Question
A travelling harmonic wave on a string is described by
$\text{y}(\text{x, t})=7.5\sin\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
Locate the points of the string which have the same transverse displacements and velocity as the x = 1cm point at t = 2s, 5s and 11s.

Answer

Propagation constant is related to wavelength as:
$\text{k}=\frac{2\pi}{\lambda}$
$\therefore\ \lambda=\frac{2\pi}{\lambda}=\frac{2\times3.14}{0.0050}$
$=1256\text{cm}=12.56$
Therefore, all the points at distances $\text{n}_\lambda,$ (n = ± 1, ± 2.... and so on) i.e. ± 12.56m, ± 25.12m, … and so on for x = 1cm, will have the same displacement as the x = 1cm points at t = 2s, 5s, and 11s.

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