= Solution
{c}
Assume $\rho$ is <mass> per unit equilibrium <length>, the <tension> has constant magnitude $T$, the string has negligible bending stiffness, and there is no damping or external transverse <force>. Assume small slopes $|y_x|\ll1$ and purely transverse motion. Then a short segment has <mass> $\rho\Delta x$ to first order. If $\alpha$ is the angle of its tangent, the small-slope approximation gives $\sin\alpha\simeq\tan\alpha=y_x$.
The vertical components of <tension> at the two ends give net transverse <force> $T[y_x(x+\Delta x,t)-y_x(x,t)]$. <Newton's second law> therefore gives
$$
\rho\Delta x\,y_{tt}
=T\bigl[y_x(x+\Delta x,t)-y_x(x,t)\bigr]
$$
to first order. Dividing by $\Delta x$ and taking its <limit> to zero yields
$$
\boxed{y_{tt}=c^2y_{xx},\qquad c=\sqrt{\frac{T}{\rho}}.}
$$
Thus $c$ is the propagation <speed> of small transverse <waves>.
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