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 300 Hz. the object is immersed in water so that one half o its volume is submerged. the new fundamental frequency in Hz is-........

Answer 1

Fundamental frequency, f₁= (1/2L)√(T/m),

where T= Tension in air medium

Or we can write   f₁== (1/2L)√(Vρg/m)

where V= volume, ρ= specific gravity, g= acceleration due to gravity

m= mass

After immersing in water, Tension in the wire, T₁= T- upward thrust

T₁= T- (V/2)ρg

Since density of water ρ= 1 g/cm³

T₁= Vρg- (V/2)g

  =Vg (ρ - 1/2)

  New fundamental frequency is

f₂= (1/2L)√(T₁/m)

 = (1/2L)√{Vg(ρ- 1/2)/m}

Now, f₂÷f₁= √{Vg(ρ- 1/2)}/ √Vρg

=>f₂/300= {ρ-(1/2)}/ ρ

=> f₂= 300×(2ρ-1)/2ρ

Answer 2

Answer:

look below

Explanation: