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Question

A harmonic wave is travelling on string 1. At the junction with string 2, it is partly reflected and partly transmitted. The linear mass density of string 2 is 9 times that of string 1, and that the boundary between the two strings is at x=0. If the expression for the incident wave is
yi=(3 cm)cos[(3.14 cm1)x(314 rad s1)t]. What is the expression for the transmitted waves?

A
yt=(3 cm)cos[(3.14 cm1)x(314 rad s1)t)]
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B
yt=(1.5 cm)cos[(3 cm1)x(9.42 rad s1)t]
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C
yt=(3 cm)cos[(9.42 cm1)x(3.14 rad s1)t]
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D
yt=(1.5 cm)cos[(9.42 cm1)x(314 rad s1)t]
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Solution

The correct option is D yt=(1.5 cm)cos[(9.42 cm1)x(314 rad s1)t]
Given equation of incident wave
yi=3 cmcos[3.14 cm1)x(314 rad s1)t]
So, Ai=3 cm, k1=3.14 cm1, ω1=314 rad s1
and
Linear mass density (μ) and tension T on the string
μ2=9μ1 and T1=T2=T
We know,
Wave speed (v)=Tμ
So, v1=Tμ1 & v2=Tμ2=T9μ1=v13

As we know that, frequency of wave does not change
i.e ω1=ω2=ω=314 rad s1
Wave number (k)=ωv
k2=ω2v2=ω1v13[v2=v13]
k2=3ω1v1=3k1=3×3.14=9.42 cm1
Now, amplitude (At) of transmitted wave
At=2v2v1+v2Ai=2v23v2+v2(3 cm)=1.5 cm

Hence, equation of transmitted wave can be written as
yt=(1.5 cm)cos[9.42 cm1)x(314 rad s1)t]

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