= Solution
Use transverse displacement $y(x,t)$, small slopes and constant <tension> $T$. For a short string element of length $dx$, <Newton's second law> gives $\rho_L dx\,y_{tt}=T[y_x(x+dx)-y_x(x)]$. Hence the <wave equation on a string> is
$$
y_{tt}=c_L^2y_{xx},\qquad c_L=\sqrt{\frac T{\rho_L}},
$$
where $\rho_L$ is the <linear mass density>, in $\mathrm{kg\,m^{-1}}$.
The <mechanical impedance> of a rightgoing <travelling wave> is the transverse driving <force> divided by the transverse <velocity> at the driven section. Choose the <force> exerted by the material to the left on the material to the right, $F_y=-Ty_x$. For $y=f(x-c_Lt)$, the transverse <velocity> is $v_y=y_t=-c_Ly_x$. Thus the positive <characteristic impedance of a stretched string> is
$$
\boxed{Z_L=\frac{F_y}{v_y}=\frac T{c_L}=\rho_Lc_L=\sqrt{T\rho_L},\qquad [Z_L]=\mathrm{N\,s\,m^{-1}}=\mathrm{kg\,s^{-1}}.}
$$
A leftgoing component has $F_y/v_y=-Z_L$ with this fixed orientation of <force>. Specifying the orientation avoids confusing that propagation sign with a negative material <mechanical impedance>.
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