A mass of 2 kg is attached to two identical springs each with stiffness k = 40 kN/m as shown in the figure. Under frictionless condition, the natural frequency of the system in Hz is close to

This question was previously asked in

ISRO Scientist ME 2017 Paper

- 32
- 16
- 24
- 200

Option 1 : 32

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ISRO Scientist ME 2020 Paper

1621

80 Questions
240 Marks
90 Mins

__Concept:__

Natural frequency of the system

\({\omega _n} = \sqrt {\frac{{{k_e}}}{m}}~rad/s\) and \(f_n=\frac{\omega_n}{2\pi}~Hz\)

where ke = equivalent stiffness, m = mass attach to the system

__Calculation:__

Given:

k = 40 kN/m = 40000 N/m, m = 2 kg

As the arrangement of springs are in parallel,

therefore, ke = k + k = 2 k = 80000 N/m

\({\omega _n} = \sqrt {\frac{{80 \times 1000}}{2}}= 200\) rad/s

And \(f_n=\frac{200}{2\pi}=31.83~Hz\approx32 ~Hz\)

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