# Q.15.22 A travelling harmonic wave on a string is described by$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$(b) Locate the points of the string which have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s.

The wavelength of the given wave is

$\\\lambda =\frac{2\pi }{k}\\ \lambda =\frac{2\pi }{0.005 cm^{-1}}\\ \lambda =1256cm\\ \lambda =12.56m$

The points with the same displacements and velocity at the same instant of time are separated by distances $n\lambda$.

The points of the string which have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s would be at a distance of

$\pm \lambda ,\pm 2\lambda ,\pm 3\lambda ...$ from x = 1cm.

$\lambda =12.56m$

Therefore all points at distances $\pm 12.56m,\pm 25.12m,\pm 37.68m$ from the point x=1cm would have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s.

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