Question
An object of specific gravity $$\rho $$ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is $$300\,Hz.$$ The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency in $$Hz$$ is
A.
$$300{\left( {\frac{{2\,\rho - 1}}{{2\,\rho }}} \right)^{\frac{1}{2}}}$$
B.
$$300{\left( {\frac{{2\,\rho }}{{2\,\rho - 1}}} \right)^{\frac{1}{2}}}$$
C.
$$300\left( {\frac{{2\,\rho }}{{2\,\rho - 1}}} \right)$$
D.
$$300\left( {\frac{{2\,\rho - 1}}{{2\,\rho }}} \right)$$
Answer :
$$300{\left( {\frac{{2\,\rho - 1}}{{2\,\rho }}} \right)^{\frac{1}{2}}}$$
Solution :
$$\eqalign{
& {\text{In air : }}T = mg = \rho Vg \cr
& \therefore \,\,f = \frac{1}{{2\,\ell }}\sqrt {\frac{{\rho Vg}}{m}} \,\,\,.....\left( {\text{i}} \right) \cr
& {\text{In water : }}T = mg - {\text{upthrust}} \cr
& = V\rho g - \frac{V}{2}{\rho _\omega }g = \frac{{Vg}}{2}\left( {2\,\rho - {\rho _\omega }} \right) \cr
& \therefore \,\,f' = \frac{1}{{2\,\ell }}\sqrt {\frac{{\frac{{Vg}}{2}\left( {2\,\rho - {\rho _\omega }} \right)}}{m}} \cr
& = \frac{1}{{2\,\ell }}\sqrt {\frac{{Vg\,\rho }}{m}} \sqrt {\frac{{\left( {2\,\rho - {\rho _\omega }} \right)}}{{2\,\rho }}} \cr
& \frac{{f'}}{f} = \sqrt {\frac{{2\,\rho - {\rho _\omega }}}{{2\,\rho }}} \cr
& \Rightarrow \,\,f' = f{\left( {\frac{{2\,\rho - {\rho _\omega }}}{{2\,\rho }}} \right)^{\frac{1}{2}}} \cr
& = 300{\left[ {\frac{{2\,\rho - 1}}{{2\,\rho }}} \right]^{\frac{1}{2}}}Hz \cr} $$