 Solution

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 $|y_x|\ll 1$ 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 givesto first order. Dividing by $\Delta x$ and taking its limit to zero yieldsThus $c$ is the propagation speed of small transverse waves.