Question

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).

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.

Answer 1

Correct Answer: 1000

Step 1: Think in terms of flexibility instead of stiffness.
The flexibility (compliance) of an elastic element is how much it stretches per unit force, $\delta/F = 1/k$. When two elements carry the same force one after another (in series, as the bar and spring do here), their total flexibility just adds up:
\[ \frac{1}{k_{eq}} = \frac{1}{k_{bar}} + \frac{1}{k_{spring}} \]

Step 2: Find each flexibility.
Bar stiffness: $k_{bar} = EA/L = (200\times10^9)(100\times10^{-6})/(0.1) = 2\times10^8$ N/m, so its flexibility is $1/(2\times10^8) = 5\times10^{-9}$ m/N.
Spring stiffness: $k_{spring} = 200$ kN/mm $= 2\times10^8$ N/m, so its flexibility is also $5\times10^{-9}$ m/N.

Step 3: Add the flexibilities and invert.
\[ \frac{1}{k_{eq}} = 5\times10^{-9} + 5\times10^{-9} = 1\times10^{-8} \text{ m/N} \]
\[ k_{eq} = 1\times10^8 \text{ N/m} \]

Step 4: Natural frequency from the equation of motion.
For the mass hanging from this combined spring, Newton's second law gives $M\ddot{x} + k_{eq} x = 0$, whose angular frequency is $\omega_n = \sqrt{k_{eq}/M}$.
\[ \omega_n = \sqrt{\frac{10^8}{100}} = \sqrt{10^6} \]

Final Answer:
\[ \boxed{\omega_n = 1000 \text{ rad/s}} \]