Question

A long string of mass per unit length 0.2 $Kg/m$ is stretched to a tension of 500 $N$ and a traverse wave of amplitude $A=10mm$ and $\lambda =0.5m$ is travelling on this string. If this string is joined to another string of same cross section and mass per unit length 0.8 $Kg/m$, the fraction of average power carried by the wave in the first string to that of average power transmitted to the second string is K/9. Then the value of K is. (Answer upto two digits after the decimal point)

Answer 1

Given,
${\mu }_1=0.2Kg/m={\mu }_0$
Speed of transverse wave in string-I
$V_1=\sqrt{\frac{T}{{\mu }_1}}=\sqrt{\frac{T}{{\mu }_0}}$
${\mu }_2=0.8Kg/m=4{\mu }_0$

Speed of transverse wave in string-II
$V_2=\sqrt{\frac{T}{{\mu }_2}}=\sqrt{\frac{T}{4{\mu }_0}}=\frac{V_1}{2}$

Amplitude of transmitted wave;
$A_t=\left(\frac{2V_2}{V_1+V_2}\right)A_i=A_i\left(\frac{2\frac{V_1}{2}}{V_1+\frac{V_1}{2}}\right)=\frac{2}{3}A$

Power carried by the first string =
$E_i=\frac{1}{2}{\mu }_0{\omega }^2{A}^2{V}_1$

Power carried by the second string =
$E_t=\frac{1}{2}\left(4{\mu }_0\right){\omega }^2{A}_t^2{V}_2=\frac{1}{2}\left(4{\mu }_0\right){\omega }^2\frac{4}{9}{A}^2\frac{V_1}{2}$

$=\left(\frac{1}{2}{\mu }_0{\omega }^2{A}^2{V}_1\right)\frac{16}{18}$

$=\frac{8}{9}\left(\frac{1}{2}{\mu }_0{\omega }^2{A}^2{V}_1\right)=\frac{8}{9}{E}_i=\frac{K}{9}{E}_i$

$⟹K=8$