Question

An object of specific gravity $\rho$ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is $300Hz$. The object is immersed in water, so that one half of its volume is submerged. The new fundamental frequency (in $Hz$) is

• A. $300{\left(\frac{2\rho -1}{2\rho }\right)}^{1/2}$
• B. $300{\left(\frac{2\rho }{2\rho -1}\right)}^{1/2}$
• C. $300\left(\frac{2\rho }{2\rho -1}\right)$
• D. $300\left(\frac{2\rho -1}{2\rho }\right)$

Answer 1

Answer: (A)

The diagramatic representation of the given problem is shown in figure.
The expression of fundamental frequency is
$v=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
In air $T=mg=(V\rho )g$
$\therefore v=\frac{1}{2l}\sqrt{\frac{A\rho g}{\mu }}...(i)$
When the object is half immersed in water
$T^{\prime }=mg$ -upthrust $=V\rho g-\left(\frac{V}{2}\right){\rho }_{w}g$
$=\left(\frac{V}{2}\right)g(2\rho -{\rho }_w)$
The new fundamental frequency is
$v^{\prime }=\frac{1}{2l}\times \sqrt{\frac{T^{\prime }}{\mu }}$
$=\frac{1}{2l}\sqrt{\frac{(Vg/2)(2\rho -{\rho }_w)}{\mu }}...(ii)$
$\therefore \frac{v^{\prime }}{v}=\left(\frac{2\rho -{\rho }_w}{2\rho }\right)$
or $v^{\prime }=v{\left(\frac{2\rho -{\rho }_w}{2\rho }\right)}^{1/2}$
$=300{\left(\frac{2\rho -1}{2\rho }\right)}^{1/2}Hz$