Equation of Stationary Wave on a Stretched String

To derive the exact resultant displacement, consider two simple harmonic progressive waves of EQUAL amplitude a and wavelength $\lambda$, travelling along the x-axis in OPPOSITE directions: $y_1=a\sin\left\{2\pi\left(nt-\dfrac{x}{\lambda}\right)\right\}$ (Eq. 6.10, travelling in the $+x$ direction) and $y_2=a\sin\left\{2\pi\left(nt+\dfrac{x}{\lambda}\right)\right\}$ (Eq. 6.11, travelling in the $-x$ direction). By the principle of superposition, the resultant is $y=y_1+y_2$; using the sum-to-product identity $\sin C+\sin D=2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$ gives $y=\left(2a\cos\dfrac{2\pi x}{\lambda}\right)\sin(2\pi nt)$ (Eq. 6.12). Writing the position-dependent factor as an amplitude, $A=2a\cos\dfrac{2\pi x}{\lambda}$, this is simply $y=A\sin(2\pi nt)=A\sin\omega t$ (using $\omega=2\pi n$).

This is the equation of a STATIONARY wave. Crucially, x and t appear SEPARATELY in this equation -- x only inside the amplitude term A, never combined with t as the single argument $2\pi(nt\mp x/\lambda)$ that characterises a genuinely travelling (progressive) disturbance -- so this wave is NOT progressive: it does not travel to the left or the right. Every point on the string instead oscillates with the SAME angular frequency $\omega$ (matching each of the two interfering progressive waves), but with an AMPLITUDE $A=2a\cos(2\pi x/\lambda)$ that varies PERIODICALLY with position x -- different particles genuinely have different, but individually fixed, amplitudes of oscillation.

A NODE is a point of minimum (zero) amplitude: setting $A=2a\cos(2\pi x/\lambda)=0$ requires $\cos(2\pi x/\lambda)=0$, i.e. $2\pi x/\lambda=\pi/2,\,3\pi/2,\,5\pi/2,\ldots$, giving node positions $x=\lambda/4,\,3\lambda/4,\,5\lambda/4,\ldots$, or in general $x=(2p+1)\dfrac{\lambda}{4}$ for $p=1,2,3,\ldots$. Successive nodes are therefore spaced $\lambda/2$ apart. An ANTINODE is a point of MAXIMUM amplitude, $A=\pm2a$: this requires $\cos(2\pi x/\lambda)=\pm1$, giving $2\pi x/\lambda=0,\,\pi,\,2\pi,\ldots$, i.e. antinode positions $x=0,\,\lambda/2,\,\lambda,\ldots$, or $x=p\lambda/2$ for $p=0,1,2,\ldots$ -- successive antinodes are ALSO spaced $\lambda/2$ apart, and since nodes and antinodes alternate, the spacing between any node and its NEAREST antinode is exactly $\lambda/4$. …