The natural frequency of the spring mass system shown in the figure is closest to

Concept:

When spring is connected in parallel as shown, the equivalent stiffness is the sum of all individual stiffness of spring.

k_eq = k₁ + k₂

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

\(\omega_n=\sqrt{\frac{k_{eq}}{m}}\;\;\; and\;\;\;\omega_n=2\pi f\)

k_eq = equivalent stiffness and m = mass of body.

Calculation:

Given:

k₁ = 4000 N/m, k₂ = 1600 N/m and m = 1.4 kg

Springs are in parallel correction

∴ k_eq = k₁ + k₂ = 4000 + 1600 = 5600 N/m

\(ω_n = \sqrt {\frac{k_{eq}}{{m}}}\Rightarrow \sqrt {\frac{{5600}}{{1.4}}} = \sqrt {4000} \)

\(f = \frac{ω_n}{{2\pi }} \Rightarrow \frac{{\sqrt {4000} }}{{2\pi }} \approx 10\;Hz\)