Question

A composite string is made by joining two strings of different masses per unit length $\mu$ and $4\mu$. The composite string is under the same tension. A transverse wave pulse. $y=\left(6\text{mm}\right)\sin (5t+40x)$, where '$t$' is in seconds and '$x$' is in meters, is sent along the lighter string towards the joint. The joint is at $x=0$. The equation of the wave pulse reflected from the joint is $y=\left(2\text{mm}\right)\sin (5t-40x+\pi )$. The percentage of the power transmitted to the heavier string through the joint is approximately

• A. $33\%$
• B. $89\%$
• C. $67\%$
• D. $75\%$

Answer 1

The correct option is B $89\%$

We know that,
wave speed on a stretched string is given by
$v=\sqrt{\frac{T}{\mu }}$
On the lighter string,
$v_1=\sqrt{\frac{T}{\mu }}$
On the heavier string,
$v_2=\sqrt{\frac{T}{4\mu }}$
Clearly, we can say that, $v_2<v_1$
and, $v_2=\frac{v_1}{2}..........\left(1\right)$
Amplitude of the reflected wave $A_r=\left|\frac{v_1-v_2}{v_1+v_2}\right|A_i\Rightarrow {\left(\frac{A_r}{A_i}\right)}^2={\left|\frac{v_1-v_2}{v_1+v_2}\right|}^2$
$\Rightarrow 1-{\left(\frac{A_r}{A_i}\right)}^2=\frac{4v_1{v}_2}{(v_1+v_2{)}^2}$

Substituting $\left(1\right)$ in the above equation, we get

$1-{\left(\frac{A_r}{A_i}\right)}^2=\frac{8}{9}$

Since, $I\propto A^2$ we can say that, $P\propto A^2$ where, $P$ is the power transmitted.

Power of incident wave $P_i=P_r+P_t$
where $P_r$ and $P_t$ are the powers of reflected and transmitted waves.

Thus, fraction of power transmitted to the heavier string is given by

$\frac{P_i-P_r}{P_i}\times 100=\frac{A_i^2-A_r^2}{A_i^2}\times 100$
$\Rightarrow (1-\frac{A_r^2}{A_i^2})\times 100=\frac{8}{9}\times 100\approx 89\%$