= Solution
At the same <tension>, $k\propto\sqrt\rho$, so $k_2/k_1=3$, $t=1/2$ and $r=-1/2$. The <transmitted power> fraction is not $t^2$, because the two strings have different <wave impedances>. Using the <energy flux of a transverse string wave>,
$$
\boxed{\mathcal T=\frac{\rho_2c_2}{\rho_1c_1}|t|^2
=\sqrt{\frac{\rho_2}{\rho_1}}\,\frac14=\frac34.}
$$
The reflected fraction is $\mathcal R=|r|^2=1/4$, so $\mathcal R+\mathcal T=1$. The transmitted displacement amplitude is $2.5\,\mathrm{mm}$ and its <wavelength> is $0.5/3\,\mathrm m$. The negative $r$ indicates a reflected phase reversal; it does not mean negative reflected energy.
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