Question

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.
• B.
• C. Zero
• D.

Answer 1

The correct option is D

Let the phase of second particle be $\varphi$. Hence phase difference between two particles is $△\varphi =\frac{2\pi }{\lambda }△x$

$\Rightarrow ⟮\varphi -\frac{\pi }{3}⟯=\frac{2\pi }{60}\times 15\Rightarrow \varphi -\frac{\pi }{3}=\frac{\pi }{2}\Rightarrow \varphi =\frac{5\pi }{6}$