$ \Rightarrow \frac{{3v}}{{4{L_1}}} = \frac{v}{{{L_2}}}$
$ \Rightarrow \frac{3}{{4{L_1}}}\sqrt {\frac{{\gamma P}}{{{\rho _1}}}} $$ = \frac{1}{{{L_2}}}\sqrt {\frac{{\gamma P}}{{{\rho _2}}}} $
$\left[ {\because v = \sqrt {\frac{{\gamma P}}{\rho }} } \right]$
$ \Rightarrow {L_2} = \frac{{4{L_1}}}{3}\sqrt {\frac{{{\rho _1}}}{{{\rho _2}}}} = \frac{{4L}}{3}\sqrt {\frac{{{\rho _1}}}{{{\rho _2}}}} $
$y = 0.02sin\left[ {2\pi \left( {\frac{t}{{0.04\left( s \right)}} - \frac{x}{{0.50\left( m \right)}}} \right)} \right]m$ The tension in the string is .... $N$
