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 Youngs 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 :

                \[\ell =60\,\,cm\] \[\rho =2.7\times {{10}^{3}}kg/{{m}^{3}}\] \[y=9.27\times {{10}^{10}}Pa\] For fundamental frequency\[\frac{\lambda }{2}=\ell \Rightarrow \lambda =2\ell \] \[f=\frac{v}{\lambda }\Rightarrow f=\frac{1}{2\ell }\sqrt{\frac{Y}{\rho }}\] \[f=\frac{1}{2\times 06}\sqrt{\frac{9.27\times {{10}^{10}}}{2.7\times {{10}^{3}}}}\] \[f\approx 5kHz\]