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Question: At a moment in a progressive wave, the phase of a particle executing S.H.M. is \(\frac{\pi}{3}\). Th...

At a moment in a progressive wave, the phase of a particle executing S.H.M. is π3\frac{\pi}{3}. Then the phase of the particle 15 cm ahead and at the time T2\frac{T}{2} will be, if the wavelength is 60 cm

A

π2\frac{\pi}{2}

B

2π3\frac{2\pi}{3}

C

Zero

D

5π6\frac{5\pi}{6}

Answer

5π6\frac{5\pi}{6}

Explanation

Solution

Let the phase of second particle be φ\varphi. Hence phase difference between two particles isΔφ=2πλΔx\Delta\varphi = \frac{2\pi}{\lambda}\Delta x

⇒ (φ−π3)=2π60×15\left( \varphi - \frac{\pi}{3} \right) = \frac{2\pi}{60} \times 15 ⇒φ−π3=π2⇒φ=5π6\Rightarrow \varphi - \frac{\pi}{3} = \frac{\pi}{2} \Rightarrow \varphi = \frac{5\pi}{6}