A 30.0 cm long brass rod is suspended at the ends by long strings. The rod is struck at one end parallel to the rod and causes the rod to vibrate longitudinally with the ends free. (Y = 9.00 × 10¹⁰ N/m², density = 8,470 kg/m³.) Assuming the rod vibrates in its fundamental mode, the frequency of the vibration is

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The frequency of the fundamental mode of a rod vibrating longitudinally with free ends is given by $ f = \frac{1}{2L} \sqrt{\frac{Y}{\rho}} $. Substituting the given values: $ L = 0.3 $ m, $ Y = 9.00 \times 10^{10} $ N/m², and $ \rho = 8,470 $ kg/m³, we find $ f \approx 5,430 $ Hz.

A 300 Cm Long Brass Rod Is Suspended at the Ends