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Question

Two strings 1 and 2 make a junction of partly reflected and partly transmitted waves. The linear mass density of string 2 is four times that of string 1, and the boundary between the two strings is at x=0. The expression for the incident wave is yi=Aicos(k1xω1t). Which of the following relations is correct, if average power carried by the incident wave, transmitted wave and reflected wave are Pi,Pt & Pr respectively? [Assume tension in string 1 and 2 to be the same]

A
Pi=PtPr
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B
Pi=PiPr
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C
Pi=Pt×Pr
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D
Pi=Pt+Pr
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Solution

The correct option is D Pi=Pt+Pr
Give that
Amplitude of Incident wave = Ai
Linear mass density =μ
Tension in the string =T
From the question
μ2=4μ1 and T1=T2=T

Wave speed (v)=Tμ
So, v1=Tμ1;v2=Tμ2=T4μ1=v12
and as we know that frequency dose not change with medium,
ω1=ω2=ω

Transmitted amplitude (At)=(2v2v1+v2)Ai=⎢ ⎢ ⎢ ⎢2(v12)v1+(v12)⎥ ⎥ ⎥ ⎥Ai=23Ai
and reflected amplitude (Ar)=(v2v1v1+v2)Ai=⎜ ⎜ ⎜ ⎜(v12)v1v1+(v12)⎟ ⎟ ⎟ ⎟Ai=Ai3
Now, average power of a wave on a string is given by
P=12ρA2ω2δv=12A2ω2μv [as ρδ=μ]

Now, Pi=12A2iω21μ1v1=12A2iω2μ1v1(1)

Pt=12A2tω22μ2v2=12×49A2iω2(4μ1)v12=49A2iω2μ1v1(2)

and Pr=12(Ar)2ω21μ1v1=12(Ai3)2×ω2×μ1v1=118A2iω2μ1v1(3)
So from (1),(2) and (3), we get
Pi=Pt+Pr

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