= Solution
{c}
Let the uniform string have constant <tension> $\tau>0$ and <linear mass density> $\mu>0$. Assume small transverse motion with $|u_x|\ll1$, negligible longitudinal motion, and no external transverse force. If the string makes angle $\beta$ with the horizontal, then $\tan\beta=u_x$, and the <small-angle approximation> gives $\sin\beta\simeq u_x$ and $ds\simeq dx$ to first order.
The net upward force on the segment from $x$ to $x+\Delta x$ is therefore
$$
\tau\bigl(u_x(x+\Delta x,t)-u_x(x,t)\bigr).
$$
By <Newton's second law>, its mass $\mu\Delta x$ times its transverse <acceleration> equals this force, to the order retained in the small displacement model. Divide by $\Delta x$ and let $\Delta x\to0$ to obtain
$$
\mu u_{tt}=\tau u_{xx}.
$$
Thus the <wave equation on a string> is
$$
\boxed{u_{tt}=c^2u_{xx},\qquad c=\sqrt{\frac\tau\mu}},
$$
where $c$ is the <wave speed>. The horizontal force balance makes the horizontal component of <tension> constant; to first order it equals $\tau$.
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