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
$(a) \frac{\pi}{2} \quad (b) \frac{2\pi}{3} \quad (c) \text{Zero} \quad (d) \frac{5\pi}{6}$
Text Solution
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Let the phase of second particle be $\phi$ . Hence phase difference between two particles is $\Delta \varphi = \frac{2 \pi}{\lambda} \Delta x$
⇒ ⇒ $\left(\varphi - \frac{\pi}{3}\right) = \frac{2\pi}{60} \times 15$ $\Rightarrow \varphi - \frac{\pi}{3} = \frac{\pi}{2} \Rightarrow \varphi = \frac{5\pi}{6}$
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