Question

# Two strings are attached to each other as shown in the figure. The linear mass density of the second string is four times that of the first string and the boundary between the two strings is at x=0. The wave equation of the incident wave at the boundary is yi=Aicos(k1x−ω1t). Then, which of the following correctly represents the equation of the transmitted wave? Assume, tension in the two strings are same. yt=23Aicos(k1x−ω1t)yt=32Aicos(k1x−ω1t)yt=32Aicos(2k1x−ω1t)yt=23Aicos(2k1x−ω1t)

Solution

## The correct option is D yt=23Aicos(2k1x−ω1t)Given that ,  yi=Aicos(k1x−ω1t) is the equation of incident wave. Let the linear mass density of the 1st string be μ and of the 2nd string be 4μ. Wave speed in 1st stretched string (v1)=√Tμ Wave speed in 2nd stretched string (v2)=√T4μ Tension in the two strings are same ⇒v2=v12......(1) The frequency of the wave depends on the vibrating source only. So,  ω1=ω2=ω (say)......(2) Angular wave number on 1st string (k1)=ω1v1 Angular wave number on 2nd string (k2)=ω2v2 Using (1) in the above formula we get, k2=2k1    ......(3) Amplitude of the transmitted wave is given by At=(2v2v1+v2)Ai From the data obtained, we can rewrite the above equation as At=⎛⎜ ⎜ ⎜ ⎜⎝2(v12)v1+(v12)⎞⎟ ⎟ ⎟ ⎟⎠Ai ⇒At=23Ai    ...........(4) Now from equations (1) , (2) , (3) and (4) equation of the transmitted wave can be written as yt=23Aicos(2k1x−ωt) Hence option (d) is the correct answer.

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