Question

An object of specific gravity ρ 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 A $300{\left[\frac{2\rho -1}{2\rho }\right]}^{1/2}$

The steel wire is first stretched by an object of specific gravity $\rho$ in air. Then the object is half
submerged in water. The stretching force diminishes due to upthrust of water on the object. Let $\sigma$
denote specific gravity of water. Weight of the object $=V\rho g$
Upthrust of water on object $=\frac{v}{2}\sigma g$
$\therefore Tension{T}^{\prime }=V\rho g-\frac{V\sigma g}{2}or{T}^{\prime }=Vg\left(\frac{2\rho -\sigma }{2}\right)\upsilon =\frac{1}{2l}\sqrt{\frac{T}{\pi }}whereT=V\rho g\upsilon =\frac{1}{2l}\sqrt{\frac{T}{\mu }}\therefore \frac{{\upsilon }^{\prime }}{\upsilon }=\sqrt{\frac{T^{\prime }}{T}}or\frac{{\upsilon }^{\prime }}{\upsilon }=\sqrt{\frac{Vg(2\rho -1)}{2}\times \frac{1}{Vg\rho }}$
$or{\upsilon }^{\prime }=\upsilon \sqrt{\frac{2\rho -1}{2\rho }}or{\upsilon }^{\prime }=300{\left[\frac{2\rho -1}{2\rho }\right]}^{1/2}$