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 $60 cm$
Medium
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(d) Let the phase of second particle be $\phi $.
Hence phase difference between two particles is $\Delta \phi = \frac{{2\pi }}{\lambda }\Delta x$
==> $\left( {\phi - \frac{\pi }{3}} \right) = \frac{{2\pi }}{{60}} \times 15$
$ \Rightarrow \phi - \frac{\pi }{3} = \frac{\pi }{2} \Rightarrow \phi = \frac{{5\pi }}{6}$
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