At a moment in a progressive wave, the phase of a particle executing S.H.M. is π3. Then the phase of the particle 15 cm ahead and at the time T2 will be, if the wavelength is 60 cm

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π2

2π3

Zero

5π6

answer is D.

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Detailed Solution

Let the phase of second particle be φ. Hence phase difference between two particles is Δφ=2πλΔx⇒φ−π3=2π60×15⇒φ−π3=π2⇒φ=5π6

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