What will be the natural frequency of a vibratory system shown in figure having the mass (m) of 10 kg and stiffness as k₁= k₂= 2N/mm?

Question

Question

This question was previously asked in

BHEL Engineer Trainee Mechanical 24 Aug 2023 Official Paper

Answer 1

Concept:

In a vibratory system with two springs in parallel, the equivalent stiffness is the sum of the individual stiffnesses.

The natural frequency of vibration is given by:

\( f = \frac{1}{2\pi} \sqrt{\frac{k_{eq}}{m}} \)

Calculation:

Given:

Mass, \( m = 10~kg \)

Stiffness of each spring, \( k_1 = k_2 = 2~N/mm = 2000~N/m \)

Since the springs are in parallel:

\( k_{eq} = k_1 + k_2 = 2000 + 2000 = 4000~N/m \)

Now, apply the formula for natural frequency:

\( f = \frac{1}{2\pi} \sqrt{\frac{4000}{10}} = \frac{1}{2\pi} \sqrt{400} = \frac{20}{2\pi} = \frac{10}{\pi}~Hz \)