Question

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

Answer 1

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$\delta _{st}$

where;

m = mass of the object

g = acceleration due to gravity

K = spring constant

$\delta _{st}$ = static deflection of the string

Given that:

m = 30 kg

g = 9.81 m/s²

$\delta _{st}$ = 50 mm = 50 × $\frac{1 \ m}{1000 \ m}$

= 0.05 m

Then;

$30 * 9.81= k * 0.05\\k = \frac{30*9.81}{0.05} \\k = 5886 N/m$

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:

$\omega_{n} = \sqrt{\frac{k}{m}}$

$\omega_{n} = \sqrt{\frac{5886}{30}}$

$\omega_{n} = \sqrt{196.2}$

$\omega_{n} = 14.007141 \ \ rad/s$

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

$f_n = \frac{\omega_n}{2 \ \pi }$

$f_n = \frac{14.007141}{2 \ \pi }$

$f_n$= 2.229305729

$f_n$ ≅ 2.23 Hz

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