Physics - Wave on String Question with Solution | TestHub

The fundamental frequency of a sonometer wire is given by $n=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$, where $T$ is the tension and $\mu$ is the linear mass density.

When the metal bob is in air, the tension is $T_1=V\rho g$, where $V$ is the volume and $\rho$ is the density of the metal.

$n_1=\frac{1}{2L}\sqrt{\frac{V\rho g}{\mu }}$

When the bob is completely immersed in water, the tension decreases due to the buoyant force. The new tension is $T_2=V\rho g-V{\rho }_{w}g=V(\rho -{\rho }_w)g$, where ${\rho }_w$ is the density of water.

$n_2=\frac{1}{2L}\sqrt{\frac{V(\rho -{\rho }_w)g}{\mu }}$

Squaring and dividing the two frequency equations, we get:

$\frac{n_1^2}{n_2^2}=\frac{V\rho g}{V(\rho -{\rho }_w)g}=\frac{\rho }{\rho -{\rho }_w}$

Dividing the numerator and the denominator by ${\rho }_w$:

$\frac{n_1^2}{n_2^2}=\frac{\frac{\rho }{{\rho }_w}}{\frac{\rho }{{\rho }_w}-1}$

Let the relative density of the metal be $d=\frac{\rho }{{\rho }_w}$.

$\frac{n_1^2}{n_2^2}=\frac{d}{d-1}$

$n_1^2(d-1)=n_2^{2}d$

$n_1^{2}d-n_1^2=n_2^{2}d$

$d(n_1^2-n_2^2)=n_1^2$

$d=\frac{n_1^2}{n_1^2-n_2^2}$

Answer: $\frac{{\mathrm{n}}_1^2}{{\mathrm{n}}_1^2-{\mathrm{n}}_2^2}$