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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
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B
π2
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C
5π6
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D
2π3
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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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