Q.What happens if a simple pendulum is pulled aside and released?

Each loop of a stationary wave is half a wavelength long, and a node–antinode gap is a quarter wavelength.

A string of length $L = 2\text{ m}$ vibrating in 2 loops means each loop has length $L/2 = 1\text{ m}$. Since each loop corresponds to half a wavelength ($\lambda/2$),

$\dfrac{\lambda}{2} = 1\text{ m} \;\Rightarrow\; \lambda = 2\text{ m}$

The distance between a node and the adjacent antinode in a stationary wave is always $\lambda/4$:

$d = \dfrac{\lambda}{4} = \dfrac{2}{4} = 0.5\text{ m}$

Spatial phase between a node and the neighbouring antinode in a stationary wave.

For a stationary wave $y = 2A\cos(kx)\sin(\omega t)$, a node occurs where $kx=0$ and the adjacent antinode where $kx=\pi/2$ (i.e. a quarter-wavelength away, $x=\lambda/4$). The spatial phase difference between these two points is therefore $\pi/2$ radians.

Adjacent nodes (or adjacent antinodes) are $\lambda/2$ apart; a node and its neighbouring antinode are half of that, $\lambda/4$ apart.

A stationary wave is described by $y = 2a\cos(kx)\sin(\omega t)$, where $k = 2\pi/\lambda$. Nodes occur where $\cos(kx)=0$, i.e. at $x = \lambda/4, 3\lambda/4, 5\lambda/4,\dots$, and antinodes occur where $|\cos(kx)|=1$, i.e. at $x=0,\lambda/2,\lambda,\dots$. The separation between any node and its immediately adjacent antinode is therefore $\lambda/4$, while the separation between two successive nodes (or two successive antinodes) is $\lambda/2$.