-mass system, the gravitational force of mass m will generate a static deflection of the spring such that we have; kδ = mg
Where;
k is spring constant
δ is static deflection
m is mass
g is acceleration due to gravity.
Now we are given;
Mass of cylinder; m = 30 kg
Static deflection; δ = 50 mm = 0.05 m
Thus from kδ = mg, we have;
k = mg/δ
k = (30 × 9.8)/0.05
k = 5880 N/m
Now, we can get the frequency from the relationship;
2πf = √(k/m)
2πf = √(5880/30)
f = (1/2π)√(5880/30)
fn = 2.228 Hz
The image of this cylinder system has been attached to this answer.
Read more at; https://brainly.com/question/15313050
