Question

An object of specific gravity $p$ 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. $300{\left(\frac{2p-1}{2p}\right)}^{1/2}$
• B. $300{\left(\frac{2p}{2p-1}\right)}^{1/2}$
• C. $300\left(\frac{2p}{2p-1}\right)$
• D. $300\left(\frac{2p-1}{2p}\right)$

Answer 1

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

Case I : In air

Fundamental frequency, $f=\frac{1}{2L}\sqrt{\frac{T}{M}}=\frac{1}{2L}\sqrt{\frac{pVg}{M}}$

Where, $T\to$ Tension in air

$V\to$ Volume

$p\to$ Specific gravity

$g\to$ Gravitational acceleration

$M\to$ Linear density

Case II: Immersed in water to half of the volume

New Tension, $T^{\prime }=T-\text{upthrust force}$

$\Rightarrow T^{\prime }=T-\frac{V}{2}\rho g$

$\Rightarrow T^{\prime }=Vpg-\frac{V}{2}g$ $[{\rho }_{water}=1g/cc]$

$\Rightarrow T^{\prime }=Vg(p-\frac{1}{2})$

New fundamental frequency, $f^{\prime }=\frac{1}{2L}\sqrt{\frac{T^{\prime }}{M}}=\frac{1}{2L}\sqrt{\frac{Vg(p-\frac{1}{2})}{M}}$

$\therefore \frac{f^{\prime }}{f}=\frac{\frac{1}{2L}\sqrt{\frac{Vg(p-\frac{1}{2})}{M}}}{\frac{1}{2L}\sqrt{\frac{Vpg}{M}}}$

$\Rightarrow \frac{f^{\prime }}{300}=\frac{1}{2L}\sqrt{\frac{Vg(p-\frac{1}{2})}{M}}\times \sqrt{\frac{M}{Vpg}}$

$\Rightarrow f^{\prime }=300\left(\frac{p-\frac{1}{2}}{p}\right)$

$\Rightarrow f^{\prime }=300\left(\frac{2p-1}{2p}\right)$