Question

Two wires having different mass densities are soldered together end to end, then stretched under tension $T$. The wave speed in the first wire is twice that in the second wire and the amplitude of incident wave is $a$. Assuming no energy losses in the wire, find the fraction of the incident power that is reflected.

• A. $\frac{1}{2}$
• B. $\frac{1}{5}$
• C. $\frac{8}{9}$
• D. $\frac{1}{9}$

Answer 1

The correct option is D $\frac{1}{9}$

Given that,
Tension in string is $T$ and amplitude of incident wave is $a$
Also, $v_1=2v_2$

Amplitude of the reflected wave
$a_r=|\frac{v_2-v_1}{v_1+v_2}|a_i$
From the given data,
$a_r=|\frac{v_2-2v_2}{2v_2+v_2}|a=\frac{a}{3}$

We know that, $I\propto a^2$
Since, Intensity is defined as power per unit area, we can say that, $\left(\frac{P}{A}\right)\propto a^2$
As the incident and reflected waves are on the same string,
$\frac{P_r}{P_i}={\left(\frac{a_r}{a_i}\right)}^2={\left(\frac{\frac{a}{3}}{a}\right)}^2=\frac{1}{9}$

Thus, option (d) is the correct answer.