1. The problem is to derive the one-dimensional wave equation, which describes how waves propagate along a string or in a medium. 2. Start with the physical setup: consider a string stretched along the x-axis under tension $T$ with linear mass density $\mu$. 3. The vertical displacement of the string at position $x$ and time $t$ is $u(x,t)$. 4. The net vertical force on a small segment $\Delta x$ of the string is due to the tension forces at both ends, which can be approximated by the difference in the slope of the string: $$ F = T \left( \frac{\partial u}{\partial x}(x+\Delta x,t) - \frac{\partial u}{\partial x}(x,t) \right) $$ 5. Using the definition of the second derivative, this force can be written as: $$ F = T \frac{\partial^2 u}{\partial x^2} \Delta x $$ 6. According to Newton's second law, the net force equals mass times acceleration: $$ F = m \frac{\partial^2 u}{\partial t^2} $$ where the mass of the segment is $m = \mu \Delta x$. 7. Substitute $F$ and $m$: $$ T \frac{\partial^2 u}{\partial x^2} \Delta x = \mu \Delta x \frac{\partial^2 u}{\partial t^2} $$ 8. Cancel $\Delta x$ from both sides: $$ T \cancel{\Delta x} \frac{\partial^2 u}{\partial x^2} = \mu \cancel{\Delta x} \frac{\partial^2 u}{\partial t^2} $$ 9. Rearranging, we get the standard form of the 1D wave equation: $$ \frac{\partial^2 u}{\partial t^2} = \frac{T}{\mu} \frac{\partial^2 u}{\partial x^2} $$ 10. Define the wave speed $c$ as: $$ c = \sqrt{\frac{T}{\mu}} $$ 11. Thus, the wave equation is: $$ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} $$ This equation describes how waves propagate along the string with speed $c$.