A vibrating system consists of a mass of 200 kg, spring stiffness of 80 N/mm. The natural frequency of vibration of the system is

Concept:

The natural frequency of a spring-mass system is given by,

\(f_n = \frac {\omega_n}{2\pi}\;\) and \(\omega_n = \sqrt {\frac km}\;\)

where k = spring stiffness (N/m) and m = mass of the body (kg)

Calculation:

Given:

Mass of body (m) = 200 kg, spring stiffness (k) = 80 N/mm = 80 × 10³ N/m

The natural frequency of vibration is:

\(\omega_n = \sqrt {\frac km}\;\)

\(\omega_n = \sqrt {\frac{80\;\times\;10^3}{200}}=\sqrt {400}=20\;rad/s\)

Confusion Points

Based on the units of answer we have to use the formula. If we use \(f_n = \frac {\omega_n}{2\pi}\;\) the answer will be in Hertz or s^-1. 

Mistake Points

Use SI unit while calculating natural frequency.

As per the given units, the answer will not come so first we have to convert it into SI unit.