Q 11-14-137JEE MainJEE Main 2018 (8 Apr)Medium
A granite rod of $60$ cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7\times10^3\ \text{kg m}^{-3}$ and its Young's modulus is $9.27\times10^{10}$ Pa. What will be the fundamental frequency of the longitudinal vibrations?
Answer: (B) $5$ kHz
Speed of longitudinal waves:
$$v = \sqrt{\frac Y\rho} = \sqrt{\frac{9.27\times10^{10}}{2.7\times10^3}} \approx 5.86\times10^3\ \text{m s}^{-1}$$
Clamped at the middle (node) with free ends (antinodes), the fundamental has $L = \dfrac\lambda2$:
$$f = \frac{v}{2L} = \frac{5.86\times10^3}{1.2} \approx 4.9\ \text{kHz} \approx 5\ \text{kHz}$$
Solution by Sreeraj P, M.Sc Physics