At a moment in a progressive wave, the phase of a particle executing $S.H.M.$ is $\frac{\pi }{3}$. Then the phase of the particle $15 cm$ ahead and at the time $\frac{T}{2}$ will be, if the wavelength is $60 cm$
A$\frac{\pi }{2}$
B$\frac{{2\pi }}{3}$
C
Zero
D$\frac{{5\pi }}{6}$
Medium
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D$\frac{{5\pi }}{6}$
d (d) Let the phase of second particle be $\phi $.
Hence phase difference between two particles is $\Delta \phi = \frac{{2\pi }}{\lambda }\Delta x$
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