At a moment in a progressive wave- the phase of a particle executing SHM is -3- Then the place of the particle 15 cm ahead and at time T2 will be- if the wavelength is 60 cm-

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Phase Difference

At a moment i...

Question

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

Zero

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\frac{π}{2}

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\frac{5 π}{6}

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\frac{2 π}{3}

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Solution

The correct option is C $\frac{5 π}{6}$
Let the phase of second particle is $ϕ$ . Hence the phase difference two particles is $△ ϕ = \frac{2 π}{λ} △ x$
$(ϕ - \frac{π}{3}) = \frac{2 π}{60} \times 15$
$ϕ = \frac{π}{2} + \frac{π}{3}$
$ϕ = \frac{5 π}{6}$ .

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At a moment in a progressive wave, the phase of a particle executing SHM is π3. Then the place of the particle 15 cm ahead and at time T2 will be, if the wavelength is 60 cm.

The correct option is C 5π6Let the phase of second particle is ϕ. Hence the phase difference two particles is △ϕ=2πλ△x(ϕ−π3)=2π60×15ϕ=π2+π3ϕ=5π6.

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

The correct option is C $\frac{5 π}{6}$
Let the phase of second particle is $ϕ$ . Hence the phase difference two particles is $△ ϕ = \frac{2 π}{λ} △ x$
$(ϕ - \frac{π}{3}) = \frac{2 π}{60} \times 15$
$ϕ = \frac{π}{2} + \frac{π}{3}$
$ϕ = \frac{5 π}{6}$ .