Question

A harmonic wave is travelling on string 1. At a junction with string 2 it is partly reflected and partly transmitted. The linear mass density of the second string is four times that of the first string, and that the boundary between the two strings is at x=0. If the expression for the incident wave is y_i = A_i cos (k_1 x-omega_1t) (a) What are the expressions for the transmitted and the reflected waves in terms of A_i, k_1 and omega_1? (b) Show that the average power carried by the incident wave is equal to the sum of the average power carried by the transmitted and reflected waves.

Answer 1

Answer is A1/3cos(k1x+w1t+π)

Since v :√T/muT2=T1 and mu2:4mu1

We have V2:v1/2.....(1)

The frequency does not change i.e.

W1=W2..(2)

Also because k=w/v the wave no of the harmonic waves in the two string are related by

K2=w2/V2=W1/v1/2=2w1/v1=2k1....(3)

The amplitude are 2/3A1...(4)

And A1/3....(5)

Now with equation 2,3,4,5=

The reflected wave =A1/3cos(k1x+w1t+π)

The transmitted wave =2a1/3cos(2k1x-w1t)

Answer 2

The  average power carried by the incident wave is equal to the sum of the average power carried by the transmitted and reflected wave.

Explanation:

(a) Since, v = √T/μ , T− 2 = T − 1 and μ2 = 4μ1

We have, v2 = v12 ………..(1)

From table 18.2, we can see that the frequency does not change, that is

ω1 = ω2 ……(2)

Also, because k=ωv, the wave numbers of the harmonic waves in the two strings are related by

k2=ω2 / v2 = (ω1 / v1 ÷ 2) = 2(ω1 / v1) = 2k1 .......(3)

The amplitudes are At  = (2v2 / v1+v2) Ai = [2(v1/2) v1 + (v1/2)] = 23 Ai  ...(4)

and Ar = (v2 − v1 / v1 + v2) Ai = [(v1/2) − v1 / v1 + (v1 / 2)] Ai = − Ai / 3 ...(5)

Now with Eqs. (ii),(iii) and (iv), the transmitted wave can be written as

yt = 2 / 3 Ai cos (2k1x − ω1t)

Similarly the reflected wave can be expressed as

yr = −Ai / 3 cos (k1x + ω1t)

= Ai / 3 cos(k1x + ω1t + π)

(b) The average power of a harmonic wave on a string is given by

P=1 / 2 ρA^2 ω^2 Sv = 1 / 2A^2ω^2 μv (as ρS = μ)

Now, Pi = 1/2 ω^(2)(Ai)^2 mu1v1      ----(4)

Pt  = 1 / 2(ω1^2)(2 / 3 Ai)^2 (4μ1)(v1 / 2) = 4 / 9(ω1^2)(Ai^2)μ1v1   .........(6)

and Pr = 1 / 2(ω2^2)(−Ai / 3)^2(μ1)(v1) = 1 / 18(ω^2)(Ai^2)μ1v1  .....(7)

From Equations. (vi), (vii) and (viii), we can show that

Pi = Pt + Pr

Hence the  average power carried by the incident wave is equal to the sum of the average power carried by the transmitted and reflected wave.