Physics - Wave on String Question with Solution | TestHub

The frequency of a sonometer wire is given by $n=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$, which implies $n\propto \sqrt{T}$.

Let $V$ be the volume of the weight and ${\rho }_w$ be the density of water. Since the specific gravity of the weight is $d$, its density is $d{\rho }_w$.

The initial tension in the wire is $T_1=Vd{\rho }_{w}g$.

The initial frequency is $n_1=n\propto \sqrt{Vd{\rho }_{w}g}$.

When the weight is completely immersed in water, it experiences an upward buoyant force equal to the weight of the displaced water, $F_B=V{\rho }_{w}g$.

The new tension in the wire is $T_2=T_1-F_B=Vd{\rho }_{w}g-V{\rho }_{w}g=V{\rho }_{w}g(d-1)$.

Since the tension decreases, the new frequency $n_2$ will be less than $n$. Given that $x$ beats are produced per second, the new frequency is $n_2=n-x$.

Taking the ratio of the frequencies:

$\frac{n_1}{n_2}=\sqrt{\frac{T_1}{T_2}}$

$\frac{n}{n-x}=\sqrt{\frac{Vd{\rho }_{w}g}{V{\rho }_{w}g(d-1)}}$

$\frac{n}{n-x}=\sqrt{\frac{d}{d-1}}$

Answer: $\frac{n}{n-x}=\sqrt{\frac{d}{d-1}}$