A system comprising a bar, spring and mass is shown in the figure below. The bar, having negligible mass, is made of a material having Young's modulus \(E = 200\) GPa, cross-sectional area \(A = 100\) mm\(^2\), and length \(L = 100\) mm. The spring stiffness \(k = 200\) kN/mm and the mass \(M = 100\) kg. The natural frequency of free vibration of the system is _______ rad/s (rounded off to the nearest integer).
Show Hint
The massless bar behaves like an axial spring of stiffness \(EA/L\). It sits in series with the given spring, so add their compliances (not their stiffnesses) to get the equivalent stiffness.
Step 1: Model the bar as a spring.
A massless elastic bar of length \(L\), area \(A\), and modulus \(E\), loaded along its axis, stretches like a spring of stiffness
\[ k_{bar} = \frac{EA}{L} \]
Here \(E = 200 \times 10^9\) Pa, \(A = 100 \times 10^{-6}\) m\(^2\), \(L = 0.1\) m.
\[ k_{bar} = \frac{(200\times10^9)(100\times10^{-6})}{0.1} = 2\times10^8 \text{ N/m} \]
Step 2: Convert the given spring stiffness to consistent units.
\(k = 200\) kN/mm. Since 1 kN/mm \(= 10^6\) N/m,
\[ k = 200 \times 10^6 = 2\times10^8 \text{ N/m} \]
Step 3: Combine the bar and the spring.
Looking at the figure, the fixed support holds the bar, the bar connects to the spring, and the spring connects to the mass. So the bar and spring carry the same force one after another, which is a series connection. For springs in series the equivalent stiffness is
\[ \frac{1}{k_{eq}} = \frac{1}{k_{bar}} + \frac{1}{k} = \frac{1}{2\times10^8} + \frac{1}{2\times10^8} = \frac{1}{10^8} \]
\[ k_{eq} = 1\times10^8 \text{ N/m} \]
Step 4: Natural frequency.
For a single mass on an equivalent spring,
\[ \omega_n = \sqrt{\frac{k_{eq}}{M}} = \sqrt{\frac{1\times10^8}{100}} = \sqrt{1\times10^6} \]
Final Answer:
\[ \boxed{\omega_n = 1000 \text{ rad/s}} \]
