Question

Sol. Equation of graph:

y = A sin (–kx) = – A sin kx

In +ve x direction, Equation:

y = – A sin (kx – ωt)

for t = $\frac{T}{2}$, shift the origin

⇒ y = – A sin$\left\lbrack kx - \omega\left( t - \frac{T}{2} \right) \right\rbrack$

= – A sin (kx – ωt + π) = A sin (kx – ωt)

Similarly

In –ve direction, equation:

y = –A sin $\left\lbrack kx + \omega\left( t - \frac{T}{2} \right) \right\rbrack$

= – A sin (kx + ωt – π) = A sin (kx + ωt)