Question

Two strings $1$ and $2$ make a junction of partly reflected and partly transmitted waves. The linear mass density of string $2$ is four times that of string $1$, and the boundary between the two strings is at $x=0$. The expression for the incident wave is $y_i=A_i\cos (k_{1}x-{\omega }_{1}t)$. Which of the following relations is correct, if average power carried by the incident wave, transmitted wave and reflected wave are $P_i,P_t\&P_r$ respectively? [Assume tension in string $1$ and $2$ to be the same]

• A. $P_i=P_t-P_r$
• B. $P_i=\frac{P_i}{P_r}$
• C. $P_i=P_t\times P_r$
• D. $P_i=P_t+P_r$

Answer 1

The correct option is D $P_i=P_t+P_r$

Give that
Amplitude of Incident wave = $A_i$
Linear mass density $=\mu$
Tension in the string $=T$
From the question
${\mu }_2=4{\mu }_1$ and $T_1=T_2=T$

Wave speed $\left(v\right)=\sqrt{\frac{T}{\mu }}$
So, $v_1=\sqrt{\frac{T}{{\mu }_1}};v_2=\sqrt{\frac{T}{{\mu }_2}}=\sqrt{\frac{T}{4{\mu }_1}}=\frac{v_1}{2}$
and as we know that frequency dose not change with medium,
${\omega }_1={\omega }_2=\omega$

Transmitted amplitude $\left(A_t\right)=\left(\frac{2v_2}{v_1+v_2}\right)A_i=\left[\frac{2\left(\frac{v_1}{2}\right)}{v_1+\left(\frac{v_1}{2}\right)}\right]A_i=\frac{2}{3}{A}_i$
and reflected amplitude $\left(A_r\right)=\left(\frac{v_2-v_1}{v_1+v_2}\right)A_i=\left(\frac{\left(\frac{v_1}{2}\right)-v_1}{v_1+\left(\frac{v_1}{2}\right)}\right)A_i=-\frac{A_i}{3}$
Now, average power of a wave on a string is given by
$P=\frac{1}{2}\rho A^2{\omega }^2\delta v=\frac{1}{2}{A}^2{\omega }^2\mu v[\text{as}\rho \delta =\mu ]$

Now, $P_i=\frac{1}{2}{A}_i^2{\omega }_1^2{\mu }_1{v}_1=\frac{1}{2}{A}_i^2{\omega }^2{\mu }_1{v}_1\dots \left(1\right)$

$P_t=\frac{1}{2}{A}_t^2{\omega }_2^2{\mu }_2{v}_2=\frac{1}{2}\times \frac{4}{9}{A}_i^2{\omega }^2\left(4{\mu }_1\right)\frac{v_1}{2}=\frac{4}{9}{A}_i^2{\omega }^2{\mu }_1{v}_1\dots \left(2\right)$

and $P_r=\frac{1}{2}(A_r{)}^2{\omega }_1^2{\mu }_1{v}_1=\frac{1}{2}{(-\frac{A_i}{3})}^2\times {\omega }^2\times {\mu }_1{v}_1=\frac{1}{18}{A}_i^2{\omega }^2{\mu }_1{v}_1\dots (3)$
So from $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get
$P_i=P_t+P_r$