Question

An automobile having a mass of 2000 kg deflects its suspension springs 0.02 m under static conditions. Determine the nafural frequency of the automobile in the vertical direction by assuming damping to be negligible.

Answer 1

Answer:Frequency = 3.525 Hertz

Explanation:In static equilibrium, kd =mg

Where k= effective spring constant of the spring.

mg= The weight of the car.

d= static deflection.

Therefore, w =SQRTg/d

w = SQRT 9.81/0.02

w= 22.15 rad/sec

Converting to Hertz unit for frequency

1 rad/s = 0.1591

22.15rad/s=?

22.15 × 0.1591= 3.525 hertz

Answer 2

The natural frequency of the automobile in the vertical direction is 3.52 Hertz

Static equilibrium

Static equilibrium refers to the physical state in which the components of a system are at rest and the net force acting through the system is equal to zero.

In static equilibrium:

kd = mg

where;

k = effective spring constant of the spring.

mg = the weight of the car

d = static deflection.

also K = mω²

where ω is angular velocity

m is mass

Thus, mω²d = mg

ω²d = g

ω = √g/d

ω = g/d

ω  = √9.8/0.02

ω = 22.15 rad/sec

Converting to Hertz unit for frequency

1 rad/s = 0.1591 Hertz

22.14 rad/s = 22.14 * 0.1591

22.14 rad/s = 3.525 hertz

Therefore, the natural frequency of the automobile in the vertical direction is 3.52 Hertz

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