Question

An object of density $\rho =500{\text{kg/m}}^3$ is hanging from a thin steel wire. The fundamental frequency of transverse standing wave on the wire is $200\text{Hz}$. The set up is now immersed in water, so that one third of the object's volume is submerged. The new fundamental frequency (in $\text{Hz}$) is: [Given : Density of water $\left({\rho }_w\right)=1000{\text{kg/m}}^3$]

• A. $100$
• B. $200$
• C. $200\sqrt{3}$
• D. $\frac{200}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{200}{\sqrt{3}}$

Density of water $\left({\rho }_w\right)=1000{\text{kg/m}}^3$



Fundamental frequency on a stretched thin wire
$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

In air : $T=mg=V\rho g$ [ $V$ is the volume of object]
$\therefore f=\frac{1}{2l}\sqrt{\frac{V\rho g}{\mu }}.........\left(1\right)$

When object is one third immersed in water, then
$T^{\prime }=mg-\text{upthrust}=V\rho g-\frac{V}{3}{\rho }_{w}g=\frac{V}{3}g(3\rho -{\rho }_w)$
The new fundamental frequency is
$f^{\prime }=\frac{1}{2l}\sqrt{\frac{T^{\prime }}{\mu }}=\frac{1}{2l}\sqrt{\frac{\frac{Vg}{3}(3\rho -{\rho }_w)}{\mu }}.......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\frac{f^{\prime }}{f}=\sqrt{\frac{\frac{Vg}{3}(3\rho -{\rho }_w)}{V\rho g}}=\sqrt{(1-\frac{{\rho }_w}{3\rho })}$

From the data given,

$f^{\prime }=\sqrt{(1-\frac{1000}{3\times 500})}\times 200=\frac{200}{\sqrt{3}}\text{Hz}$