Question

Two springs of stiffness 100 N/m each are connected in series and support a mass of 2 kg. The natural frequency of the system will be
1. \(\frac{{\pi}}{{2.5}}\) Hz
2. \(\frac{{2.5}}{{\pi}}\) Hz
3. \(\frac{{\sqrt{50}}}{{2\pi}}\) Hz
4. \(\sqrt{50}\) Hz

Answer 1

Correct Answer - Option 2 : \(\frac{{2.5}}{{\pi}}\) Hz

Concept:

When springs are connected in series then, Equivalent stiffness,

 \(\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}\)

When springs are connected in parallel then, Equivalent stiffness,

keq = k1 + k2

Natural frequency of vibration is given by,

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

Calculation:

Given:

k₁ = 100 N/m, k₂ = 100 N/m, m = 2 kg

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}=\frac{{1}}{{100}}~+~\frac{{1}}{{100}}\) , k_eq = 50 N/m

\(f=\frac{{1}}{{2\pi}}\times\sqrt{\frac{{k_{eq}}}{{m}}}=\frac{{1}}{{2\pi}}\times\sqrt{\frac{{50}}{{2}}}=\frac{{2.5}}{{\pi}}\) Hz