Wave Motion 2 Question 22

22. An object of specific gravity ρ 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(2ρ12ρ)1/2

(b) 300(2ρ2ρ1)1/2

(c) 300(2ρ2ρ1)

(d) 300(2ρ12ρ)

(1995, 2M)

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Answer:

Correct Answer: 22. (a)

Solution:

  1. The diagramatic representation of the given problem is shown in figure. The expression of fundamental frequency is

 In air T=mg=(Vρ)gv=12lAρgμ

When the object is half immersed in water

T=mg upthrust =Vρg(V2)ρwg=(V2)g(2ρρw)

The new fundamental frequency is

v=12l×Tμ=12l(Vg/2)(2ρρw)μvv=(2ρρw2ρ) or v=v(2ρρw2ρ)1/2=300(2ρ12ρ)1/2Hz



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