An object of specific gravity is hung from a thin steel wire. The...

Let $V$ be the volume of the object. The initial tension in the wire is $T_1 = V \rho g$, where $\rho$ is the density of the object.

The fundamental frequency is given by $f_1 = \dfrac{1}{2L} \sqrt{\dfrac{T_1}{\mu}} = 300 \text{ Hz}$.

When the object is half submerged in water, the buoyant force acting on it is $F_B = \dfrac{V}{2} \rho_w g$. Since the density of water is $\rho_w = 1 \text{ g/cc}$, $F_B = \dfrac{Vg}{2}$.

The new tension in the wire becomes $T_2 = V \rho g - \dfrac{Vg}{2} = Vg \left(\dfrac{2\rho - 1}{2}\right).<br><br>The new fundamental frequency is$f_1' = $\dfrac{1}{2L} \sqrt{\dfrac{T_2}{\mu}} =$ f_1 $\sqrt{\dfrac{T_2}{T_1}}.<br><br>Substituting$ the values of T_1 and T_2$, we get:<br><br>$f_1' = 300 $\sqrt{\dfrac{Vg \left(\dfrac{2\rho - 1}{2}\right)}{V \rho g}} =300 \left(\dfrac{2\rho - 1}{2\rho}\right)^{1/2}<br><br>The second harmonic frequency is twice$ the fundamental frequency, so f_2' = 2f_1'$.<br><br>$f_2' = 600 \left(\dfrac{2\rho - 1}{2\rho}\right)^{1/2}

Answer: $600\left(\dfrac{2\rho-1}{2\rho}\right)^{1/2}$