An object of specific gravity is hung from a thin steel wire. The fundamental frequency for the transverse stationary waves is 300 Hz. The object is immersed in water, so that one half of its volume is submerged. The second harmonic frequency will be ( density of water = 1 g / cc)

Step-by-step solution

Let V be the volume of the object. The initial tension in the wire is T₁ = V g , where is the density of the object. The fundamental frequency is given by f₁ = 1 2L T₁ = 300 Hz . When the object is half submerged in water, the buoyant force acting on it is F_B = V 2 _w g . Since the density of water is _w = 1 g/cc , F_B = Vg 2 . The new tension in the wire becomes T₂ = V g - Vg 2 = Vg ( 2 - 1 2 ) . The new fundamental frequency is f₁' = 1 2L T₂ = f₁ T₂ T₁ . Substituting the values of T₁ and T₂ , we get: f₁' = 300 V