Question

# At a moment in a progressive wave, the phase of a particle executing SHM is $$\dfrac {\pi}{3}$$. Then the place of the particle $$15\ cm$$ ahead and at time $$\dfrac {T}{2}$$ will be, if the wavelength is $$60\ cm$$.

A
Zero
B
π2
C
5π6
D
2π3

Solution

## The correct option is C $$\dfrac {5\pi}{6}$$Let the phase of second particle is $$\phi$$. Hence the phase difference two particles is $$\triangle \phi = \dfrac {2\pi}{\lambda} \triangle x$$$$\left (\phi - \dfrac {\pi}{3}\right ) = \dfrac {2\pi}{60}\times 15$$$$\phi = \dfrac {\pi}{2} + \dfrac {\pi}{3}$$$$\phi = \dfrac {5\pi}{6}$$.Physics

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