Q.Find the distance between two successive nodes in a stationary wave on a string vibrating with frequency 64 Hz. The velocity of the progressive wave that resulted in the stationary wave is 48 m s⁻¹.

A stationary (or standing) wave is the special resultant produced when TWO waves of EQUAL amplitude and EQUAL frequency travel through the SAME medium along the SAME path but in EXACTLY OPPOSITE directions -- the case arising, for instance, when a wave reflects off a fixed boundary and interferes with its own oncoming, still-arriving self. Adding $y_1=a\sin\{2\pi(nt-x/\lambda)\}$ and $y_2=a\sin\{2\pi(nt+x/\lambda)\}$ via the sum-to-product identity gives $y=\left(2a\cos\dfrac{2\pi x}{\lambda}\right)\sin(2\pi nt)$ -- since position x and time t appear SEPARATELY here (x only inside the amplitude factor, never combined with t into one travelling argument), this is NOT a progressive wave: every particle oscillates with the same frequency n, but with a fixed AMPLITUDE that varies periodically with its own position, $A(x)=2a\cos(2\pi x/\lambda)$. …