In the equilibrium position, the 30-kg cylinder causes a static defl ection of 50 mm in the coiled spring. If the cylinder is depressed an additional 25 mm and released from rest, calculate the result-ing natural frequency ƒn of vertical

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

where;

m = mass of the object

g = acceleration due to gravity

K = spring constant

= static deflection of the string

Given that:

m = 30 kg

g = 9.81 m/s²

= 50 mm = 50 ×

= 0.05 m

Then;

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:

Finally; the natural frequency can be calculated by using the equation

= 2.229305729

≅ 2.23 Hz

Thus; the resulting natural frequency of the vertical vibration of the cylinder = 2.23 Hz

its a 5.5 n

glad that