An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency (in Hz) is 300Hz. The object is immersed in water with specific gravity ρw, so that one half its volume is submerged. The new fundamental frequency (in Hz) is
300(2ρ−ρw2ρ)12
The diagrammatic representation of the given problem is shown in figure. The expression of fundamental frequency is v=12l√Tμ
In air T=mg=(Vρ)g
∴ v=12l√vρgμ …(1)
When the object is half immersed in water
T′=mg−upthurst=Vρg−(V2)ρwg
=(V2)g(2ρ−ρw)
The new fundamental frequency is
v′=12l√(Vg2)(2ρp−ρw)μ …(2)
∴ v′v=(√(2ρ−ρw)2ρ)
or v′=v(2ρ−ρw2ρ)12=300(2ρ−ρw2ρ)12 Hz