Two particles of $P$ and $Q$ describe S.H.M of same amplitude $a$ , frequency $υ$ along the same straight line. The maximum distance between the two particles is $a \sqrt{2}$ . What is the initial phase difference between the particles?

zero

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\frac{π}{2}

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\frac{π}{6}

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\frac{π}{3}

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Solution

The correct option is B $\frac{π}{2}$
Let the equation of motion of the two particles be
$y_{1} = a$ sin $2 π n t$
$y_{2} = a$ sin $(2 π n t + ϕ)$
where $ϕ$ is the phase difference
$y = y_{2} - y_{1} = a [$ sin $(2 π n t + ϕ)$ -sln $2 π n t]$
$= 2 a sin \frac{ϕ}{2} cos (2 π n t + \frac{ϕ}{2})$
Maximum value of y = 2a sin $\frac{ϕ}{2}$
. $\cdot . 2 a$ sin $\frac{ϕ}{2} = a \sqrt{2}$ or sin $\frac{ϕ}{2} = \frac{1}{\sqrt{2}}$
. $\cdot . ϕ = \frac{π}{2}$

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