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

A, B, C

Explanation

(a) Since, `v=sqrt T//mu, T-2 =T-1` and `mu_2 = 4mu_1`
We have, `v_2 = v_1/2` ………..(i)
From table 18.2, we can see that the frequency does not change, that is
`omega_1 = omega_2` ……(ii)
Also, because `k=omega/v,` the wave numbers of the harmonic waves in the two strings are related by
`k_2 = omega_2/v_2 = (omega_1/(v_1//2)) = 2 (omega_1/v_1) = 2k_1` .......(iii)
The amplitudes are `A_t = ((2v_2)/(v_1+v_2)) A_i = [(2(v_1//2))/(v_1+(v_1//2))] = 2/3A_i`..........(iv)
and `A_r = ((v_2-v_1)/(v_1+v_2)) A_i = [((v_1//2)-v_1)/(v_1+(v_1//2))]A_i= -A_i/3` ........(v)
Now with Eqs. (ii),(iii) and (iv), the transmitted wave can be written as
`y_t = 2/3 A_i cos(2k_1x- omega_1t)`
Similarly the reflected wave can be expressed as
`y_r = -A_i/3 cos (k_1x + omega_1t)`
`=A_i/3 cos (k_1x + omega_1t+ pi)`
(b) The average power of a harmonic wave on a string is given by
`P=1/2 rhoA^2omega^2 Sv = 1/2A^2omega^2muv` (as `rhoS= mu`)
Now, P_i=1/2omega_(1)^(2)A_(i)^(2)mu_(1)v_1`.........(vi)`
` P_t=1/2 (omega_(1)^2) (2/3 A_i)^2 (4mu_1)(v_1/2) = 4/9 (omega_(1)^2)(A_(i)^2) mu_1 v_1.........(vii)`
and `P_r = 1/2 (omega_(2)^2) (-A_i/3)^2 (mu_1)(v_1) = 1/18 (omega_(1)^2)(A_(i)^2) mu_1 v_1......... (viii)`
From Eqs. (vi), (vii) and (viii), we can show that
`P_i = P_t + P_r`.