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Yanka [14]
3 years ago
6

In the equilibrium position, the 30-kg cylinder causes a static defl ection of 50 mm in the coiled spring. If the cylinder is de

pressed an additional 25 mm and released from rest, calculate the result-ing natural frequency ƒn of vertical

Physics
1 answer:
MArishka [77]3 years ago
3 0

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

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