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 300 Hz. 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

The correct option is D $300\left(\frac{2\rho -1}{2\rho }\right)$

Case $1$ (air)

Fundamental frequency:

$n=\frac{1}{2L}\sqrt{\frac{T}{m}}=\frac{1}{2L}\sqrt{\frac{V\rho }{m}}$

where

$\rho$= specific gravity

$V=$ volume

$m=$ linear density

Case $II$ (immersed in water to half of the volume)

New tension:

$T^{{}^{\prime }}=T-upthrustforce=T-\frac{V}{2}\rho g=Vg(\rho -\frac{1}{2})\left\{as\rho \right(water)=1g/c.c\}$

$\therefore$ new fundamental frequency

$n^{{}^{\prime }}=\frac{1}{2L}\sqrt{\frac{T^{{}^{\prime }}}{m}}=\frac{1}{2L}\sqrt{\frac{Vg(\rho -\frac{1}{2})}{m}}\therefore \frac{n^{{}^{\prime }}}{n}=\frac{1}{2L}\sqrt{\frac{Vg(\rho -\frac{1}{2})}{m}}÷\frac{1}{2L}\sqrt{\frac{V\rho g}{m}}or,\frac{n^{{}^{\prime }}}{300}=\sqrt{\frac{\rho -\frac{1}{2}}{\rho }}\therefore n^{{}^{\prime }}=300\left(\frac{2\rho -1}{2\rho }\right)$