Question

A long string of mass per unit length 0.2 Kg/m is stretched to a tension 500 N. A transverse wave of amplitude A = 10 mm and $\lambda$ = 0.5 m is travelling on this string. This string is joined to another string of same cross section and mass per unit length $0.8$ kg/m. The fraction of the average power carried by the wave in the first string that is transmitted to the second string is K/9. Find K.

Answer 1

${\mu }_1=0.2kg/m={\mu }_0,$ $V_1=\sqrt{\frac{T}{{\mu }_0}}$
${\mu }_2=0.8kg/m=4{\mu }_0,$ $V_2=\sqrt{\frac{T}{4{\mu }_0}}=\frac{V_1}{2}$
$A_t=\left(\frac{2V_2}{V_1+V_2}\right)A_1$
$=A_1\left(\frac{2\frac{V_1}{2}}{V_1+\frac{V_1}{2}}\right)=\frac{2}{3}{A}_1=\frac{2}{3}A$
$E_1=\frac{1}{2}{\mu }_0{\omega }^2{A}^2{V}_1$
$E_2=\frac{1}{2}\left(4{\mu }_0\right){\omega }^2{A}_t^2{V}_2$
$=\frac{1}{2}\left(4{\mu }_0\right){\omega }^2\frac{4}{9}{A}^2\frac{V_1}{2}$
$=\left(\frac{1}{2}{\mu }_0{\omega }^2{A}^2{V}_1\right)\frac{16}{18}$
$=\frac{8}{9}\left(\frac{1}{2}{\mu }_0{\omega }^2{A}^2{V}_1\right)$
$=\frac{8}{9}{E}_1$