question_answer

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is \[2.7\times {{10}^{3}}\text{ }kg/{{m}^{3}}\] and its Young's modulus is\[\text{9}\text{.27}\times \text{1}{{\text{0}}^{\text{10}}}\text{ Pa}\]. What will be the fundamental frequency of the longitudinal vibrations?                                                                                      [JEE Main Online 08-04-2018]

A)  \[10\,\,kHz\]                 

B)  \[7.5\,\,kHz\]

C)  \[5\,\,kHz\]       

D)       \[2.5\,\,kHz\]

Correct Answer: C

Solution :

(c)

\[\ell =60\,\,cm\]

\[\rho =2.7\times {{10}^{3}}kg/{{m}^{3}}\]

\[y=9.27\times {{10}^{10}}Pa\]

For fundamental frequency\[\frac{\lambda }{2}=\ell \Rightarrow \lambda =2\ell \]

\[f=\frac{v}{\lambda }\Rightarrow f=\frac{1}{2\ell }\sqrt{\frac{Y}{\rho }}\]

\[f=\frac{1}{2\times 06}\sqrt{\frac{9.27\times {{10}^{10}}}{2.7\times {{10}^{3}}}}\]

\[f\approx 5kHz\]