Concept of Wave motion

Question

The rate of energy transfer in incident wave, P1 and transmitted wave, P2. Then:

(a) !!P_2 = 0.5 P_1!!

(b) !!P_2 = 0.25 P_1!!

(c) !!P_2 = 0.75 P_1!!

(d) !!P_2 = P_1!!

Solution

Suppose the equation of incident wave,

!!y_i = A_1 e^(i(kx-wt))!!

Equation of transmitted wave,

!!y_t = A_2 e^(-i(k^'x- wt))!!

We know that,

Rate of energy transfer, !!P= 1/2 mu A^2 omega^2 v!!

!!P_1 = 1/2 mu_1 A_1^2 omega^2 v_1!! and

!!P_2 = 1/2 mu_2 A_2^2 omega^2 v_2!!.

!!v_1= sqrt(T/mu_1)!! and !!v_2= sqrt(T/mu_2)!!

!!P_1/P_2 = (1/2 sqrt mu_1 A_1^2 omega^2)/(1/2 sqrt mu_2 A_2^2 omega^2)!!

!!=> P_1/P_2 = (sqrtmu_1 A_1^2)/(sqrtmu_2 A_2^2)...(1)!!

Given that, !!mu_1= mu!! and !!mu_2= 9mu!!

!!A_t = (2Z_1)/(Z_1+Z_2) A_i!!, !!Z_1= sqrt(mu_1T)!! and !!Z_2= sqrt(mu_2T)!!

!!=> A_2 = (2sqrt(mu_1T))/(sqrt(mu_2T)+ sqrt(mu_1T)) A_1!!

!!=> A_2 = 2/(3+1) A_1 = 1/2 A_1!!

From equation !!(1)!!, we will get

!!P_1/P_2= (mu/(3mu))xx((2A_2)/A_2)^2!!

!!=> P_1/P_2= 4/3!!

!!=> P_2 = 3/4 P_1 = 0.75P_1!!

Therefore, option (c) is correct.

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