Physics - Wave on String Question with Solution | TestHub

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 $f_1=\frac{1}{2L}\sqrt{\frac{T_1}{\mu }}=300\text{ Hz}$.

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

The new tension in the wire becomes $T_2=V\rho g-\frac{Vg}{2}=Vg\left(\frac{2\rho -1}{2}\right)$.

The new fundamental frequency is $f_1^{\prime }=\frac{1}{2L}\sqrt{\frac{T_2}{\mu }}=f_1\sqrt{\frac{T_2}{T_1}}$.

Substituting the values of $T_1$ and $T_2$, we get:

$f_1^{\prime }=300\sqrt{\frac{Vg\left(\frac{2\rho -1}{2}\right)}{V\rho g}}=300{\left(\frac{2\rho -1}{2\rho }\right)}^{1/2}$

The second harmonic frequency is twice the fundamental frequency, so $f_2^{\prime }=2f_1^{\prime }$.

$f_2^{\prime }=600{\left(\frac{2\rho -1}{2\rho }\right)}^{1/2}$

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