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(D) The phase of a wave is given by $\phi = (\omega t - kx + \phi_0)$.The phase difference due to a spatial separation $\Delta x$ is $\Delta \phi_x =

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$\frac{5\pi}{6}$

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(D) The phase of a wave is given by $\phi = (\omega t - kx + \phi_0)$.The phase difference due to a spatial separation $\Delta x$ is $\Delta \phi_x = \frac{2\pi}{\lambda} \Delta x$.Given $\Delta x = 15 \ cm$ and $\lambda = 60 \ cm$,the phase change due to distance is $\Delta \phi_x = \frac{2\pi}{60} 	imes 15 = \frac{\pi}{2}$.Since the particle is ahead in the direction of wave propagation,its phase is $\phi_1 = \phi_0 - \Delta \phi_x = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}$.The phase change due to a time interval $\Delta t = \frac{T}{2}$ is $\Delta \phi_t = \omega \Delta t = \frac{2\pi}{T} 	imes \frac{T}{2} = \pi$.The new phase is $\phi_{new} = \phi_1 + \Delta \phi_t = -\frac{\pi}{6} + \pi = \frac{5\pi}{6}$.

At a moment in a progressive wave,the phase of a particle executing $S.H.M.$ is $\frac{\pi}{3}$. What will be the phase of a particle $15 \ cm$ ahead at a time $\frac{T}{2}$ later,if the wavelength is $60 \ cm$?

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