Answer: \(y = 2A\cos kx\,\sin\omega t\); nodes and antinodes each occur at intervals of \(\lambda/2\).
Consider two simple harmonic progressive waves of the same amplitude \(A\), wavelength \(\lambda\) and frequency travelling along a stretched string in opposite directions:
\(y_1 = A\sin(\omega t - kx)\) (along +x), \(y_2 = A\sin(\omega t + kx)\) (along −x, reflected wave), where \(k = \dfrac{2\pi}{\lambda}\).
By the principle of superposition, \(y = y_1 + y_2 = A[\sin(\omega t - kx) + \sin(\omega t + kx)]\).
Using \(\sin C + \sin D = 2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\): \(y = 2A\sin\omega t\cos kx\), i.e.
\(y = R\sin\omega t\), where \(R = 2A\cos kx = 2A\cos\dfrac{2\pi x}{\lambda}\) is the amplitude, which depends on position \(x\). This is the equation of a stationary wave.
Nodes (amplitude zero): \(\cos\dfrac{2\pi x}{\lambda} = 0\Rightarrow \dfrac{2\pi x}{\lambda} = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2},\ldots\), so \(x = \dfrac{\lambda}{4}, \dfrac{3\lambda}{4}, \dfrac{5\lambda}{4},\ldots\). Distance between successive nodes \(= \dfrac{3\lambda}{4} - \dfrac{\lambda}{4} = \dfrac{\lambda}{2}\).
Antinodes (amplitude maximum \(2A\)): \(\cos\dfrac{2\pi x}{\lambda} = \pm1\Rightarrow \dfrac{2\pi x}{\lambda} = 0, \pi, 2\pi,\ldots\), so \(x = 0, \dfrac{\lambda}{2}, \lambda, \dfrac{3\lambda}{2},\ldots\). Distance between successive antinodes \(= \dfrac{\lambda}{2}\).
