Q. At a moment is a progressive wave, the phase of a particle executing SHM is $\frac{\pi}{3}$ Then the phase of the particle 15 cm ahead and at the $\frac{T}{2}$ Will be, if the wavelength 60 cm
A) $\frac{\pi}{2}$
B) $\frac{2\pi}{3}$
C) Zero
D) $\frac{5\pi}{6}$
Explanation
(d) $\mathrm{\Delta}\Phi_{1}$ due to path difference $= \pi/2$
$\mathrm{\Delta}\Phi_{2}$ after time
$\frac{T}{2} = \pi$
$\Longrightarrow \Phi = \frac{\pi}{3} - \frac{\pi}{2} + \pi = \frac{5\pi}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.