Discussion
The boundary condition of a rod under longitudinal vibration is changed from fixed-fixed to fixed-free. The fundamental natural frequency of the rod is now k times the original frequency, where k is
Answer: A
The fundamental frequency becomes 1/2 of the original, so k = 1/2.
For longitudinal vibration of a uniform rod with wave speed \(c = \sqrt{E/\rho}\), the fundamental frequencies are
\[ \begin{aligned} f_{\text{fixed-fixed}} &= \frac{c}{2L} \\ f_{\text{fixed-free}} &= \frac{c}{4L} \end{aligned} \]
A fixed-fixed rod vibrates in its first mode as half a wavelength, a fixed-free rod as a quarter wavelength, so the wavelength doubles and the frequency halves.
The values 2 and \(\sqrt{2}\) wrongly say the frequency rises when a support is removed; removing a constraint always lowers the fundamental frequency. \(1/\sqrt{2}\) would apply if stiffness halved with mass unchanged, which is not what happens here.
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