Q. An object of specific gravity $\rho $ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is $300\, Hz$. The object is immersed in water, so that one half of its volume is submerged. The new fundamental frequency (in $Hz$) is
IIT JEEIIT JEE 1995Waves
Solution:
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
$T'=mg$ -upthrust $=V\rho g-\left (\frac {V}{2}\right )\rho_wg $
$= \left(\frac {V}{2}\right )g(2\rho-\rho_w) $
The new fundamental frequency is
$v'=\frac {1}{2l}\times \sqrt {\frac {T'}{\mu}}$
$=\frac {1}{2l}\sqrt {\frac {(Vg/2)(2\rho-\rho_w)}{\mu}} ...(ii) $
$\therefore \frac {v'}{v}= \left (\frac {2 \rho-\rho_w}{2 \rho}\right)$
or $v'=v \left (\frac {2 \rho-\rho_w}{2 \rho}\right)^{1/2} $
$=300 \left (\frac {2 \rho-1}{2 \rho}\right )^{1/2}Hz $
