GATE 2022 AE – Question 63
A uniform rigid prismatic bar of mass m is suspended by two identical springs at its ends, as shown. If $\omega_1<\omega_2$ are the two natural frequencies, $\omega_2/\omega_1$ is (one decimal place).
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Correct answer: 1.69 to 1.71
Explanation
The rigid bar of mass $m$ and length $l$ is suspended on two identical springs of stiffness $k$ at its ends. It has two degrees of freedom: vertical translation and rotation.
**Mode 1: translation (bounce).** Both springs stretch equally, so the stiffness is $2k$:
$$\omega_1^2=\frac{2k}{m}.$$
**Mode 2: rotation (pitch).** For a small rotation $\theta$, each spring end moves $\pm\theta l/2$, giving a restoring moment $2\cdot k\,(\theta l/2)\cdot(l/2)=\dfrac{kl^2}{2}\theta$. The moment of inertia about the centre is $I=\dfrac{ml^2}{12}$:
$$\omega_2^2=\frac{kl^2/2}{ml^2/12}=\frac{6k}{m}.$$
**Ratio:**
$$\frac{\omega_2}{\omega_1}=\sqrt{\frac{6k/m}{2k/m}}=\sqrt3=\mathbf{1.7}.$$